Algebra 1: 2000 and Beyond - Ector County Independent School

Transcription

Algebra 1:
2000 and Beyond
Mathematics Institute
http://www.utdanacenter.org/ssi/projects/texteams
Dwight D. Eisenhower Professional Development Program, Title II, Part B
Texas Education Agency
Texas Statewide Systemic Initiative in Mathematics, Science, and Technology Education
Charles A. Dana Center, The University of Texas at Austin
Permission is given to any person, group, or organization to copy and distribute Texas
Teachers Empowered for Achievement in Mathematics and Science (TEXTEAMS)
materials for noncommercial educational purposes only, so long as the appropriate credit is
given. This permission is granted by The Charles A. Dana Center, a unit of the College of
Natural Sciences at The University of Texas at Austin.
Acknowledgements
The TEXTEAMS Algebra I: 2000 and Beyond institute was developed under the direction
and assistance of the following:
Academic Advisors/Reviewers
Paul Kennedy
Texas Christian University
Anne Papakonstantinou
Rice University
Writer
Pam Harris
Consultant
Advisory Committee
Linda Antinone
Kathy Birdwell
Kathi Cook
Eva Gates
Juan Manuel Gonzalez
Susan Hull
Paul Kennedy
Laurie Mathis
Diane McGowan
Bonnie McNemar
Barbara Montalto
Anne Papakonstantinou
Cindy Schimek
Jane Silvey
JoAnn Wheeler
Susan Williams
Fort Worth ISD
New Braunfels ISD
Dana Center, University of Texas at Austin
Consultant
Laredo ISD
Southwest Texas State University
Consultant
Texas Education Agency
Rice University
Katy ISD
ESC VII
ESC IV
University of Houston
TEXTEAMS Algebra I: 2000 and Beyond Institute
Table of Contents
About TEXTEAMS Institutes......................................................................................vii
Institute Introduction.....................................................................................................viii
Institute Overview............................................................................................................. ix
Section Overviews............................................................................................................x
Materials List .......................................................................................................................x v
I. Foundations for Functions
1 Developing Mathematical Models
1.1 Variables and Functions .....................................................................................1
Activity 1: Examples of Dependent Relationships .....................................11
Activity 2: Independent and Dependent Variables.....................................12
Reflect and Apply ............................................................................................13
1.2 Valentine’s Day Idea ........................................................................................14
Activity 1: Valentine’s Day Idea .....................................................................24
Activity 2: Using Tables to Find the More Economical Offer.......................25
Activity 3: Using Graphs to Find the Better Offer.........................................27
Activity 4: New Rose Offers ..........................................................................29
Activity 5: Using Tables for New Rose Offers.............................................30
Activity 6: Using Graphs for New Rose Offers............................................31
Student Activity: Investigate Recursively ...................................................36
2 Using Patterns to Identify Relationships
2.1 Identifying Patterns ...........................................................................................41
Activity 1: Painting Towers..............................................................................51
Activity 2: Building Chimneys ........................................................................54
Activity 3: Constructing Trucks........................................................................57
Activity 4: Generating Patterns.......................................................................60
Student Activity: Perimeter of Rectangles ....................................................63
2.2 Identifying More Patterns.................................................................................67
Activity 1: Building Blocks...............................................................................72
Activity 2: Starting Staircases .........................................................................75
Activity 3: Too Many Triangles.......................................................................78
3 Interpreting Graphs
3.1 Interpreting Distance versus Time Graphs.....................................................82
Activity 1: Walking Graphs .............................................................................87
Activity 2: Walking More Graphs...................................................................88
Student Activity: Walk This Way ...................................................................91
3.2 Interpreting Velocity versus Time Graphs .....................................................97
Activity 1: Matching Velocity Graphs.......................................................... 103
Activity 2: Connecting Distance and Velocity Graphs .............................. 105
Reflect and Apply ......................................................................................... 107
TEXTEAMS AlgebraI: 2000 and Beyond
iii
Table of Contents
II. Linear Functions
1 Linear Functions
1.1 The Linear Parent Function............................................................................ 108
Activity 1: ACT Scores................................................................................ 115
Activity 2: Temperatures.............................................................................. 116
Activity 3: Symbolic ..................................................................................... 117
Student Activity 1: Age Estimates.............................................................. 119
Student Activity 2: Sales Goals.................................................................. 128
1.2 The Y- Intercept............................................................................................. 132
Activity 1: The Birthday Gift......................................................................... 139
Activity 2: Spending Money....................................................................... 142
Activity 3: Money, Money, Money............................................................ 145
Student Activity: Show Me the Money! .................................................... 148
1.3 Exploring Rates of Change.......................................................................... 153
Activity 1: Wandering Around ..................................................................... 158
Activity 2: Describe the Walk ...................................................................... 160
Student Activity: What’s My Trend? .......................................................... 163
1.4 Finite Differences............................................................................................ 170
Activity 1: Rent Me!...................................................................................... 176
Activity 2: Guess My Function .................................................................... 177
Activity 3: Finite Differences......................................................................... 179
Student Activity: Graphs and Tables ......................................................... 181
2 Interpreting Relationships Between Data Sets
2.1 Out for a Stretch ............................................................................................. 183
Activity 1: Stretch It....................................................................................... 190
Activity 2: Comparing Graphs .................................................................... 194
Student Activity 1: Have You Lost Your Marbles?.................................. 196
Student Activity 2: Unidentified Circular Objects (UCO’s)....................... 206
Student Activity 3: Going to Great Depths................................................ 215
Student Activity 4: Height versus Arm Span ............................................ 224
2.2 Linear Regression.......................................................................................... 232
Activity 1: Sum of Squares ......................................................................... 245
Activity 2: Lines of Best Fit.......................................................................... 246
Activity 3: The Correlation Coefficient ........................................................ 248
3 Linear Equations and Inequalities
3.1 Solving Linear Equations .............................................................................. 251
Activity 1: Concrete Models........................................................................ 259
Activity 2: Using Concrete Models............................................................. 262
3.2 Stays the Same............................................................................................. 265
Activity: Stays the Same............................................................................. 274
3.3 Solving Linear Inequalities ............................................................................ 279
Activity 1: Linear Inequalities in One Variable............................................ 288
Activity 2: Linear Inequalities in Two Variables.......................................... 290
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Table of Contents
3.4 Systems of Linear Equations and Inequalities............................................ 295
Activity 1: Using a Table.............................................................................. 302
Activity 2: Solve the System Graphically.................................................. 303
Activity 3: Solve the System Symbolically............................................... 305
Student Activity: Concrete Models and Systems of Linear Equations .. 307
III. Nonlinear Functions
1 Quadratic Functions
1.1 Quadratic Relationships................................................................................. 319
Activity 1: Building a Sandbox.................................................................... 329
Activity 2: Projectile Motion.......................................................................... 332
1.2 Transformations.............................................................................................. 336
Activity 1: Investigating the Role of a ......................................................... 344
Activity 2: Investigating the Role of k.......................................................... 345
Activity 3: Investigating the Role of h ......................................................... 346
Activity 4: Transformations .......................................................................... 347
1.3 Lines Do It Too .............................................................................................. 351
Activity 1: Exploring Slope ......................................................................... 361
Activity 2: Exploring Vertical Shifts ............................................................. 362
Activity 3: Exploring Horizontal Shifts......................................................... 363
Activity 4: A Different Perspective.............................................................. 365
2 Quadratic Equations
2.1 Connections ................................................................................................... 367
Activity 1: Roots, Factors, x-intercepts, Solutions..................................... 374
Activity 2: Which Form?............................................................................... 377
Activity 3: Jump!........................................................................................... 378
2.2 The Quadratic Formula................................................................................... 380
Activity 1: Programming the Quadratic Formula......................................... 384
Activity 2: Hang Time................................................................................... 385
Student Activity: Investigate Completing the Square.............................. 388
3 Exponential Functions and Equations
3.1 Exponential Relationships............................................................................ 392
Activity 1: Paper Folding.............................................................................. 405
Activity 2: Measure with Paper ................................................................... 407
Activity 3: Regions ...................................................................................... 409
Activity 4: How Big is a Region?................................................................ 411
Student Activity: Recursion Again............................................................... 414
3.2 Exponential Growth and Decay................................................................... 420
Activity 1: Exponential Growth.................................................................... 427
Activity 2: Exponential Decay..................................................................... 428
Student Activity: On the Wall ...................................................................... 430
3.3 Exponential Models...................................................................................... 434
Activity 1: Population Growth...................................................................... 441
Activity 2: Cooling Down............................................................................. 443
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4.1 Bounce It!........................................................................................................ 445
Activity 1: Collect the Data........................................................................... 452
Activity 2: A Bounce..................................................................................... 453
Activity 3: Bounce Height versus Bounce Number.................................. 454
Activity 4: Bounce Height versus Drop Height ......................................... 455
Student Activity 1: Pattern Blocks............................................................... 457
Student Activity 2: Throw Up!..................................................................... 463
Student Activity 3: Radioactive Decay....................................................... 467
Student Activity 4: Pendulum Decay.......................................................... 474
Calculator Programs................................................................................................... 477
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About TEXTEAMS Institutes
TEXTEAMS Philosophy
• Teachers at all levels benefit from extending their own mathematical knowledge
and understanding to include new content and new ways of conceptualizing the
content they already possess.
• Professional development experiences, much like the school mathematics
curriculum itself, should focus on few activities in great depth.
• Professional development experiences must provide opportunities for teachers
to connect and apply what they have learned to their day-to-day teaching.
Features of TEXTEAMS Institute Materials
Multiple representations (verbal, concrete, pictorial, tabular, symbolic, graphical)
Mathematical ideas will be represented in many different formats. This helps both teachers
and students understand mathematical relationships in different ways.
Integration of manipulative materials and graphing technology
The emphasis of TEXTEAMS Institutes is on mathematics, not on learning about particular
manipulative materials or calculator keystrokes. However, such tools are used in various
ways throughout the institutes.
Rich Connections within and outside mathematics
Institutes focus on using important mathematical ideas to connect various mathematical
topics and on making connections to content areas and applications outside of mathematics.
Questioning strategies
A variety of questions are developed within each activity that help elicit deep levels of
mathematical understanding and proficiency.
Hands-on approach with “get-up-and-move” activities
Institutes are designed to balance intense thinking with hands-on experiences.
Math Notes and Reflect and Apply
A feature called Math Notes includes short discussions of mathematical concepts
accompanying the learning activities. Similarly, the Reflect and Apply feature is designed to
extend and apply participants’ understanding of the mathematical concepts.
The Charles A. Dana Center is approved by the State Board for Educator Certification as a registered Continuing
Professional Education (CPE) provider. Hours received in TEXTEAMS institutes may be applied toward the required training
for gifted and talented in the area of curriculum and instruction. Individual district/ campus acceptance of these hours for
gifted and talented certification is a local decision.
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Introduction
The Algebra I: 2000 and Beyond Institute is based on the groundbreaking work of
the 1996 TEXTEAMS Algebra I Institute. Both institutes assert that “Algebra for All”
is a realistic and attainable goal. To teach “Algebra for All” will require educators to
possess a deep understanding of mathematical content, pedagogy, and strategies
to meet the needs of diverse student populations.
This institute is not meant as a scope and sequence for the Algebra I course, nor is
it a set of student activities for use in a classroom without careful thought and
modification on the part of a knowledgeable teacher. The Algebra I: 2000 and
Beyond Institute is a rich, carefully designed professional development tool which is
intended to deepen teacher content knowledge. It is also intended to model the
importance of content depth by focusing on carefully selected activities that are few
in number and grounded in the mathematics necessary to support teacher and
student learning. The concepts and ideas explored within this institute are
connected to classroom instruction and key assessments.
In this institute, not all Algebra I topics can be addressed. It is important for
teachers to develop a deep and powerful understanding of the concepts and ideas
of algebra. This requires educators to understand the mathematics in a different
capacity from that of the student. Therefore, content within the institute is
approached from a more complex perspective and some topics are addressed at
a level that is deeper than would be typical for Algebra I students. Much of the
institute is built upon learning experiences that develop and promote the power of
using concrete experiences to introduce and build mathematical concepts.
Throughout the institute, multiple representations are utilized as a powerful strategy
to assist educators and students in making connections, understanding
mathematical concepts, and reasoning in meaningful and complex ways.
The institute assumes some prior participant knowledge. If teachers are unfamiliar
with the following, presenters may need to supplement with introductory materials.
• Representing, adding and subtracting polynomials with algebra tiles (area
model)
• Using algebra tiles to model monomial and binomial multiplication
• Modeling factoring trinomials with algebra tiles
• Graphing calculators
The Algebra I; 2000 and Beyond Institute draws on the work of the 1996
TEXTEAMS Algebra I Institute; Principles and Standards for School Mathematics,
NCTM, 2000; Discovering Algebra, Key Curriculum Press, 2000; and Dr. Paul
Kennedy, Southwest Texas State University.
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Institute Overview
1 Developing Mathematical Models
2 Using Patterns to Identify Relationships
3 Interpreting Graphs
1 Linear Functions
3 Linear Equations and Inequalities
1 Quadratic Functions
2 Quadratic Equations
3 Exponential Functions and Equations
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I. Foundations of Functions
Section Overviews
I. Foundations of Functions Overview
Activity
Overview
Developing Mathematical Models
1.1 Variables and
Participants discuss the concept of function in
Functions
the context of a school fund-raising venture.
1.2 Valentine’s Day Participants investigate functional
Idea
relationships for a given problem situation
using tables, graphs, and algebraic
representations.
Using Patterns to Identify Relationships
2.1 Identifying
Participants represent linear relationships
Patterns
among quantities using concrete models,
tables, diagrams, written descriptions, and
algebraic forms.
2.2 Identifying More Participants represent non-linear
Patterns
relationships among quantities using
concrete models, tables, diagrams, written
descriptions, and algebraic forms.
Interpreting Graphs
3.1 Interpreting
Participants use motion detectors to
Distance versus
investigate distance over time graphs. This
Time Graphs
lays the groundwork for graph reading and
for work with rates of change.
3.2 Interpreting
Velocity versus
Time Graphs
Participants use motion detectors to
investigate velocity over time graphs. This
continues to build graph-reading skills and to
build understanding for rates of change.
Materials
butcher paper,
masking tape
graphing
calculators
building blocks,
color tiles,
graphing
calculators
building blocks,
graphing
calculators
motion detector
connected to an
overhead
calculator, motion
detectors, data
collection devices,
graphing
calculators, 2 or 3
transparencies cutto-fit on the
overhead
calculator screen
motion detector
connected to a
overhead
calculator, motion
detectors, data
collection devices,
graphing
calculators
x
Section Overviews
II. Linear Functions Overview
Activity
Linear Functions
1.1 The Linear
Parent Function
1.2 The YIntercept
1.3 Exploring
Rates of Change
Overview
Using contextual situations, participants
investigate the linear parent function, the
line y = x.
Participants use “canned” real life
experiences to build the concepts of yintercept as the starting point and slope as
a rate of change, both with contextual
significance.
Participants use real data from a motion
detector to model motion at a constant rate
over time. Participants translate among
algebraic, tabular, graphical, and verbal
descriptions of linear functions.
1.4 Finite
Differences
Participants use their cumulative concrete
experiences with the linear model to build
to the abstract symbolic representations of
slope. Finite differences are use to find
linear models and to discover what makes
data linear.
Interpreting Relationships Between Data Sets
2.1 Out for a
Participants investigate the relationship
Stretch
between the “stretch” of a rubber band
attached to a container and the number of
marbles in the container.
Materials
colored pencils or
pens, pieces of flat
spaghetti, graphing
calculators
Student Activity:
markers, 1” grid
paper
graphing calculators
motion detector
connected to an
overhead calculator,
motion detectors,
graphing calculators,
data collection
devices
Student Activity:
calculator programs
Styrofoam cups, 3”
long thin rubber
bands, marbles of
the same size, large
paper clips, tape,
meter sticks, graphing
calculators
Student Activities:
blocks, PVC pipe,
marbles, tape
measure
flashlights, rulers,
cylinders, uniform
objects, water
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2.2 Linear
Regression
Section Overviews
Participants write a program to find a least
squares linear function to model data. They
use the program and calculator regression
to find linear models for data and they
discuss the correlation coefficient, r.
Linear Equations and Inequalities
3.1 Solving Linear Participants solve linear equations with
Equations
concrete models and make connections
between the concrete model, abstract, and
symbolic representations.
3.2 Stays the
Participants solve linear equations in one
Same
variable, making connections between
algebraic solution steps, algebra tile
solution steps, and graphical solution
steps.
3.3 Solving Linear Participants use problem situations and
Inequalities
technology to explore linear inequalities.
3.4 Systems of
Linear Equations
and Inequalities
Participants use a table to develop a
system of linear inequalities. They solve
the system using various methods and
make connections between a system of
inequalities and a system of equations.
graphing calculator,
dynamic geometry
program with a
prepared
demonstration of a
linear least squares fit
for data, computer
with a projection
device.
algebra tiles,
overhead algebra
tiles
algebra tiles,
overhead algebra
tiles, graphing
calculators, 1” grid
paper, markers
transparencies of the
Student Activity:
Age Estimates from
2.1.1 The Linear
Parent Function,
Student Activity:
algebra tiles
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Section Overviews
III. Nonlinear Functions Overview
Activity
Quadratic
Functions
1.1 Quadratic
Relationships
Overview
Participants use lists to develop a quadratic
function representing the volume of a
sandbox with a fixed depth. Using the
quadratic function, participants solve
quadratic equations numerically and
graphically.
1.2
Participants investigate the effects of
Transformations
changing the parameters of quadratic
function of the form y = ax2 + c. They
apply this understanding by fitting a
quadratic to real data. Participants extend
their understanding and investigate the
effects of changing the parameter h in
2
quadratic functions of the form y = ( x − h) .
1.3 Lines Do It Too Participants connect their knowledge of
transformations with quadratic functions with
the equations of lines. The point-slope
form of a line is looked at from a
transformational perspective.
Quadratic Equations
2.1 Connections
Participants make connections between the
roots of quadratic functions and the
solutions to quadratic equations and the
factors of quadratic polynomials and the xintercepts of a parabola. They connect this
understanding to the vertex, polynomial,
and factored form of the equation of a
parabola. Using this understanding,
participants model a vertical jump, finding
the height of the jump.
2.2 The Quadratic
Participants program the quadratic formula
Formula
into the graphing calculator and use the
program to solve quadratic equations at
appropriate times.
Materials
pieces of lumber or
cardboard to simulate
lumber
patty paper or blank
transparencies
patty paper or blank
transparencies
data collection
devices, light
sensors, laser
pointers or flashlights
1” graph paper,
markers, meter sticks
Student Activity:
algebra tiles
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Exponential Functions and Equations
3.1 Exponential
Participants explore exponential growth
Relationships
and decay situations. Using “canned”
situations, participants develop the ideas
of the common multiplier or ratio as the
base of an exponential function and the
starting point as the y-intercept of an
exponential function.
3.2 Exponential
Participants find models for exponential
Growth and Decay growth and decay situations.
3.3 Exponential
Participants find exponential models for
Models
given data sets.
Interpreting Relationships Between Data Sets
4.1 Bounce It!
Collecting three sets of data from a
bouncing ball experiment, participants find
appropriate models and justify their
choices.
Section Overviews
sheets of blank
paper
Student Activity:
sticky notes, poster
boards, large blank
paper, markers, tape
balls, data collection
devices, motion
detectors, graphing
calculators
Student Activity:
pattern blocks, balls,
stop watches,
soda cans, string,
meter sticks
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Materials List
Materials List
Consumables
butcher paper
masking tape
blank transparencies (some cut to fit
the overhead graphing calculator)
colored pencils or pens
pieces of flat spaghetti
markers
1” grid paper
Styrofoam cups
3” thin rubber bands
string
marbles of the same size
large paper clips
water
pieces of lumber or cardboard
patty paper or blank transparencies
blank paper
sticky notes
poster boards (large construction
paper)
balls (racket, basketball, golf)
soda cans
Non-Consumables
building blocks
meter sticks
color tiles
tape measure
PVC pipe
algebra tiles
overhead algebra tiles
flashlights
cylinders
uniform objects (to put in
cylinders)
Technology
overhead graphing calculator
data collection devices
motion detectors
light sensors
computer with projection device
laser pointers
stop watches
Software
dynamic geometry program (w/
least squares demo)
Calculator Programs:
LINEGRPH
LINETBL
ACT
CMOVE
JUMPIT
PENDULUM
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1.1 Variables and Functions: Leaders’ Notes
1.1 Variables and Functions
Overview:
Participants discuss the concept of function in the context of a school fundraising venture.
Objective:
Algebra I TEKS
(b.1) The student understands that a function represents a dependence of one
quantity on another and can be described in a variety of ways.
Terms:
variable, function, independent variable, dependent variable, mathematical
modeling
Materials:
butcher paper, masking tape
Procedures:
Participants should be seated at tables in groups of 3 – 4.
Traditionally algebra has been taught as an abstract set of algorithms based on
definitions, properties, and theorems that lead students to “an answer.”
Finding x , sketching a graph, factoring an expression, and solving a word
problem were the orders for the day. Devoid of context, these problems asked
students to make a jump from their understanding of numbers and operations
to the abstract notions of variables, equations, functions, etc. With contextrich problem situations, and using manipulatives and technology
appropriately, we can bridge the gap between students’ concrete thinking and
the abstract world of algebra.
Transparency #1: Valentine’s Day Idea
Introduce the school’s drill team money making project.
• What factors might effect the success of the project? [Record at least
one idea from each group.]
Transparency #2: Variables and Functions
Discuss list of factors that was acquired from field testing.
Introduce the terms variables and functions.
Transparency #3: Non-Mathematical Definition of Function
Activity 1: Examples of Dependent Relationships
Read the definition of function and have participant groups list at least 7
examples of dependent relationships on Activity 1. Have each group write
examples on butcher paper and post on the walls of the room. Encourage
participants to use varied language, not only “is a function of” and “depends
on,” but also examples like, “The colder it gets, the more I shiver.”
Have participants do a gallery tour and record two or three favorite examples.
TEXTEAMS Algebra I: 2000 and Beyond
Spring 2001
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Transparency #4: Examples of Dependent Relationships
Discuss examples on butcher paper and on Transparency #4.
• Which examples show a clear dependent relationship?
• Which examples, if any, do not show a clear dependent relationship?
Activity 2: Independent and Dependent Variables
Have the groups write some of their examples from Activity 1 and some of the
examples from Transparency 4 in the chart as cause-and-effect statements.
Some participants may word these in an if-then format. If no one does, help
participants word a few in an if-then format. As you are discussing their
examples, write the word “IF” under “Cause” and “THEN” under “Effect” in
the table.
Ask:
• Which of these columns depends on the other? [The Effect column
depends on the Cause column]
Write “Dependent” under “THEN”
• If “dependent” describes the second column, what word can we use to
describe the first column? [Independent]
Write “Independent” under “IF”
So the table will look as follows:
Cause
IF
Independent
If I get more sleep at night,
Effect
THEN
Dependent
then I wake up faster.
Math Note:
Not all dependent relationships can be written in an if-then format. Consider
the statement: “How I feel depends on how I eat.” You could infer meaning
and rewrite it as “If I eat chocolate, then I will be happy.” Or “If I eat quickly,
I will feel indigestion.”
Also, not all relational statements should imply causation. Just because things
are related does not necessarily mean that one causes the other. “Height is
related to shoe size.” “People who spend the most time on the Internet have
high rates of depression” Does that mean that time on the Internet causes
depression or that depressed people gravitate toward the Internet? Be careful
in the training to not infer causation just because there is relation.
Sometimes determining which is the cause and which is the effect in a relation
can be open to interpretation. The time and distance you fly in a plane are
related. Does the distance you fly in an airplane depend on the time it takes to
fly somewhere? Does the time it takes to fly somewhere depend on how far
away the place is?
The idea is to not get caught up being too picky. The focus of the activities is
to connect participant’s previous understanding of cause and effect to the
mathematical definition of function, with independent and dependent
variables. Choose some good cause and effect relationships that illustrate
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independent and dependent variables and don’t get hung up on those that do
not.
Transparency #5: Mathematical Definition of Function
Traditionally many math teachers have introduced functions with the
mathematical definitions. Students understand the concept of cause and
effect. We use that understanding in Activities 1 – 2 to bridge to this
mathematical definition.
Introduce the mathematical definition of independent and dependent variables
and discuss cause and effect relationships with variables.
Transparency 6: Stages of Mathematical Modeling Process
As with the Valentine’s Day idea, every business begins with a product or
service that the owners believe will produce a profit. Creating the business
idea is only the beginning stage. Soon decisions have to be made about how
to best deliver the product or service so that the most money can be made.
Several issues involve numbers, but simple arithmetic is seldom sufficient to
handle all the varying possibilities.
Algebraic methods will allow you (just as they allow businesses) to
mathematically define some of the varying situations. You can put together
variables and relationships into a mathematical model that describe a
business situation and study questions (i.e., What would happen if one factor
is changed?) so that optimum solutions can be identified.
Use the following statements in discussing the stages of modeling.
Businesses:
⇓ begin with a product or service that they believe will produce $$$$$,
⇓ identify issues that involve numbers and the use of algebraic methods to
mathematically define some of the varying situations,
⇓ put together variables and relationships into a mathematical model that
describes the business situation,
⇓ analyze the model by studying questions (i.e., What would happen if one
factor is changed?) so that optimum solutions can be identified,
⇓ interpret the result within the context of the business situation,
⇓ use the information to formulate conclusions and to make informed
decisions that will profit the business, and
⇓ implement plans based on the information gained through the use of the
mathematical modeling process.
Within real world situations, you must learn to:
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(1) identify problems and discern the important factors (variables/
parameters) that affect the problem,
(2) determine the relationships among those factors (variables) and
describe them mathematically,
(3) analyze the model by applying appropriate mathematical techniques
and draw mathematical conclusions,
(4) interpret the results in context,
(5) formulate conclusions and predictions so that optimum choices can be
made, and
(6) apply decisions to the real world situation.
Answers to Reflect and Apply
Often we ask students to match situations with graphs with little prior
experience in making, reading, and interpreting such graphs. Ask participants
to reflect on these Exercises as they progress through the institute. Ask
participants to consider how students may be able to better match these graphs
after the experiences suggested in the institute.
1.
2.
3.
4.
5.
Summary:
e
d.
c
a
b
Building on the non-mathematical definitions of function and variables,
participants are introduced to mathematical functions and mathematical
modeling.
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1.1 Variables and Functions: Transparency 1
Transparency 1: Valentine’s Day Idea
The school’s drill team has
decided on a money-making
project for February. They plan
to take orders (and money) for
roses in advance and deliver
them to the designated students
on Valentine’s Day.
What factors might affect the success of this moneymaking project?
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Transparency 2: Variables and Functions
The success of the money-making project depends on
many factors such as:
• publicity,
• appeal of project to the students,
• selling price of roses,
• cost of roses,
• willingness of the flower distributors to work
with the student group, and
• cooperation of the administration.
These factors are called variables because they can
change regularly. It is appropriate to say that the success
of the money-making project is a function of those
variables.
Spring 2001
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Transparency 3: Non-Mathematical Definition of Function
The American Heritage Dictionary of the English
Language defines function as something closely related to
another thing and dependent upon it for its existence,
value, or significance.
Examples of Function
1. How fast I wake up in the morning depends on (is a
function of) how much sleep I get.
2.
3.
4.
5.
6.
7.
8.
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Transparency 4: Examples of Dependent Relationships
1. How fast I wake up in the morning depends on (is a
function of) how much sleep I get.
2. Our height is a function of our age.
3. The amount of change in my pocket depends on (is a
function of) the type of coins found there.
4. The amount of studying determines the grade we make.
5. As my car gets older, it is worth less.
6. The grass gets greener as I put more fertilizer on it.
7. The amount of money I make at my job depends on (is
a function of) the number of hours I work.
Spring 2001
8
Transparency 5: Mathematical Definition of Function
A mathematical function expresses a dependency
relationship: one quantity depends in a systematic way on
another quantity.
For Example:
y = 2 x + 1 is a function and expresses a dependency
relationship.
The value of y depends on the value of x.
The variable x is called the input or independent variable.
The variable y is called the output or dependent variable.
In an applied mathematics setting, we must decide which
decisions (input) influence or produce which results
(output). In many cases this is a cause and effect
relationship where the cause is the independent variable
and the effect is the dependent variable.
Spring 2001
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Transparency 6: Modeling with Mathematics
Stages of Mathematical
Modeling Process
Entry
Real world
Situation
Identify
Apply
Problem,
Variables,
Constraints
Create
Conclusions,
Predictions,
Decisions
Mathematical
Model
Analyze/
Summarize
Formulate
Problem
Situation
Interpret
Spring 2001
Mathematical
Conclusions
10
1.1 Variables and Functions: Activity 1
Activity 1: Examples of Dependent Relationships
1. How fast I wake up in the morning depends on (is a function
of) how much sleep I get.
2.
3.
4.
5.
6.
7.
8.
Spring 2001
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1.1 Variables and Functions: Activity 2
Activity 2: Independent and Dependent Variables
Identify the independent and dependent variables from some of
the dependent relationship examples.
Cause
Effect
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1.1 Variables and Functions: Reflect and Apply
Reflect and Apply
Match the following descriptions with a graph:
a.
___ 1. The volume of
popcorn popping over
time.
___ 2. Phone deal: $0.50 for
the first 20 minutes and
$0.07 per minute after.
b.
___ 3. The worth of my car
over time.
c.
___ 4. He walked up and then
down a hill, speed
versus time.
d.
___ 5. She walked away and
then walked back,
distance versus time.
e.
Spring 2001
13
1.2 Valentine’s Day Idea: Leaders’ Notes
1.2 Valentine’s Day Idea
Overview:
Participants investigate functional relationships for a given problem situation
using tables, graphs, and algebraic representations.
Objective:
Algebra I TEKS
(b.3) The student understands algebra as the mathematics of generalization and
recognizes the power of symbols to represent situations.
Terms:
function, independent variable, dependent variable, pattern
Materials:
Procedures:
Participants should be seated in groups of 3 – 4.
This entire activity is an overview problem to give participants a feel for a
function based approach to algebra and for the power of multiple
representations. Thus participants do not need to gain mastery at this point,
but work to give them exposure to many of the goals and concepts of the
institute.
Activity 1: Valentine’s Day Idea
Introduce the problem situation to the whole group.
• Which seems better:
75 cents per rose or 50 cents per rose?
$20 fixed cost or $60 fixed cost?
75 cents per rose plus $20 fixed cost or 50 cents per rose plus $60
fixed cost?
Because students have a tendency to focus on certain parts of a problem
situation while not attending to other parts, this line of questioning assists
them in considering multiple variables at one time.
• What kind of customer would order from Roses-R-Red? [A customer
who needs a relatively small amount of flowers.]
• What kind of customer would order from The Flower Power? [A
customer who needs a relatively large amount of flowers.]
Student Activity: Investigate Recursively
Do the student activity with your participants as appropriate. Help
participants generate the recursive routines, first without braces and then with
braces.
Note that both the original and the new rose offer are explored.
Have participants point out questions that ask for an output value given an
input value, and questions that ask for an input value given an output value.
Spring 2001
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Doing activities like this helps students gain confidence in reading a math
word problem.
Math Note: Recursion is a process where each successive term is based on the
previous term.
Activity 2: Using Tables to Find the More Economical Offer
Have participants complete the chart numerically first, looking for patterns.
Develop the patterns live on the transparency of Activity 2 or use
Transparency 1. Using the patterns, develop the function rules representing
the cost from each flower shop.
Ask participants to demonstrate the validity of their function rules for each
flower shop.
Have half of the group use the statistics capability of graphing calculators to
enter the number of flowers sold into a list. Then enter the formula
20 + 0.75( List1) into the second list. Enter the formula 60 + 0.5( List1) into a
third list. Check that the values generated match the values on the activity
sheet.
Have the other half of each group use the table building capability of graphing
calculators to enter y = 20 + 0.75 x and y = 60 + 0.5 x .
•
What would you expect 240 roses to cost from Roses-R-Red? [If
participants respond $220, then they do not understand that the
“doubling effect” does not work here because this is not a proportional
relation. A proportional relation is in the form y = mx .]
Work through the rest of the Activity, using the following to discuss.
1. The cost of roses from Roses-R-Red is $20 and add $0.75 for every rose.
Cost is 20 + 0.75( Number of Roses) , C = 20 + 0.75r .
2. The cost of roses from Flower Power is $60 and add $0.50 for every rose.
Cost is 60 + 0.50( Number of Roses) , C = 60 + 0.50r .
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15
3. Answers will vary, including increasing, increasing at different rates,
increasing by constant amounts.
4. The cost increases as the number of roses purchased increases. The
graph is linear with a positive slope.
5. Write the equation 65 = 20 + 0.75r , then look for the answer in the table.
Write r = 60. Do not solve the equation algebraically at this time.
6. Write the equation 65 = 60 + 0.5r , then look for the answer in the table.
Write r = 10 . As above, do not solve the equation algebraically at this
time. Rather, suggest that this activity could be done before students have
learned to solve one and two step equations. If students see equations like
this in context, learn how to solve them using a table (and later in the
activity, using a graph), then when students learn the algebraic steps to
solving similar equations, they will already have context and purpose.
Students will see algebraic methods as another way to find answers.
Maybe we can eradicate, “Why do we have to find x anyway?”
7. Roses-R-Red offers the better deal for n < 160, there is not difference for
n = 160 , and Flower Power offers the better deal for n > 160.
8. The point of intersection is (160, 140). This point signifies that at n=160
roses, the cost ($140) is the same with either flower dealer.
9. 20 + 0.75r = 60 + 0.50r .
Activity 3: Using Graphs to Find the Better Offer
Note that the ways in which two variables are related is not always shown
clearly by tables of input-output values. Patterns in the data may be lost amid
all the specific numbers. However, when data are displayed in a graph, it is
often much easier to see trends and therefore to make predictions and/or
informed decisions.
First, have participants make a scatter plot of the data that was entered into
lists (number of roses sold, cost from Roses-R-Red) and (number of roses
sold, cost from Flower Power).
Use the following questions to help participants graph the scatter plots and
functions on the graphing calculator.
• What do you think the graph will look like? Predict.
• What viewing window makes sense for the problem situation?
• What does the variable x represent? [Number of roses.]
• What values make sense for the number of roses sold? [See Sample
Answers below.]
• What does the variable y represent? [Cost of roses]
• What values make sense for the cost of roses? [See Sample Answers
below.]
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16
After they have graphed the scatter plots, graph the functions together with the
plots.
Make the connection between patterns seen in the tables of values and the
graphs of the same ordered pairs. As the number of roses increases, the total
cost increases.
Make connections between the patterns seen in the graphs and the
interpretation in the context. Note that to get from one point to another one,
we must move to the right (increase in quantity) and then up (increase in cost)
or we can move to the left (decrease in quantity) and then down (decrease in
cost).
Sample Answers:
xmin=0:
The variable x stands for the number of roses. It makes sense to
look at the cost if no one bought a rose.
xmax=270: We think that we might sell at most 270 roses.
ymin= 0:
The variable y stands for the cost of the roses. The cost will not
be negative.
ymax=200: This is a reasonable maximum cost for our maximum rose count
of 270.
•
What does the ordered pair (210, 165) mean on the graph? [You can
buy 210 roses for $165.]
3. The 75 cent per stem cost is the rate of change or slope of the line. The
$20 is the y-intercept and in this case, it makes sense to call it the starting
point.
4. The 50 cent per stem cost is the rate of change or slope of the line. The
$60 is the y-intercept and in this case, it makes sense to call it the starting
point.
5. The point of intersection is (160, 140), which means that it does not matter
from whom you purchase 160 stems because it will cost $140 at both
places. Show participants the following screen.
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17
6. Roses-R-Red offers the better deal when you purchase less than 160 roses.
Flower Power offers the better deal when you purchase more than 160
roses. Use the trace cursor to demonstrate.
• What are the meanings of the coordinates on the screens?
At Roses-R-Red, 110 stems cost $102.50.
At Flower Power, 110 stems cost $115.
At Roses-R-Red, 197 stems cost $167.75
At Flower Power, 197 stems cost $158.50.
Discuss the meaning of slope and y-intercept within the context of this
problem. For example, the algebraic rule y = 20 + 0.75 x yields (a) y-intercept
of 20 which indicates the fixed cost that has to be paid initially and (b) slope
of 0.75 which indicates the constant rate of change or the constant increase in
cost for the purchase of each rose.
Activity 4: New Rose Offer
Introduce the new offer made by the distributors.
Activity 5: Using Tables for New Rose Offers
Encourage participants to complete the table so that comparisons between the
table in Activity 2 and this one can be made. Discuss changes from old deal
to new deal. Ask participants to share ways to assist students in making the
connection between the table values and the corresponding graphs.
Spring 2001
18
Note: Depending on the experience of your audience, you may want to
continue to have them write sentences before they write function rules for
each deal.
Transparency 2: Comparing Tables
• Looking at the table values for Roses-R-Red, what has happened to all
of the costs, the y-values? [The y-values have all been decreased by
20.]
• How has that affected the function representing the cost from Roses-RRed? [The starting point from the original offer has been decreased by
20, so the new function is 20 less than the original.]
• Looking at the table values for Flower Power, what has happened to
all of the costs, the y-values? [The y-values have also all been
decreased by 20.]
• How has that affected the function representing the cost from Flower
Power? [The starting point from the original offer has been decreased
by 20, so the new function is 20 less than the original.]
Note that the slight modification made to the old deal is reflected in the
function rules and their representations. For example,
C = 20 + 0.75n
Cn = 20 + 0.75n − 20 = 0.75n
C = 60 + 0.5n
Cn = 60 + 0.5n − 20 = 40 + 0.5n
old offer by Roses-R-Red
new offer by Roses-R-Red
old offer by Flower Power
new offer by Flower Power
Making the connections above is an important precursor for the next section,
graphing. Return to the table when you discuss the vertical translation (shift)
in Activity 6.
Activity 6: Using Graphs for New Rose Offer
1. Encourage participants to always predict what a graph will look like
before looking at the graph on the calculator.
2.
3. The graph of the new rule is the graph of the original rule translated
(shifted) down 20. Connect this with your comparisons in the table, where
all of the costs, the y-values decreased by 20.
Spring 2001
19
5.
6. The graph of the new rule is the graph of the original rule translated
(shifted) down 20. Connect this with your comparisons in the table,
where all of the costs, the y-values decreased by 20.
7. Show the following screens:
The x-coordinates are the same! In other words, the “deal” is the same in
that Roses-R-Red are still better for less than 160 stems and Flower Power
are still better for over 160 stems. The costs are different by $20, but the
answer to the question, “Who has the better deal?” remains the same. You
can see this in the screens below, where a vertical line shows that the xcoordinates of the intersection points are the same.
8. A graph can be used in comparing the two dependent relations. We can
see from the graphs that the two flower distributors charge the same
amount when 160 roses are purchased. Use the trace cursor on both
graphs to see that Roses-R-Red is less expensive when n < 160and Flower
Power is less expensive when n > 160.
Note the relationship between the graphical representation of the:
20 + 0.75 x = 60 + 0.50 x .
original offers
0.75 x = 40 + 0.50 x .
new offers
Although both functions shift 20 units down the y-axis, the intersection of the
transformed graphs has the same input value as the original graphs.
• Why does this make sense algebraically? [Subtract 20 from both sides
of the equation and the result is an equivalent equation, thus having the
same solution. In both equations, x = 160 .]
Spring 2001
20
Extension: In the new rose deal, we looked at the effect of subtracting $20
from the cost of the roses from both flower dealers. Essentially, the deal
remained the same in that Roses-R-Red were better for less than 160 stems
and Flower Power were better for over 160 stems. Now, try looking at the
effect of leaving the fixed costs the same and changing the amount of money
per stem. Will the deal remain the same?
Answers to Reflect and Apply:
1. Flowers-R-Us sells roses for $1.00 a stem. y = x
2. All Occasion Roses sells roses for $10.00 plus $0.50 per stem. y = 10 + .5 x
3. The point of intersection (20, 20) tells us that we can buy 20 roses for $20
from either place. Buying less than 20 stems is a better deal from
Flowers-R-Us and buying more than 20 stems is a better deal from All
Occasion Roses.
4. You can purchase roses from R for $15.00 plus $1.25 per stem.
5. You can purchase roses from S for $35 plus $1.00 per stem.
6. The point of intersection (80, 115) tells us that we can buy 80 roses for
$115 from either place. Buying less than 80 stems is a better deal from
Flowers-R-Us and buying more than 80 stems is a better deal from All
Occasion Roses.
7. With the coupon, roses are $2.00 plus $0.50 per stem. y = 2 + .5 x
8. Lilies cost $5.00 per stem.
9. Daisies cost $5.00 plus $0.25 per stem.
10.
Summary:
Fund-raising by selling roses for Valentines Day is the context for a rich
problem where participants make connections between verbal, numerical,
graphical, and algebraic representations of a linear situation. Participants
build intuition for constant rates of change and y-intercepts of lines.
Spring 2001
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1.2 Valentine’s Day Idea: Transparency 1
Transparency 1: Using Tables to Find the More Economical
Offer
From the description of the two offers, complete
the chart to find an algebraic rule that will
determine the cost of n roses.
Number
of Roses
Process
Column
10
20
30
60
90
120
150
180
210
240
1000
20 + 0.75(10)
20 + 0.75(20)
20 + 0.75(30)
20 + 0.75(60)
20 + 0.75(90)
20 + 0.75(120)
20 + 0.75(150)
20 + 0.75(180)
20 + 0.75(210)
20 + 0.75(240)
20 + 0.75(1000)
n
20 + 0.75n
Cost at
Process
RosesColumn
R-Red
$27.50
60 + 0.50(10)
$35.00
60 + 0.50(20)
$42.50
60 + 0.50(30)
$65.00
60 + 0.50(60)
$87.50
60 + 0.50(90)
$110.00 60 + 0.50(120)
$132.50 60 + 0.50(150)
$155.00 60 + 0.50(180)
$177.50 60 + 0.50(210)
$200.00 60 + 0.50(240)
$770.00 60 + 0.50(1000)
Cost at
Flower
Power
$65.00
$70.00
$75.00
$90.00
$105.00
$120.00
$135.00
$150.00
$165.00
$180.00
$560.00
60 + 0.50n
Spring 2001
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1.2 Valentine’s Day Idea: Transparency 2
Transparency 2: Comparing Tables
Original Offer
Number
of Roses
Process
Column
30
60
90
120
150
180
210
240
n
20 + 0.75(30)
20 + 0.75(60)
20 + 0.75(90)
20 + 0.75(120)
20 + 0.75(150)
20 + 0.75(180)
20 + 0.75(210)
20 + 0.75(240)
20 + 0.75n
Cost at
Process
RosesColumn
R-Red
$42.50 60 + 0.50(30)
$65.00 60 + 0.50(60)
$87.50 60 + 0.50(90)
$110.00 60 + 0.50(120)
$132.50 60 + 0.50(150)
$155.00 60 + 0.50(180)
$177.50 60 + 0.50(210)
$200.00 60 + 0.50(240)
60 + .50 n
Cost at
Flower
Power
$75.00
$90.00
$105.00
$120.00
$135.00
$150.00
$165.00
$180.00
New Offer
Number
of Roses
(new)
30
60
90
120
150
180
210
240
n
Process
Column
0.75(30)
0.75(60)
0.75(90)
0.75(120)
0.75(150)
0.75(180)
0.75(210)
0.75(240)
0.75n
Cost at
RosesR-Red
$22.50
$45.00
$67.50
$90.00
$112.50
$135.00
$157.50
$180.00
Spring 2001
Process
Column
40 + .50(30)
40 + .50(60)
40 + .50(90)
40 + .50(120)
40 + .50(150)
40 + .50(180)
40 + .50(210)
40 + .50(240)
40 + .50 n
Cost at
Flower
Power
$55.00
$70.00
$85.00
$100.00
$115.00
$130.00
$145.00
$160.00
23
1.2 Valentine’s Day Idea: Activity 1
Activity 1: Valentine’s Day Idea
The school’s drill team has contacted several flower distributors
and has narrowed the choice to two companies.
Option 1: Roses-R-Red has offered to sell its roses for a fixed
down payment of $20 and an additional charge of 75 cents per
stem.
Option 2: The Flower Power has offered to sell its roses for a
fixed down payment of $60 and an additional charge of 50 cents
per stem.
Which is the more economical offer?
Spring 2001
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Activity 2: Using Tables to Find the More Economical Offer
From the description of the two offers, complete
the chart to find an algebraic rule that will
determine the cost of n roses.
Number
of Roses
Cost at
Process
Roses-RColumn
(Roses-R-Red)
(Flower Power)
Red
Process
Column
Cost at
Flower
Power
10
20
30
60
90
120
150
180
210
240
1000
n
1. Write a sentence and a function rule for the cost of roses
from Roses-are-Red.
2. Write a sentence and a function rule for the cost of roses
from Flower Power.
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3. What patterns do you observe from the table of values?
4. What happens to the cost of the roses as the number of roses
purchased increases? What would a graph of this
relationship look like?
5. How many roses can you buy from Roses-are-Red for
$65.00?
6. How many roses can you buy from Flower Power for
$65.00?
7. Which company offers the better deal?
8. Is there a point where the two flower dealers charge the same
total amount? If so, what is the charge? If not, why do the
costs never equal?
9. Write an equation that represents the point where the two
flower shops charge the same amount.
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Activity 3: Using Graphs to Find the Better Offer
1. Find an appropriate viewing window for the graphs of both
functions. Sketch both functions here and label.
2. Justify your viewing window choice:
xmin:
xmax:
ymin:
ymax:
Spring 2001
27
3. What effect does the 75 cents per stem cost have on the graph
of the Roses-R-Red function? What effect does the $20 have
on the graph?
4. What effect does the 50 cents per stem cost have on the graph
of the Roses-R-Red function? What effect does the $60 have
on the graph?
5. What are the coordinates of the point of intersection of the
two functions? What is the significance of this point?
6. Which flower dealer offers the better deal? Justify your
answer.
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Activity 4: New Rose Offers
To entice these potential new customers, Roses-R-Red decides
to eliminate its fixed charge of $20. According to its new offer,
the drill team pays only for the roses they buy. When the
Flower Power learns about the new offer by its competitor, it
immediately enters the price war by reducing its fixed charge
also by $20.
Which new deal is the better offer?
How does the new offer compare to the original offer?
Spring 2001
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Activity 5: Using Tables for New Rose Offers
From the description of each of the two new deals, complete the
chart and write new algebraic rules that will determine the cost
of n roses.
Cost at
Process
Cost at
Number
Process
Roses-RColumn
Flower
of Roses
Column
(Roses-R-Red)
(Flower Power)
Red
Power
10
20
30
60
90
120
150
180
210
240
270
300
n
1. What patterns do you observe in the new table of values?
2. Compare the costs on this chart to the costs on the first chart.
What changes do you observe? Predict what the graphs will
look like.
3. Which company offers the better deal?
4. Is there a point where the two flower dealers charge the same
amount? If so, what is the charge?
5. Write an equation that represents the point where the two
flower shops charge the same amount.
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Activity 6: Using Graphs for New Rose Offers
1. On your graphing calculator graph only the original Roses-RRed function, y = 20 + 0.75 x . Now, predict what you think
the graph of the new offer, y = 0.75 x , will look like.
2. Graph both the original offer, y = 20 + 0.75 x , and the new
offer, y = 0.75 x , together.
3. What effect does subtracting $20 from the old rule have on
the new graph of the Roses-R-Red function?
4. Turn off the above two graphs. Graph only the original
Flower Power function, y = 60 + 0.5 x . Now, predict what
you think the graph of the new offer will look like,
y = 40 + 0.5 x .
5. Graph both the original offer, y = 60 + 0.5 x , and the new
offer, y = 40 + 0.5 x , together.
6. What effect does subtracting $20 from the old rule have on
the new graph of the Flower Power function?
7. Graph all four functions at the same time. What are the
coordinates of the point where the two new functions
intersect? What is the significance of this point?
8. Which flower dealer now offers the better deal?
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9. Sketch the graphs of all four functions and label the relevant points
of intersection.
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1.2 Valentine’s Day Idea: Reflect and Apply
Reflect and Apply
1. Describe the rose deal represented by
the graph with a sentence and a
function.
Flowers-R-Us
Cost
$10
5
Number of Roses
All Occasion Roses
2. Describe the rose deal represented by
the graph with a sentence and a
function.
Cost
$10
5
Number of Roses
3. Using your graphing calculator, graph the above two rose
deals together in the same window, find the point of
intersection and discuss what the point means in the fundraising context. Sketch the window with the two graphs.
Spring 2001
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4. Describe the rose deal represented by the function:
R( x ) = 15 + 1.25 x
5. Describe the rose deal represented by the function:
S( x ) = 35 + x
6. Sketch the graphs of the above two rose deals below, find the
point of intersection and discuss what the point means in the
fund-raising context. Use your graphing calculator to
confirm.
7. The drill team has a coupon for All Occasion Roses for $8
off the purchase of roses. Find a function to represent the
cost of buying roses from All Occasion Roses if there was no
minimum purchase required. Write the function and sketch
the graph.
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8. Describe the flower deal represented by the table.
Number of Lilies
0
1
2
3
4
Cost
$0.00
$5.00
$10.00
$15.00
$20.00
9. Describe the flower deal represented by the table.
Number of Daisies
0
1
2
3
4
Cost
$5.00
$5.25
$5.50
$5.75
$6.00
10. Sketch the graphs of the functions in Exercises 8 and 9 on
the grid below.
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1.2 Valentine’s Day Idea: Student Activity
Overview:
Using the power of the graphing calculator, students explore problems
recursively.
Objective:
Algebra I TEKS
(b.3) The student understands algebra as the mathematics of generalization and
recognizes the power of symbols to represent situations.
Terms:
Recursion
Materials:
Graphing calculators
Procedures:
Do exercises 1 – 6 with students, using an overhead calculator to demonstrate.
Then have students complete the New Rose Offers exercises.
1. Help students generate the recursive routines as follows, first without
braces and then with braces.
2.
3. $44.00
4. 23 roses
5. $76.00
6. 43 roses
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New Rose Offers
1.
2.
3.
4.
5.
6.
7. A fixed charge of $50 with $1.00 per rose.
9. No fixed charge with $0.90 per rose.
10. No fixed charge with $0.50 per rose.
Note: A recursively defined sequence is defined by a starting value and a
rule. You generate the sequence by using the rule on the starting value and
then again on the resulting value and so on. Traditionally we have focused on
functions written in closed form in a first year algebra class. In the institute,
we use recursion to lead students to writing equations to model situations.
Summary:
Students naturally operate recursively. We capitalize on this and use the
power of the graphing calculator to explore some linear situations, looking for
both input and output answers. Students build confidence in reading and
understanding word problems. Students build intuition for constant rates of
change and y-intercepts in linear situations.
Spring 2001
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Roses-R-Red has offered to
sell its roses for a fixed down
payment of $20 and an
additional charge of 75 cents
per stem.
The Flower Power has offered
to sell its roses for a fixed
down payment of $60 and an
additional charge of 50 cents
per stem.
1. Generate a recursive routine to investigate the Roses-R-Red
offer.
2. Generate a recursive routine to investigate the Flower Power
offer.
Using your recursive routine, answer the following:
3. How much would it cost to order 32 roses from Roses-RRed?
4. How many roses can you order from Roses-R-Red for
$37.25?
5. How much would it cost to order 32 roses from Flower
Power?
6. How many roses can you order from Flower Power for
$81.50?
Spring 2001
38
New Rose Offers
To entice these potential new customers, Roses-R-Red decides
to eliminate its fixed charge of $20. According to its new offer,
the drill team pays only for the roses they buy. When the
Flower Power learns about the new offer by its competitor, it
immediately enters the price war by reducing its fixed charge
also by $20.
1. Generate a recursive routine to investigate the new Roses-RRed offer.
2. Generate a recursive routine to investigate the new Flower
Power offer.
Using your recursive routine, answer the following:
3. Now how much would it cost to order 43 roses from RosesR-Red?
4. Now how many roses can you order from Roses-R-Red for
$27.00?
5. How much would it cost to order 43 roses from Flower
Power?
6. How many roses can you order from Flower Power for
$58.00?
Spring 2001
39
Students decided to check around at other shops. Based on the
screen shots below, what kind of deals did they find?
7.
8.
9.
10.
11.
Spring 2001
40
2.1 Identifying Patterns: Leaders' Notes
2.1 Identifying Patterns
Overview:
Participants represent linear relationships among quantities using concrete
models, tables, diagrams, written descriptions, and algebraic forms.
Objective:
Algebra I TEKS
quantity on another and can be described a variety of ways.
(b.2.B) For a variety of situations, the student identifies the mathematical
domains and ranges and determines reasonable domain and range values for
given situations.
(b.2.C) The student interprets situations in terms of given graphs or creates
situations that fit given graphs.
(b.3) The student understands algebra as the mathematics of generalization
and recognizes the power of symbols to represent situations.
(c.1.B) The student determines the domain and range values for which linear
functions make sense for given situations.
Terms:
function, independent variable, dependent variable
Materials:
building blocks, color tiles, graphing calculators
Procedures:
Do the Student Activity depending on the level of participants.
Activity 1: Painting Towers
Do the Activity together as a whole group, bringing out the following points
and asking the indicated questions.
1. Encourage participants to write how they found the number of faces to
paint in the process column. This can often be done in several ways,
which will lead to different, yet equivalent algebraic expressions. This is a
desired outcome. Possible equivalent expressions include:
4 +1
2( 4 ) + 1
3( 4) + 1
n( 4) + 1 = 4n + 1
2 + 2 +1
2(2) + 2(2) + 1
3(2) + 3(2) + 1
n(2) + n(2) + 1
5
5+4
5 + 2( 4)
5 + (n − 1)( 4)
Ask participants to use the cubes to physically demonstrate the algebraic rules
they found in the table. In this example, they will mainly be pointing to faces
on the cubes and relating them to the heights of the towers.
Note: encourage participants to obtain the equivalent expressions from the
model, not by simplifying the algebraic expressions.
Spring 2001
41
2.
Discuss with participants that choosing appropriate windows and then
justifying window choices is a precursor to students learning about domain
and range.
Graph the function over the scatter plot to verify as shown above.
3.
4.
4 x + 1 = 25
4 x = 24
x=6
•
•
What does the ordered pair (8, 33) mean? [For term number 8
(figure 8 or a tower 8 cubes high), the numerical term value
(number of faces to paint) is 33.]
Does the ordered pair (12,50) belong to this graph? Why or why
not? [A simple answer might be that there will always be that
extra top face to paint, and therefore the number of faces to paint
will be odd. Thus it is not possible for there to be 50 faces to paint.
Also, the symbolic rule suggests odd numbers.]
5. Examples:
6+2
2(6) + 2
3(6) + 2
n(6) + 2 = 6n + 2
2(3 + 1)
2(2(3) + 1)
2(3(3) + 1)
2(n(3) + 1) =
2(3n + 1)
Spring 2001
42
6. Sample answer: For 35 cubes in each column: you need to paint 6 faces
times 35 plus the top 2 faces.
7. F = 6n + 2
9.
•
What changed in the rule? The figure? The graph? The domain?
The range? Compare.
Use the following questions to compare the single tower situation and the
double tower situation.
• What represents the “changing quantity” and what represents the
“fixed quantity” in the pattern? [The addition of 4 faces with each
additional cube represents the changing quantity in the single
tower. The addition of 6 faces with each additional level
represents the changing quantity in the double tower. The faces on
tops of the towers represents the unchanging one and two faces.]
• What number in the rule affects the slope, steepness, of the line?
[The coefficient of x . In this case, it is the number 4 in the single
tower and the number 6 in the double tower.]
• What number in the rule affects the starting point for the scatter
plot (y-intercept for the line)? [The constant. In the first case, it is
the number 1, and in the second case, the number 2.]
Underline the constants in both of the functions, y = 1 + 4 x and
y = 2 + 8x .
• What do the constants represent in the functions, y = 1 + 4 x and
y = 2 + 8 x ? [faces to paint]
Circle the coefficient of x in both of the functions, y = 1 + 4 x and
y = 2 + 8x .
• What do the coefficients of x represent? [faces to paint per tower
height or faces to paint per figure number.]
Repeat the above questions for Activities 2 – 3.
Activity 2: Building Chimneys
Have participants do Activity 2 together in groups. Discuss as a whole group,
asking several participants to share their different methods of arranging the
cubes to find appropriate expressions.
1. Encourage participants to write how they found the number of blocks in
the process column.
Spring 2001
43
For example:
6 + 2(1)
6 + 2(2)
6 + 2(3)
6 + 2n
2 + 2(2 + 1)
2 + 2(2 + 2)
2 + 2(2 + 3)
2 + 2(2 + n)
2. Graph the function over the scatter plot to verify as shown.
A reasonable domain for the situation is 0 to 10 blocks and a reasonable
range is 0 to 26 blocks.
3. You need 6 blocks for the base and then 23 rows of 2 blocks for the
chimney. The ordered pair is (23, 52).
4. You need 28 blocks to build a house with a chimney 11 blocks high. The
ordered pair is (11, 28).
5. No, the ordered pair (13, 34) does not belong to the graph because if you
were building figure 13, you would need 13(2) + 6 = 32 blocks, not 34.
6. Examples:
9+1
9+2
9+3
9+n
3 + 3 + (3 + 1)
3 + 3 + (3 + 2)
3 + 3 + (3 + 3)
3 + 3 + (3 + n)
7. Sample answer: The total number of blocks equals 9 plus the number of
blocks in the chimney.
Spring 2001
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9.
•
What changed in the rule? The figure? The graph? Compare.
Activity 3: Constructing Trucks
1. Encourage participants to write how they found the number of blocks in
the process column.
For example:
3+2
2+2+1
3(2) + 2
3+3+2
3(3) + 2
4+4+3
3n + 2
2(n+1) + n
2. Graph the function over the scatter plot to verify as shown.
3. You need 152 blocks for the 50th figure.
4. If you use 242 blocks, you are on the 80th figure (term number).
2 + 3n = 242
3n = 240
n = 80
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5. Examples:
3+1
3(2) + 1
3(3) + 1
3n + 1
2+2
2(2) + 3
2(3) + 4
2n + (n + 1)
6. Sample answer: The total number of blocks equals 1 plus 3 times the
figure number.
8. The second scatter plot, with one block on top of each truck starts (the yintercept is) lower than the previous plot, with two blocks on the top of
each truck.
10. The graph starts higher than the original because now you have a constant
4 blocks on top of each truck. The graph is steeper than the original
because you are now adding 6 blocks each time, instead of 3.
Compare Activities 1 – 3
Use the following questions to compare the previous activities.
• What changed in the rule? The figure? The graph? Compare.
• What represents the “changing quantity” and what represents the
“fixed quantity” in each of the patterns?
• What number in the rule changes the slope of the line?
• What number in the rule affects the starting point for the line?
On graphs of the lines generated in the Activities, draw triangles to show the
idea that all of these rules have a constant rate of change, each time the same
thing was changing. See below for an example.
Spring 2001
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Activity 4: Generating Patterns
Have half of the groups generate patterns that model surface area, similar to
Activity 1: Painting Towers. Have the other half of the groups generate
patterns that model volume (number of cubes), similar to Activity 2: Building
Chimneys and Activity 3: Constructing Trucks.
Sample answers:
1. How many faces to paint:
Term
Number
1
Visual
(Figure)
Written
Description
Paint 4 lateral
faces and the top
and bottom.
Paint 4 lateral
faces twice and the
top and bottom.
Paint 4 lateral
faces three times
and the top and
bottom.
2
3
Process
Column
4+2
Total Faces
to Paint
6
2( 4) + 2
10
3( 4) + 2
14
4n + 2
n
How many blocks to build:
Term
Number
1
Visual
(Figure)
Written
Description
Base of 4 plus 2.
Process
Column
4+2
Number of
Blocks
6
2
Base grows by 4
plus the 2 on top.
2( 4) + 2
10
3
Base grows by 4
more plus the 2 on
top.
3( 4) + 2
14
4n + 2
n
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2. How many faces to paint:
Term
Number
1
Visual
(Figure)
Written
Description
Paint the two faces
on the end plus the
8 front and back
faces.
Paint the 2 new
front and back
faces plus the two
on the end plus the
original 8 lateral
faces.
Add the 2 new
faces to the
previous.
2
3
Process
Column
2+8
Faces to
Paint
10
2(2) + 8
12
3(2) + 8
14
2n + 8
n
How many blocks to build:
Term
Number
1
Visual
(Figure)
Written
Description
Base of 8 plus 2.
Process
Column
8+2
Number of
Blocks
10
2
Base of 8 plus 2
rows of 2 on top.
8 + 2( 2 )
12
3
Base of 8 plus 3
rows of 2 on top.
8 + 3(2)
14
8 + 2n
n
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20
15
10
5
1
2
3
4
Term Number
Note that both exercises in Activity 4 are examples of doing and undoing – an
important habit of mind for algebraic thinking.
Math Note: The manipulative model (using blocks to build figures to
represent patterns) has inherent domain and range restrictions. The sequences
generated in the table are for whole number input (domain) values. For
example, you would not build figures with 0.5 or 0.3 of a block. The
algebraic equations developed in this Activity are linear. The domain and
range of a line are all real numbers
1. The first set of figures show adding 3 blocks every time (3x) to a constant 6
blocks, 3 x + 6 . The second set of figures show 3 groups of adding a block
every time, x, to a constant 2 blocks, 3( x + 2) .
Spring 2001
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2. b
3. d
4. a
5. c
As an extension, ask participants to label the axes with units and explain their
reasoning.
Summary:
By using concrete models and the process column, participants model linear
patterns and explore constant rates of change. Participants model both input
and output questions with equations and solve them using tables and graphs.
Spring 2001
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2.1 Identifying Patterns: Activity 1
Activity 1: Painting Towers
Suppose you are painting a tower built from cubes, based
on the pattern below. Use the table to find the
relationship between the number of faces to paint and the
number of blocks in the tower.
(Paint only the sides and the top.)
Term
Visual
Written
Number (Figure) Description
(Number
of blocks)
A 1 cube-high
1
tower has 5
faces to paint.
2
Process
Column
Numerical
Value of Term
(Faces to Paint)
5
9
3
4
n
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1. Use the process column to write a function that expresses the
relationship between the number of faces to be painted and
the number of cubes.
2. Graph the data from your table on 1” graph paper and/or
create a scatter plot on a graphing calculator. What is a
reasonable domain for this situation? A reasonable range?
3. How many faces need to be painted for a 25 cube tower?
Explain two ways of getting an answer.
4. If the tower you paint has 25 faces, how many cubes are in
the tower? Explain two ways of getting an answer.
Spring 2001
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5. Suppose you have two adjacent columns of cubes instead of
the one column as before. Use your cubes to build the first
four figures and determine the number of faces that need to
be painted.
Term
Visual
Written
Process
Numerical
Number (Figure) Description Column
Value of Term
(Faces to Paint)
1
2
3
4
n
6. Write a rule in words to describe how to find the total
number of faces that need to be painted for two columns of
cubes with 35 cubes in each column.
7. Write a rule in symbols that expresses the relationship
between the number of cubes in each column and the total
number of faces to be painted.
8. Predict how the graph of this data differs from the graph of
the original data. Explain.
9. Graph the above data on 1” graph paper and/or create a
scatter plot on a graphing calculator and compare to the
previous graph.
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Activity 2: Building Chimneys
Suppose you are building a house with a chimney, based on the
pattern below. Use the table to find the relationship between the
number of blocks you need and the term number.
Term
Number
1
Visual
(Figure)
Written
Description
A house with a
chimney 1
block high
takes 8 blocks
to build.
2
Process
Column
Numerical
Value of Term
(number of
blocks)
8
10
3
4
n
Spring 2001
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relationship between the total number of blocks needed to
build the house and the term number.
Explain.
3. Use words to describe how to use blocks to build a house
with a total of 52 blocks. What is this ordered pair on the
graph?
4. If a house has a chimney that is 11 blocks high, how many
blocks will you need to build the house? What is this ordered
pair on the graph?
5. Does the ordered pair (13, 34) belong to this graph? How do
you know?
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6. Suppose the chimney is made of 1 block instead of two and
the house is built of three rows of 3 blocks instead of two
rows of 3 blocks. Use your cubes to build the first three
figures and record the data below.
Term
Visual
Written
Process
Numerical
Column
Value of
Term
1
2
3
4
n
7. Write a rule for this new data that expresses the relationship
between the total number of blocks and the number of blocks
in the chimney for a house.
previous graph.
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Activity 3: Constructing Trucks
Suppose you are building a truck, based on the pattern below.
Use the table to find the relationship between the number of
blocks you need and the figure number.
Term
Number
1
Visual
(Figure)
Written
Description
The truck has a
base of 3
blocks with 2
blocks on top.
2
Process
Column
Numerical
Value of
Term
5
8
3
4
n
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relationship between the total number of blocks needed to
build the truck and the term number.
3. Find the total number of blocks needed for the 50th figure.
4. If there are a total of 242 blocks, what term number is this?
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5. Suppose there is only one block on the top of each truck.
Use your cubes to build the first three figures and record the
data below.
Term
Visual
Written
Process
Numerical
Column
Value of
Term
1
2
3
4
n
between the total number of blocks and the figure/term
number.
previous graph. What effect did changing the number of
blocks on top of the truck have on the graph?
9. Suppose the original trucks (2 blocks on top) were built
“double-wide.” Predict how the graph differs from the
original.
10. Build the first three “double-wide” trucks and graph the data
on your graphing calculator. How does this graph compare
to the graph of the original? Why?
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Activity 4: Generating Patterns
1. Given the following graph, use blocks to generate a sequence
of figures that fits the data. Fill in the table and sketch the
figures.
20
15
10
5
1
2
3
4
Term Number
Term
Number
Visual
(Figure)
Written
Description
Process
Column
Numerical
Value of
Term
1
2
3
n
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2. Given the function y = 2 x + 8 , use blocks to generate a
sequence of figures that fits the function. Fill in the table,
sketch the figures, and plot the graph. Label the graph.
Term
Number
Visual
(Figure)
Written
Description
Process
Column
Numerical
Value of
Term
1
2
3
n
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2.1 Identifying Patterns: Reflect and Apply
Reflect and Apply
1. Create a physical model to demonstrate 3 x + 6 = 3( x + 2) .
Match:
2. ____ y = 3 + x
a
b
3. ____ y = x
c
4. ____ y = 3 + 2 x
d
5. ____ y = 2 x
6. Reflect on the activities. How might you adapt the activities
to use with your students?
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2.1 Identifying Patterns: Student Activity
Student Activity: Perimeter of Rectangles
Overview:
Students investigate linear relationships using concrete models, tables,
diagrams, written descriptions, and algebraic forms.
Objective:
Algebra I TEKS
quantity on another and can be described a variety of ways.
(b.3) The student understands algebra as the mathematics of generalization
and recognizes the power of symbols to represent situations.
Terms:
function, independent variable, dependent variable, pattern
Materials:
color tiles, graphing calculator
Procedures:
Students should be seated in groups of 3 – 4.
Activity : Perimeter of Rectangles
Do the activity together as a whole group, bringing out the following points
and asking the indicated questions.
1. Encourage students to write how they found the number of perimeter in
the process column. This can often be done in several ways, which will
lead to different, yet equivalent algebraic expressions. This is a desired
outcome. Possible equivalent expression include:
Sample Process
1+1+1+1
2+2+2+2
3+3+3+3
n+n+n+n
Sample Process
4(1)
4(2)
4(3)
4n
Sample Process
2. Justify: The variable x stands for the figure number and xmin=0 to
xmax=5 shows the figures 1 – 4 nicely.
The variable y stands for the perimeter and ymin= −2 to ymax=20 shows
the perimeters of 4 to 18 nicely.
3. The perimeter of figure 11 is 44. 4(11) = 44
4. Figure 12 has a perimeter of 48. 4n = 48
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Ask students to use the tiles to physically demonstrate the algebraic rules they
found in the table. In this example, they will mainly be pointing to sides on
the tiles and relating them to the numbers in the process column.
5.
Sample Process
2 +1+ 2 +1
3+2+3+2
4+3+ 4+3
(n + 1) + n + (n + 1) + n
Sample Process
2(2) + 2(1)
2(3) + 2(2)
2( 4) + 2(3)
2(n + 1) + 2(n)
Sample Process
4(1) + 2
4(2) + 2
4(3) + 2
4(n) + 2
6.
7. Figure 11 has a perimeter of 46. 4(11) + 2 = 46 .
8. Figure 13 has a perimeter of 54. 4n + 2 = 54 .
Ask students to use the tiles to physically demonstrate the algebraic rules they
found in the table. For example, the rule in the first column above is simply
adding each side in order. The rule in the second column above is noting that
there are two sides of length n+1 and two sides of length n. The rule in the
third column above is based on the idea of adding two additional sides to a
square of side n.
Ask students to compare the rules, P = 4n and P = 4n + 2, and their
respective graphs. Note that the lines have the same slope but that the line
P = 4n + 2 is the line P = 4n shifted up two. The perimeters grow by the
same amount each time you change figure numbers by one, but P = 4n + 2
starts 2 higher than P = 4n .
Summary
Using multiple representations, students gain added understanding for the
linear relationship of a rectangle’s perimeter and the length of a side.
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Student Activity: Perimeter of Rectangles
Build these squares and the next three squares in the sequence, using
color tiles.
Figure number
1,
2,
3
Figure
1. Complete the table, using the process column to write a function for
figure n, and graph the relation.
Figure
Number
(length of side)
1
2
3
4
5
Process
Perimeter
n
Length of Side
2. On your graphing calculator, make a scatter plot. Graph the function
over the scatter plot to confirm. Justify your window choice.
Answer the questions and write the equation that represents the question:
3. What is the perimeter of figure number 11?
4. What figure number has a perimeter of 48?
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Build these rectangles and the next three rectangles in the sequence,
using color tiles.
Figure number
1,
2,
3
Figure
5. Complete the table, using the process column to write a function for
perimeter of the nth figure, and graph the relation.
Figure
Number
(length of side)
1
2
3
4
5
Process
Perimeter
n
Length of Side
6. On your graphing calculator, make a scatter plot. Graph the function
over the scatter plot to confirm.
Justify your window choice:
Answer the questions and write the equation that represents the question:
7. What is the perimeter of figure number 11?
8. What figure number has a perimeter of 54?
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2.2 Identifying More Patterns: Leaders' Notes
2.2 Identifying More Patterns
Overview:
Participants represent non-linear relationships among quantities using
concrete models, tables, diagrams, written descriptions, and algebraic forms.
Objective:
Algebra I TEKS
(b.1.C) The student describes functional relationships for given problem
situations and writes equations or inequalities to answer questions arising
from the situations.
(d.2.A) The student solves quadratic equations using concrete models, tables,
graphs, and algebraic methods.
(d.3.C) The student analyzes data and represents situations involving
exponential growth and decay using concrete models, tables, graphs, or
algebraic methods.
Terms:
function, independent variable, dependent variable
Materials:
building blocks, graphing calculators
Procedures:
Have participants do Activities 1 and 2 in their groups. Again encourage
participants to do the process column in several ways, leading to different, yet
equivalent algebraic expressions. Give groups the transparencies of the
activities and have them present their results, using overhead tiles to
demonstrate different ways of physically modeling the algebraic rules. After
the assigned group has presented, ask if the other participants saw any other
ways to physically model the rules and have them demonstrate also.
Ask the following with each activity
• What does the ordered pair ( , ) mean?
• Does the ordered pair ( , ) belong to this graph? Why or why not?
• Trace to any ordered pair on the function. What meaning, if any,
do the coordinates have for the problem situation?
Activity 1: Building Blocks
This sequence of numbers is known as the square numbers (1, 4, 9, 16, 25,...)
and provides a beginning point for identifying dependent quadratic
relationships.
1.
Sample Process
1
2⋅2
3⋅3
n⋅n
Sample Process
12
22
32
n2
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Sample Process
67
2.
3.
4.
6. The fifth figure has 5 rows by 6 rows, so there are 5 ⋅ 6 = 30 cubes.
7.
Sample Process
1⋅ 2
2⋅3
3⋅ 4
n(n + 1)
Sample Process
12 + 1
22 + 2
32 + 3
n2 + n
Sample Process
9.
Activity 2: Starting Staircases
1.
Sample Process Sample Process
1⋅ 2
1
2
2⋅3
1+ 2
2
3⋅ 4
1+ 2 + 3
2
n(n + 1)
?
2
Spring 2001
Sample Process
1
1 2 1
(2 ) − 2 (2 )
2
1 2 1
(3 ) − 2 (3)
2
1 2 1
n − n
2
2
68
2.
3. A base of 15 means the 15th triangular number so it has
15(15 + 1)
= 120 blocks.
2
4. For a similar triangle with 36 blocks, the figure number is 8.
6.
Sample Process Sample Process Sample Process
2 + 2(1)
2⋅2
3 + 2(3)
3⋅3
4 + 2(6)
4⋅4
2
n(n + 1)
(n + 1)
(n + 1) + 2
2
Flip the blocks
in the back up
on top to make
a square of n+1
dimensions.
The center
column plus 2
triangular
numbers.
8.
Activity 3: Too Many Triangles
Note that TI has a Sierpinski triangle program in the manual.
Sample Process
1
1⋅ 3
1⋅ 3 ⋅ 3
1 ⋅ 3n
Sample Process
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2.
3.
4. We do not have the expectation that participants should solve this
algebraically. We want participants to understand that the exponential
model is accessible to algebra students when approached numerically and
graphically and later symbolically. We spend more time on the
exponential model at the end of the institute.
6.
Sample Process
1
1
1 
 4
1 1
1   
 4 4
n
1
1 
 4
Sample Process
8.
Wrap it up:
• Refer to the linear patterns in the previous activities. Do we have
quantities here that are the “changing quantity” and the “fixed
quantity” in these patterns? [No, not like the linear patterns. The
patterns here are a little different. We do have a “starting” point
(y-intercept, the y-value when x = 0 .]
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•
What about slope? [The main idea we want to get across here is
that the slope is not constant. The rate of change is changing.
Constant rate of change, constant slope means linear relationship.
Changing rate of change means non-linear.]
Draw triangles on the graphs to demonstrate that the rate of change is not
constant (the triangles are changing—for every constant change in x, the
change is y is different.)
Math Note: Participants may note that for the linear and quadratic patterns,
the term number started with 1, but for the exponential patterns, the term
number started with 0. These ideas will be further explored in the Linear
section and the Exponential Activities.
1a. All of the tables in 2.1 Identifying Patterns can be produced recursively
using repeated addition.
b. The graphs of repeated addition are linear.
2a. The two tables in Activity 3 in 2.2 Identifying More Patterns can be
produced recursively using repeated multiplication.
b. The graphs of repeated multiplication are not linear. They are
exponential.
Summary:
By using concrete models and the process column, participants model nonlinear patterns.
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2.2 Identifying More Patterns: Activity 1
Activity 1: Building Blocks
Suppose you are building square arrays out of cubes, as shown
below. Use the table to find the relationship between the
number of blocks needed for each figure and the dimension of
the square array.
Term
Number
Visual
Written
(dimension
(figure) Description
of square
array)
1
A 1 by 1
square has 1
cube.
2
Process
Column
Numerical
Value of Term
(number of
blocks)
1
4
3
4
n
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relationship between the term number and the number of
cubes.
3. What figure number will have 25 cubes? What is this
ordered pair on the graph?
4. How many cubes do you need for the 25th term? What is this
ordered pair on the graph?
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5. Suppose you add an extra column of cubes to the figures
above. Use your cubes to build the first four figures and
record the data below.
Term
Number
Visual
Written
Description
Process
Column
Numerical
Value of
Term
1
2
3
4
n
6. Write a rule in words to describe how to find the total
number of cubes for the fifth figure.
7. Write a rule in symbols that expresses the relationship
between the total number of cubes and the term number.
previous graph.
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Activity 2: Starting Staircases
Suppose you are building the triangular numbers from cubes,
based on the pattern below. Use the table to find the
relationship between the number of blocks and the term number.
Term
Number
1
Visual
Written
Description
The 1st staircase
needs 1 block.
2
Process
Column
Numerical
Value of
Term
1
3
3
4
n
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relationship between the total number of blocks and the term
number.
3. Use words to describe how to use blocks to build a similar
staircase with a base of 15 blocks. What is this ordered pair
on the graph?
4. If a similar staircase has 36 blocks, what is the figure
number? What is this ordered pair on the graph?
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5. Suppose the figures are as shown. Use your cubes to build
the first three figures and record the data below.
Term
Number
Visual
Written
Description
Process
Column
Numerical
Value of
Term
(number of
blocks)
1
2
3
4
n
between the total number of blocks and the term number.
previous graph.
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Activity 3: Too Many Triangles
Suppose you are drawing the fractal, based on the pattern below.
Use the table to find the relationship between the number of
upward triangles and the term number.
Term
Number
0
1
Visual
Written
Description
One new
upward triangle
Three new
upward
triangles
Process
Column
Numerical
Value of Term
(number of
new upward
triangles)
1
3
2
3
4
n
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relationship between the number of new upward triangles and
the term number.
Explain.
3. Find the total number of new triangles for term number 8.
4. If there are a total of 2187 new upward triangles, what term
number is this?
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5. Now consider the area of one of the smallest triangles in each
figure and record the data below.
Numerical
Value of
Term
Written
Process
Visual
Term (area
Number
Description
Column
of smallest
triangle)
0
1
The area of the
smallest
triangle is 1
unit
The area of one
of the smallest
triangles is
1
of a unit
4
1
1
4
2
3
n
between the area of one of the smallest triangles and the term
number.
previous graph.
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2.2 Identifying More Patterns: Reflect and Apply
Reflect and Apply
1. Refer to the tables in 2.1 Identifying Patterns.
a. Look at the column “Numerical Value of Term.” What
operation when applied to a term produces the next term?
In other words, what is happening recursively?
b. What kind of graph does this repeated operation produce?
2. Refer to the tables in Too Many Triangles.
a. Look at the column “Numerical Value of Term.” What
operation when applied to a term produces the next term?
In other words, what is happening recursively?
b. What kind of graph does this repeated operation produce?
3. Reflect on the activity. How might you adapt the activity to
use with your students?
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3.1 Interpreting Distance versus Time Graphs: Leaders’ Notes
3.1 Interpreting Distance versus Time Graphs
Overview:
Participants use motion detectors to investigate distance over time graphs.
This lays the groundwork for graph reading and for work with rates of change.
Objective:
Algebra I TEKS
Terms:
rate of change, increasing, decreasing, constant
Materials:
motion detector connected to an overhead calculator, motion detectors, data
collection devices, graphing calculators, 2 or 3 transparencies cut-to-fit on the
overhead calculator screen
Procedures:
This activity is done in an open area with room for participants to move about
in groups of 2 – 4.
Have participants complete the Student Activity: Walk This Way. As you do
the whole group introduction, sketch two of the walks of the participants. Do
this by placing a transparency, cut to fit, on the overhead calculator screen and
then sketch the walk with a transparency marker. We will use these sketches
of walks at the end of Activity 2.
Math Note: Scientific convention is to write y versus x when referring to a
situation of (x, y). In other words, the dependent variable is always listed first,
then the independent variable as in: dependent variable versus independent
variable. Thus, this activity refers to Distance versus Time graphs or Distance
over Time graphs, where time is the independent variable and distance is the
dependent variable.
Activity 1: Walking Graphs
Have participants answer the questions in their groups. Circulate and answer
questions. Ask a member of each group to present an answer for one of the
Exercises.
Answers will vary. Sample answers:
1. The y-intercept of the first segment is the distance away from the motion
detector at time = 0. This tells you where to start walking.
2. The x-axis represents time and each tick mark represents a second. The
number of tick marks on the x-axis represents the time to walk for each
segment.
3. If the segment decreases as time increases on the x-axis, then the distance
from the motion detector is decreasing. This means you should walk
toward the motion detector.
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4. If the segment increases as time increases on the x-axis, then the distance
from the motion detector is increasing. This means you should walk away
from the motion detector.
5. If the segment decreased or increased slowly (it was shallow), then that
means that the distance from the motion detector was changing slowly as
time increased. Therefore, you should walk slowly.
6. If the segment decreased or increased rapidly (it was steep), then that
means that the distance from the motion detector was changing rapidly as
time increased. Therefore, you should walk quickly.
7. If the segment was horizontal, then that means that the distance from the
motion detector was not changing as time increased. The distance was
remaining constant. Therefore you should stand still.
Put up the transparency of the student activity: Practice Walking Linear
Graphs. Draw triangles on some of the line segments to show that each line
segment has a constant rate of change, that for every increment in the xdirection, the segment increases or decreases by the same amount in the ydirection. Do not spend too much time here. Simply demonstrate that for a
given line segment, triangles drawn as shown with equal bases are congruent.
This means that the heights are equal. In the next activity, we will contrast
these constant rates of change with non-constant rates of change of non-linear
graphs.
Activity 2: Walking More Graphs
In this activity, participants walk non-linear graphs. Have participants collect
data for about 4 seconds. (This may necessitate collecting the data not in realtime. If participants prefer to collect data in real-time (seeing the data as they
collect it), simply have participants ignore the tick marks on the x-axis and
collect data for 15 seconds. Discuss the advantages and disadvantages of both
methods. In real time, participants can quickly adjust their motion in the
middle of data collection. However, that may mean that they do not analyze
and plan carefully first. Out of real time, participants must analyze and plan
carefully. Then after obtaining the resulting graph, they adjust the plan and
walk again.
The big idea in the activity is to get a feel for changing rates of change,
therefore the exact starting and stopping points are less important then the
general shape of the graph.
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Have participants do the activity and answer the questions in their groups.
Circulate and help. Ask a member of each group to present an answer for one
of the Exercises.
1. Start 6 feet from the motion detector. Stand still for less than a second.
Start walking toward the motion detector slowly at first and then speeding
up over 4 seconds.
2. Start about 2.5 feet from the motion detector. Start walking away quickly,
slowing down over 4 seconds to a dead stop at the end.
3. Start about 2 feet from the motion detector. Start walking away slowly,
speeding up over 4 seconds.
4. Start about 6 feet from the motion detector. Start walking toward the
motion detector quickly, slowing down over 4 seconds to a dead stop at
the end.
5. Start about 6 feet from the motion detector. Start walking toward the
motion detector quickly, slowing down over 2 seconds, stop briefly, then
walk away from the motion detector slowly, speeding up till the end of the
four seconds.
6. Start about 2 feet from the motion detector. Start walking away from the
motion detector quickly, slowing down over 2 seconds, stop briefly, then
walk toward the motion detector slowly, speeding up till the end of the
four seconds.
7. You know to speed up (accelerate) when the curve gets steeper, when the
change in y over an increment in x is greater than it was before.
8. You know to slow down (decelerate) when the curve gets less steep, when
the change in y over an increment in x is less than it was before.
9. e, f, g. These graphs show walking at a constant rate because the change
in the y-direction is constant for a constant change in the x-direction. The
rate of change is constant. For the horizontal line, the rate of change is a
constant zero. Draw triangles to illustrate.
For every change in x ,
the change in y is zero!
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10. a,c. These graphs show a walker speeding up because the change in y is
increasing over a constant change in x. The rate of change is increasing.
Draw triangles to illustrate.
You can also draw tangent lines to illustrate that the slopes of the tangent lines
(the instantaneous rate of change at the point of tangency) are increasing.
11. b,d. These graphs show a walker slowing down because the change in y is
decreasing over a constant change in x. The rate of change is decreasing.
Draw triangles to illustrate.
You can also draw tangent lines to illustrate that the slopes of the tangent lines
(the instantaneous rate of change at the point of tangency) are decreasing.
12. e. As shown above, the rate of change for a horizontal line is a constant
zero. The walker is standing still because for every change in x, the
change is y is a constant zero. The distance from the motion detector is
not changing.
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•
Can you produce a vertical line by walking in front of a motion
detector? [No, a vertical line suggests that at one instant in time, the
walker is at an infinite number of locations, every point on that line. A
walker can only be one distance from the motion detector at a time.]
Wrap up these activities as a whole group using the sketches of participants’
walks on the cut-to-fit transparencies from the beginning. Put a sketch on a
cut-to-fit transparency on the overhead calculator. (Consider using a walk that
has varied sections, linear and non-linear.) Ask participants to discuss in their
groups, how a walker should walk to reproduce the walk. Randomly choose a
group’s description and have a walker walk the description to confirm.
Repeat if desired.
Sample Answers to Reflect and Apply:
1. I started jogging away from my house (when I was 0.5 kilometers from
my house) and I gradually sped up to a sprint. At 15 minutes (tick marks
are at 5 minute increments), I fell on the ground, 4 kilometers from my
house. I sat there for about 20 minutes, catching my breath. I then started
gradually back home, speeding up as I went until I was again sprinting in
the door.
2. I walked from my locker to my class over two minutes. I stood there
chatting for a minute until I realized I had left my book in my friend’s
locker. I walked quickly to my friend’s locker, which was thankfully
open, grabbed the book and started quickly back to class. But I ran into
the principal who walked me back to class, slowing down as we went,
because the principal was talking to me about an upcoming event.
Summary:
The big idea here is that walking at a constant rate produces a linear distance
over time graph. The rate of change of a line is constant. Speeding up or
slowing down, non-constant rates of change, produce non-linear distance over
time graphs. The rates of change are changing.
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3.1 Interpreting Distance versus Time Graphs: Activity 1
Activity 1: Walking Graphs
Answer the following questions based on your experiences in
the Student Activity: Walk This Way.
1. How did you know where to start walking for each graph?
2. How did you know how long to walk for each segment?
3. How did you know when to walk toward the motion
detector? Use the words “time” and “distance” in your
answer.
4. How did you know when to walk away from the motion
detector? Use the words “time” and “distance” in your
answer.
5. How did you know when to walk slowly?
6. How did you know when to walk quickly?
7. How did you know when to stand still?
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Activity 2: Walking More Graphs
distance
distance
Practice walking the following graphs using a motion detector
and a graphing calculator. Describe the walk that you used to
produce each graph.
time
2.
distance
distance
1.
time
time
4.
distance
distance
3.
time
time
5.
time
6.
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7. How did you know when to speed up?
8. How did you know when to slow down?
Which graph(s) below show:
9. a constant rate? Why?
10. a walker speeding up? Why?
11. a walker slowing down? Why?
12. a walker standing still? Why?
b.
a.
e.
c.
f.
d.
g.
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3.1 Interpreting Distance versus Time Graphs: Reflect and Apply
Reflect and Apply
Make up a story for the following graphs of distance over time.
Distance over Time
1.
Distance over Time
2.
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3.1 Interpreting Distance versus Time Graphs: Student Activity
Student Activity: Walk This Way
Overview:
This is an introduction to graphs of motion data, specifically linear data.
Objective:
Algebra I TEKS
Terms:
rate of change
Materials:
motion detector connected to a viewscreen calculator, a motion detector with
graphing calculator for each group of 3-4 students, and data collection devices
Procedures:
Often students are asked to interpret distance over time graphs. In this
activity, students experience seeing many examples quickly of their own
motion graphed over time. Students gain intuition for interpreting graphs as
they make connections between their own motion and the graphs of their
motion.
The classroom should be set up with an aisle down the middle. Set up a
motion detector pointing down the aisle, connected to a viewscreen calculator,
so the class can see both the students walking down the aisle and the data
projected from the calculator on a screen in front of the room.
Explain that the motion detector sends out an ultrasonic pulse. The pulse
bounces off the walker, and the motion detector records the distance at that
time. The calculator displays the data as a graph with the distance measured
in meters and the time measured in seconds.
Run the CBR Ranger program or a similar program. Use the following
screens to set up the experiment, and then follow the instructions on the
screen.
Ask a few students to walk one at a time in front of the motion detector.
Encourage students to walk differently - slowly, quickly, standing still, toward
the motion detector, away from the motion detector, etc.
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After each walk, discuss the following:
• What does the starting point represent?
• What does a fast walk look like?
• What does a slow walk look like?
• What does a pause look like?
• What does it look like when you walk away from the motion detector
or toward the motion detector?
• Discuss a “straight” verses a “curved” graph. Point out that for
constant rates of change, we use a linear model to model the data.
Activity: Practice Walking Linear Graphs
Each group of 3 – 4 students will practice walking different linear graphs
using a motion detector and a graphing calculator.
First Time: Hold the motion detector and the calculator. Point the motion
detector at the wall and practice walking the graphs.
Second Time: Have the group hold the motion detector and calculator. Point
the motion detector at one person in the group. As a group, instruct the
walker on how to walk the graph.
Extension: Have each group come to the front and match a graph. Give the
group a minute to discuss. Then a member of the group walks to match the
graph. Have the rest of the class rate the group using the following rubric.
Have the class rate the group quickly using a show of fingers. Choose the
mode for each rating to get a quick score for the team.
Starting Point:
(1 – 5) Did the walker start at the correct point?
Rate:
(1 – 5) Did the walker walk at the correct rate?
Direction:
(1 – 5) Did the walker walk in the correct direction?
Accuracy:
(1 – 5) Was the rate correct but the distance was
incorrect?
Teamwork:
(1 – 5) Did the group work as a team well?
Assessment: Linear Motion.
Now your students should be able to complete the Assessment. Note that this
assessment is intended to provide teachers with sample assessment items. It is
not intended as a stand-alone worksheet or quiz. Teachers can use the items
as a starting point for creating meaningful assessments.
Answers:
1. B
2. C
3. A
4 – 7. Answers will vary.
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Sample answers:
8.
9.
10.
11.
Note: this activity was adapted from an activity in the Mathematical
Modeling Institute for Secondary Teachers.
Summary:
Through experience, students learn intuitive notions about distance over time
graphs. Starting points (y-intercepts), rates of change, direction, and time
intervals are among the ideas that are built.
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Activity: Practice Walking Graphs
Practice walking the following graphs using a motion detector.
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Assessment: Linear Motion
Match the description with the graph.
a.
b.
c.
_____ 1. Start one meter away from the motion detector. Walk slowly
away from the motion detector for about 3 seconds, stand still
for about 4 seconds, and then walk quickly away from the
motion detector for about 3 seconds.
_____ 2. Start 3 meters away from the motion detector, and walk away
from it at a moderate rate for about 3 seconds. Stand still for
about 4 seconds, and then walk quickly toward the motion
detector for 3 seconds.
_____ 3. Start 2 meters away from the motion detector, and walk toward
it at a moderate rate for about 3 seconds. Stand still for about 4
seconds, and then walk toward the motion detector at about the
same moderate rate as earlier for about 3 seconds.
Write a description for a walk that would produce each of these graphs.
4. ____________________
_______________________
5. ___________________________
______________________________
6. ____________________
_______________________
7. ___________________________
______________________________
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Sketch a graph that would match the description.
8. Start 4.5 meters from the motion detector, and walk quickly toward it
for 2.5 seconds. Walk slowly toward the motion detector for 3
seconds. Walk even more slowly away from the motion detector for
4.5 seconds.
9. Start 0.5 meters from the motion detector. Walk slowly away from
the motion detector for 5 seconds. Walk extremely slowly away from
the motion detector for 2 seconds. Walk quickly away from the
motion detector for 3 seconds.
10. Start one meter from the motion detector and stand still for 2
seconds. Walk away from the motion detector quickly for 5 seconds.
Walk back slowly toward the motion detector for 3 seconds.
11. Start 3 meters from the motion detector, and walk slowly away from
it for 3 seconds. Stand still for 4 seconds. Walk quickly toward the
motion detector for 3 seconds.
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3.2 Interpreting Velocity versus Time Graphs: Leaders’ Notes
3.2 Interpreting Velocity versus Time Graphs
Overview:
Participants use motion detectors to investigate velocity over time graphs.
This continues to build graph-reading skills and to build understanding for
rates of change.
Objective:
Algebra I TEKS
Terms:
velocity, speed, direction, rate of change
Materials:
motion detector connected to an overhead calculator, motion detectors, data
collection devices, graphing calculators
Procedures:
Set up the room similarly to the Student Activity: Walk This Way, with a
motion detector connected to an overhead calculator. The room should have
an aisle down the middle in front of the motion detector.
Set up the motion detector to display velocity data, collecting data for 4
seconds.
• What is velocity? [Speed and direction.]
• What are some common units for speed? [miles per hour, feet per
second, meters per sec, kilometers per hour, etc.]
• How will a graph of velocity over time differ than a graph of distance
over time? [Do not answer at this point.]
Have a volunteer walk the following instructions. Collect data for about 4
seconds. Before you display each graph, have participants sketch a prediction
of the graph. After each walk, discuss the resulting graph, using the questions
following each instruction.
1: Start 1.5 feet away from the motion detector. Stand still briefly when the
data collection starts, then walk away slowly, speeding up as you go.
Repeat if necessary.
• What does x represent? [elapsed time]
• What does y represent? [the speed of the walker]
• Why does the graph increase? [As time increased, the speed of the
walker increased.]
Trace to a value on the graph as shown.
• What is the meaning of the coordinates
shown? [After about 1.8 seconds, I was
moving at 0.2 meters per second away
from the motion detector.]
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2: Start behind the motion detector. Walk past the motion detector and start
the data collection when you are about 1.5 feet in front of it. Continue to
walk away, slowing down as you go to a complete stop by the end of the
time. Repeat if necessary.
•
•
•
What does x represent? [elapsed time]
What does y represent? [the speed of the walker]
Why does the graph decrease? [As time increased, the speed of the
walker decreased.]
shown? [After about 1 second, I was
moving at 0.5 meters per second away
from the motion detector.]
3: Start about 10 feet from the motion detector. Stand still briefly when the
data collection starts, then walk toward the motion detector slowly,
speeding up as you go. Repeat if necessary.
•
•
•
•
Why does the graph increase? [As time increased, the speed of the
walker increased.]
Why is the graph in the fourth quadrant? [Velocity is speed and
direction. The negative y-values indicate that the walker was moving
toward the motion detector. The magnitude of the y-values indicate
the rate, the speed. Magnitude is the same as the absolute value of the
number.]
shown? [After about 2 seconds, I was
moving at 0.3 meters per second toward
the motion detector.]
4: Start about 15 feet from the motion detector, walking quickly toward the
motion detector, and begin the data collection when you are about 10 feet
in front of it. Continue to walk toward the motion detector, slowing down
as you go. Repeat if necessary.
•
•
•
Why does the graph decrease? [As time increased, the speed of the
walker decreased.]
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•
Why is the graph in the fourth quadrant? [Velocity is speed and
direction. The negative y-values indicate that the walker was moving
toward the motion detector. The magnitude of the y-values indicate
the rate, the speed. Magnitude is the same as the absolute value of the
number.]
moving at 0.34 meters per second
toward the motion detector.]
5: Start right at the motion detector. Walk away from the motion detector at
a constant rate. Begin the data collection when you are at least 1.5 feet
from the motion detector or when you are walking at a constant rate.
Continue to walk at a constant rate until time is up. Repeat if necessary.
•
•
moving at 0.34 meters per second
toward the motion detector.]
•
Why is the graph so wavy? Was the rate not constant? [Trace on the
graph and note that the values are really quite close together, but the
window is probably very small, which exaggerates small changes in
the y-values. Change the window as shown below and note that the
graph now looks quite constant.]
Math Note: Velocity is speed and direction. The speed is represented by the
magnitude of the velocity. The direction is represented by the sign of the
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velocity. If the velocity at time t is –4, then −4 = 4 is the speed at time t and
the direction is toward the motion detector. If the velocity at time t is 6, then
6 = 6 is the speed and the direction is away from the motion detector.
When discussing direction and velocity, remind participants of the direction
implied in a different measure of motion, acceleration. The force of
ft
m
or −9.8
. The value of
acceleration due to gravity is −32
2
sec
sec 2
acceleration is negative which means that the object is falling toward the
earth, just as the negative in velocity means that the walker is walking toward
the motion detector.
Activity 1: Matching Velocity Graphs
Have participants do the activity in groups of 3 – 4, by walking the graphs and
then answering the questions. Ask a member of each group to present an
answer for one of the Exercises.
1. The y-values stand for velocity so when the y-values have small
magnitudes, then the rate is slower. Therefore, you should walk slower.
2. The y-values stand for velocity so when the y-values have large
magnitudes, then the rate is faster. Therefore, you should walk faster.
3. Since the y-values stand for velocity, when the graph decreases in the first
quadrant, then the velocity should decrease, or in other words, you should
slow down. In the fourth quadrant, when the graph increases, then the
velocity is decreasing (because the magnitude is getting smaller) and you
should slow down.
4. Since the y-values stand for velocity, when the graph increases in the first
quadrant, then the velocity should increase, or in other words, you should
speed up. In the fourth quadrant, when the graph decreases, then the
velocity is increasing (because the magnitude is getting larger) and you
should speed up.
5. Points in the first quadrant have positive y-values. In this case, the
positive velocity means that you were walking away from the motion
detector.
6. Points in the fourth quadrant have negative y-values. In this case, the
negative velocity means that you were walking toward the motion
detector.
7. Since the y-values stand for velocity, points along the x-axis have y-values
of zero and therefore, the velocity is zero. Therefore, you should stop.
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8. You don’t know where to start! You need to start where ever you will
have enough room to do what you need to do. If you need to move toward
the motion detector, you had better start a ways in front of it. If you need
to move away from the motion detector, you probably ought to start closer
to it.
Activity 2: Connecting Distance and Velocity Graphs
Do Exercise 1 together as a whole group. Ask a volunteer to follow the
instructions and first display the distance graph. In order to see both the
distance and the velocity graphs at the same time, do the following. Put an
overhead transparency on the overhead calculator screen and sketch the
distance over time graph. Place this transparency on another overhead
projector. Now display the velocity graph on the viewscreen calculator.
With both graphs displayed simultaneously, label them as follows, based on
the questions on the activity page.
1.
standing still
slowing down
slower rate
quicker rate
slowing down
quicker rate
standing still
speeding up
speeding up
slower rate
2.
quicker rate
quicker rate
standing still
slowing down
speeding up
slower rate
slower rate
standing still
speeding up
slowing down
Note: This activity provides participants with intuitive experience in the
calculus and physics concepts of a function (a position function) and the
function’s derivative (a velocity function). However, the main idea of this
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activity in the Algebra I Institute is to use the graphs of participant’s own
motion to build the skills of reading, interpreting, and analyzing graphs.
If participants request that you help them connect these experiences with their
previous experiences in calculus and physics, then briefly help them make
those connections if you wish. Do not suggest that the participants are to
teach calculus and physics in an algebra I course.
Answers for Reflect and Apply:
Sample Answers:
Distance over Time: In PE one day, students were running on the basketball
court. They started at the free throw line and jogged to the end of the court.
Then they stood there and took a breather. Then they sprinted back the whole
length of the court.
Velocity over Time: In PE one day, students were running on the basketball
court. They were already jogging, and when the time keeper said, “Go,” they
sped up to a sprint. They sprinted for a bit and then they slowed down (faster
than they had sped up) until they came to a stop.
Ask participants to recall the Reflect and Apply Exercises from 1.1.1
Variables and Functions.
• Do you think students would complete these Exercises differently after
the graphing experiences in this Activity?
Summary:
The big idea here is to build intuition for interpreting graphs. With the use of
technology, participants get instant feedback on distance and velocity graphs
and can see many examples rapidly. They can monitor and adjust quickly to
gain added understanding. By comparing a distance graph for a situation with
a velocity graph of the same situation, participants make valuable connections
between distance and velocity and about analyzing graphs.
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3.2 Interpreting Velocity versus Time Graphs: Activity 1
Activity 1: Matching Velocity Graphs
Practice walking the following graphs using a motion detector
and a graphing calculator. Describe the walk that you used to
produce each graph.
a.
b.
(0, 0)
Time
c.
(0, 0)
Time
(0, 0)
Time
d.
(0, 0)
Time
e.
Time
(0, 0)
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1. How did you know when to walk slowly?
2. How did you know when to walk quickly?
3. How did you know when to slow down?
4. How did you know when to speed up?
5. How did you know when to walk away from the motion
detector?
6. How did you know when to walk toward the motion
detector?
7. How did you know when to stop?
8. How did you know where to start walking for each graph?
9. How did you know what speed to start walking?
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Activity 2: Connecting Distance and Velocity Graphs
Read the following directions and sketch your prediction of the
resulting graphs. Then collect the data (for about 4 seconds.)
1. Start about 1.5 feet in front of the motion detector and walk
away quickly for about 2 seconds. Stand still for 1 second
and then walk toward the motion detector slowly for the
remaining time.
Predict :
b. Velocity versus Time
a. Distance versus Time
The actual results:
c. Distance versus Time
d. Velocity versus Time
Label the following sections in the graphs above:
A. The walker is standing still.
B. The walker is slowing down.
C. The walker is speeding up.
D. The walker is traveling at a quicker rate.
E. The walker is traveling at a slower rate.
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2. Start about 5 feet in front of the motion detector and walk
toward the motion detector slowly for about 1 second. Stand
still for 1 second and then walk away from the motion
detector quickly for the remaining time.
Predict :
a. Distance versus Time
b. Velocity versus Time
The actual results:
c. Distance versus Time
d. Velocity versus Time
Label the following sections in the graphs above:
A. The walker is standing still.
B. The walker is slowing down.
C. The walker is speeding up.
D. The walker is traveling at a quicker rate.
E. The walker is traveling at a slower rate.
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3.2 Interpreting Velocity versus Time Graphs: Reflect and Apply
Reflect and Apply
Make up two different stories based on the graph below. One
story should be based on the graph representing distance over
time. The other story should be based on the graph representing
velocity over time.
Story for distance versus time:
Story for velocity versus time:
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1.1 The Linear Parent Function: Leaders’ Notes
1.1 The Linear Parent Function
Overview:
Using contextual situations, participants investigate the linear parent function,
y = x.
Objective:
Algebra I TEKS
(b.2.A) The student identifies and sketches the general forms of linear ( y = x )
and quadratic ( y = x 2 ) parent functions.
Terms:
input, output
Materials:
colored pencils or pens, pieces of flat spaghetti, graphing calculators, program
ACTSCRS
Procedures:
Because of the role of the line y = x as the linear parent function, we begin
our work with linear functions using contextual situations to develop intuitive
notions about the line y = x .
Complete the Student Activity, Age Estimates, with participants. Talk
through the assessment.
Talk through the Student Activity, Sales Goals, with participants.
Activity 1: ACT Scores
Have participants run the program ACTSCRS. The program stores the data
on Transparency 1 into five lists in the calculator.
Create the indicated scatter plots and graph the line y = x , finding appropriate
windows to be able to see the comparisons.
Sample General Statements:
1. Most states have higher mean Mathematics scores than that state’s mean
English scores.
2. Most states have higher mean Reading scores than that state’s mean
Mathematics scores.
3. Most states have higher mean Sci Reasoning scores than that state’s mean
Mathematics scores.
4. Most states have higher mean Composite scores than that state’s mean
Mathematics scores.
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Sample Scatter Plots:
English versus Mathematics
Reading versus Mathematics
23.0
23.0
21.0
21.0
19.0
19.0
17.0
17.0
17.0
19.0
21.0
17.0
23.0
Mathematics
21.0
23.0
Mathematics
Sci Reasoning versus Mathematics
Composite versus Mathematics
23.0
23.0
21.0
21.0
19.0
19.0
17.0
17.0
19.0
17.0
19.0
21.0
23.0
17.0
19.0
21.0
23.0
Mathematics
Mathematics
There is other comparative information about ACT scores at
http://www.act.org/news/data/99/99data.html
•
•
•
Did any of the results surprise you?
How did you use the line y = x to make general statements about the
scores?
What are some of the factors that may influence the results? [Some
states have large populations of students who take the ACT, where
there are other states whose student populations primarily take the
SAT. In those SAT states, why might students choose to take the
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ACT? Perhaps those students are desiring to go out of state, looking
for specific scholarships, or going to private schools.]
Is it true that the higher the mathematics score, the higher the other test
score tend to be? In other words, is there a positive correlation
between math scores and the other scores? [Yes]
•
Activity 2: Temperatures
In the Student Activity and Activity 1, participants explore data in the first
quadrant, comparing the data to the line y = x . In this activity, participants
will explore data in all 4 quadrants, using average temperatures around the
world.
Have participants work in groups to complete the activity. Ask a group to
present their results. Use Transparency 3: World Map to determine the
location of indicated places. Note that for places north of the equator seasons
are opposite of those south of the equator.
Below find the average temperatures used to make the graph.
Place
1
2
3
4
5
6
7
8
July ( C ˚ )
35
28.5
14
13
10.2
-11
-57
-0.4
Austin, TX
Detroit, MI
Siberia
Santiago, Argentina
Nome, Alaska
Greenland icecap
Antarctica
Butlers Gorge, Australia
December (C˚)
16.7
1.8
-14
26.9
-13.9
-47
-18
5.4
Average Temperatures (C˚) in Selected Locations
60
40
4
1
20
8
-60
-40
-20
2
0
-20
7
6
20
40
60
5 3
-40
-60
Average July Temperatures (C˚)
•
How many points are in the second quadrant? [One]
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•
•
Describe the place that would have a point in the third quadrant. [The
place would have both a cold average July temperature (below
freezing) and a cold average December temperature (below freezing).
The place with the colder July temperature would need to be in the
southern hemisphere, our example is Antarctica. The place with the
colder December temperature would need to be in the northern
hemisphere, our example is the Greenland icecap.]
Describe the place that would have a point in the second quadrant.
[The place would have a cold average July temperature (below
freezing) and a warmer average December temperature. Thus, it
would need to be in the southern hemisphere, our example is Butler’s
Gorge, Australia.]
Activity 3: Symbolic
Have participants work in groups to complete the activity. Ask a group to
present their results.
1.
2.
3.
4.
5.
A, B, C, K, L
E, F, G, H, I
D, J
A, G
Any point in the second quadrant, points in the first quadrant above the
line y = x , points in the third quadrant above the line y = x .
6. Any point in the fourth quadrant, points in the first quadrant below the line
y = x , points in the third quadrant below the line y = x .
7. Points on the line y = x .
8. Points on the line y = − x .
•
•
In what ways have the concrete problem situations in the Activity
changed the way you approached these symbolic questions?
In what ways can experience with similar concrete problem situations
aid students in their understanding of similar symbolic questions?
1. The walker is moving at about 1 foot per second.
2. From 0 to 1 seconds, 4 to 5 seconds, 5 to 6 seconds
3. From 1 to 2 seconds, 2 to 3 seconds, 7 to 8 seconds
Summary:
The line y = x is the parent function for the linear function family. In this
activity we build intuition for the line y = x as an important starting point for
work with linear functions.
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1.1 The Linear Parent Function: Transparency 1
Transparency 1: ACT Scores by State 1999
State Name
Alabama
Alaska
Arizona
Arkansas
California
Colorado
Connecticut
Delaware
Florida
Georgia
Hawaii
Idaho
Illinois
Indiana
Iowa
Kansas
Kentucky
Louisiana
Maine
Maryland
Massachusetts
Michigan
Minnesota
Mississippi
Missouri
Montana
Nebraska
Nevada
New
Hampshire
English
Math
Reading
20.3
20.1
20.7
20.6
20.5
20.9
21.2
19.7
20.0
19.3
20.3
20.7
20.9
20.6
21.5
21.0
19.9
19.7
21.4
20.2
21.7
20.6
21.4
18.8
21.3
20.9
21.3
20.7
21.7
19.5
21.0
21.4
19.2
21.9
21.1
21.7
20.1
20.5
20.0
22.7
20.8
21.4
20.9
21.6
21.0
19.3
18.9
21.8
20.9
22.0
21.1
22.0
17.9
20.9
21.2
21.4
21.3
22.0
20.5
21.7
21.9
20.7
21.5
21.9
21.8
21.1
21.0
20.3
21.5
22.1
21.6
21.7
22.2
21.9
20.6
19.7
22.8
21.3
22.5
21.5
22.4
18.9
22.0
22.5
21.9
22.1
23.0
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Sci
Reasoning
19.9
21.1
21.2
20.1
20.8
21.7
21.1
20.4
20.5
20.0
21.4
21.4
21.3
21.2
22.1
21.4
20.2
19.5
21.8
20.8
21.4
21.5
22.3
18.6
21.5
21.9
21.7
21.4
21.7
Composite
20.2
21.1
21.4
20.3
21.3
21.5
21.6
20.5
20.6
20.0
21.6
21.4
21.4
21.2
22.0
21.5
20.1
19.6
22.1
20.9
22.0
21.3
22.1
18.7
21.6
21.8
21.7
21.5
22.2
112
1.1 The Linear Parent Function: Transparency 2
Transparency 2: ACT Scores by State 1999 (cont.)
New Jersey
New Mexico
New York
North Carolina
North Dakota
Ohio
Oklahoma
Oregon
Pennsylvania
Rhode Island
South Carolina
South Dakota
Tennessee
Texas
Utah
Vermont
Virginia
Washington
Washington, DC
West Virginia
Wisconsin
Wyoming
20.1
19.5
20.8
18.6
20.6
20.8
20.4
21.8
20.8
22.5
18.6
20.5
19.8
19.7
21.0
21.3
20.2
21.9
18.0
20.1
21.6
20.7
20.8
19.5
22.2
19.4
21.2
21.1
19.8
22.1
21.2
22.3
19.0
20.9
19.1
20.2
20.7
21.5
20.4
22.2
18.8
19.0
22.2
20.9
20.9
20.7
22.2
19.6
21.7
21.9
21.0
23.5
21.9
23.4
19.3
21.5
20.4
20.6
22.0
22.7
21.0
23.4
18.7
20.9
22.4
22.0
Spring 2001
20.4
20.3
22.2
19.5
21.6
21.4
20.5
22.4
21.3
21.9
19.2
21.5
19.8
20.4
21.3
21.8
20.4
22.3
18.5
20.3
22.4
21.6
20.7
20.1
22.0
19.4
21.4
21.4
20.6
22.6
21.4
22.7
19.1
21.2
19.9
20.3
21.4
21.9
20.6
22.6
18.6
20.2
22.3
21.4
113
1.1 The Linear P
Transparency 3: World Map
6
5
3
2
1
4
7
1
2
Detroit, Michigan
3
4
5
Nome, Alaska
6
Austin, Texas
Siberia
Santiago, Argentina
Greenland
7
Antarctica
8
Butler’s Gorge, Australia
Spring 2001
1.1 The Linear Parent Function: Activity 1
Activity 1: ACT Scores
Run the program ACTSCRS. The program stores the following
ACT test score data for 1999 for each state in the US and the
District of Columbia into lists in your calculator: the mean
scores for Mathematics, Reading, English, Sci Reasoning and
the mean composite score.
•
•
•
•
Set up the scatter plots 1 – 4 listed below.
Graph the line y = x over each scatter plot.
Find windows that will help you compare the scores.
Make a general statement about each plot, comparing the two
scores.
• Identify at least four coordinate pairs to justify your summary.
1. English versus Mathematics
2. Reading versus Mathematics
3. Sci Reasoning versus Mathematics
4. Composite versus Mathematics
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Activity 2: Temperatures
Average Temperatures (C˚) in Selected Locations
Average December Temperatures (C˚)
60
40
20
-60
-40
-20
0
20
40
60
-20
-40
-60
Average July Temperatures (C˚)
1.
2.
3.
4.
5.
6.
7.
8.
Austin, Texas
Detroit, Michigan
Siberia
Santiago, Argentina
Nome, Alaska
Greenland Icecap
Antarctica
Australia
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Activity 3: Symbolic
For the following points (x, y), choose a point or points that
match the description.
1. x < y
2. x > y
3. x = y
4. x = − y
B
C
A
D
L
E
F
K
J
I
H
G
Describe where the point (x, y) appears in general when:
5. x < y
6. x > y
7. y = x
8. y = − x
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1.1 The Linear Parent Function: Reflect and Apply
Reflect and Apply
The scatter plot represents a walker’s distance from a motion
detector.
8
6
4
2
2
4
6
8
Time (sec)
1. Judging by the graph, about how fast do you think the walker
was moving?
2. Which one-second interval(s) show the walker moving faster
than that rate?
3. Which one-second interval(s) show the walker moving
slower than that rate?
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1.1 The Linear Parent Function: Student Activity 1
Student Activity 1: Age Estimates
Overview:
Students develop intuitive notions about the line y = x .
Objective:
Algebra I TEKS
b.2.a The student identifies and sketches the general forms of linear ( y = x )
b.2.c The student interprets situations in terms of given graphs or creates
Terms:
trend line
Materials:
colored pencils or pens, pieces of flat spaghetti or brightly colored string,
graphing calculator
Procedures:
Students should be seated at tables in groups of 3 – 4.
Activity: How Old?
Read 12 names from the list below and ask students to guess the person’s
current age, filling in the first and second columns of the table. It is important
that you list a variety of ages, from old to young.
Have students fill in the third column as you read the actual ages based on the
birth dates below.
Name
Julie Andrews
Bill Gates
Ronald Reagan
George W. Bush
Shania Twain
LeAnn Rimes
Sophia Loren
Bill Cosby
Britney Spears
McCaughey septuplets
Jennifer Love Hewitt
Jennifer Aniston
Charlton Heston
Leonardo DiCaprio
Harrison Ford
Tim Allen
Oprah Winfrey
Michelle Pfeiffer
Michael J. Fox
Jodie Foster
Birthdate
10-01-1935
10-28-1955
02-06-1911
07-06-1946
08-28-1965
08-28-1982
09-20-1934
07-12-1937
12-02-1981
11-19-1997
02-21-1979
02-11-1969
10-04-1924
11-11-1974
07-13-1942
06-12-1953
01-29-1954
04-29-1958
06-09-1961
11-19-1962
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Ben Affleck
Drew Barrymore
Frankie Muniz
Mary Kate and Ashley Olsen
Haley Joel Osment
Jonathan Lipnicki
08-15-1972
02-22-1975
12-05-1985
06-13-1986
04-10-1988
10-22-1990
• How well did you guess?
• How do you think we could judge if you are a good guesser?
• How do you think we could tell if you are an “over” guesser?
• How do you think we could tell if you are an “under” guesser?
Have students plot the data by hand on the grid and label the axes and a few
ordered pairs.
Tell students that we want to use a trend line to get a feel for how good they
estimate ages.
Instruct students to use a piece of spaghetti or string to sketch a trend line for
their data. (Use the same color for the trend line as the color of the data
points.) Do the same on the transparency of the Activity.
Instruct groups to compare their graphs within their group.
Based on the graphs:
• Who is the better guesser?
• Who is more of an “under” guesser? How can you tell?
• Who is more of an “over” guesser? How can you tell?
Instruct students to use a different color and sketch a “perfect-guess” line, a
line that represents perfect guessing.
Ask students to write a sentence to describe the perfect-guess line and then
translate to symbols. An example:
“All my guesses are the same as the actual ages”
Guess = Actual
G = A or y = x
Check that the line students sketched as their “good-guess” line is indeed the
line y = x .
2. Have students create a scatter plot (guess, actual).
Sample graph:
3. Sample answer: The variable x stands for the guessed age so x min is 0
years and x max is 100 years. The variable y stands for the actual age so
ymin is 0 years and y max is 100 years.
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4.
5. If the student’s points lie mostly above the line, they are an under-guesser.
If the student’s points lie mostly below the line, they are an over-guesser.
Over or Under
Have students complete the activity and then discuss as a group.
1.
50
50
40
40
30
perfect guess
30
20
20
10
10
10
20
30
40
perfect guess
50
10
guess
20
30
40
50
guess
2. Underguess because all of the guesses are smaller when the actual ages are
larger.
3. Overguess because all of the guesses are larger when the actual ages are
smaller.
4.
50
40
30
20
10
10
guess
20
30
40
50
guess
5. Accept answers between 30 and 40 years old.
6. Accept answers between 20 and 30 years old.
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7.
guess
Sample Assessment:
1. You guessed B’s age better.
2. You overguessed C’s age.
3. y = x .
4. [0, 55], [0, 45]
5. The variable x stands for my partner’s guesses of people’s ages, so [0, 40]
shows all of the guesses listed. The variable y stands for the actual ages of
the people listed, so [0, 45] is a reasonable choice for those ages. This
choice allows one to see the origin, which may be helpful in orienting the
reader to whether the student is a good age estimator.
Note: In this activity students draw trend lines and a “perfect guess” line.
The trend lines are approximations or estimates of their guessing, showing the
general trend of how they guessed. Refrain from calling trend lines “lines of
best fit.” The “perfect guess” line is the line y = x .
Summary:
The line y = x is the parent function for the linear function family. In this
activity intuition for the line y = x is developed as an important starting point
for work with linear functions.
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Student Activity 1: How Old?
Name
Guess (Age)
Actual (Age)
1. Sketch a graph of the data.
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2. Using your data, create a scatter plot on your graphing
calculator. Sketch it here.
3. Justify your viewing window choice.
4. Using your calculator, graph the “perfect-guess” line over the
scatter plot.
5. Are you an “over-guesser” or an “under-guesser”? Explain.
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Over or Under?
Graph #1
Graph #2
50
50
40
40
30
30
20
20
10
10
10
20
30
40
50
10
guess
20
30
40
50
guess
1. For each graph above, sketch in the “perfect-guess” line,
actual = guess, and label it.
2. For graph #1, did the person over-guess or under-guess?
Explain.
3. For graph #2, did the person over-guess or under-guess?
Explain.
4. For each graph above, sketch a trend line for the data and
label it “my trend”.
5. Use your trend line on graph #1. If I guessed an age of 24,
what is the actual age of the person?
(24, ______)
6. Use your trend line on graph #1. If the person really was 36
years old, what did I probably guess? (______, 36)
7. On the back, create a scatter plot of a person who guessed
really well. Label the axes.
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Sample Assessment
1. Consider A and B. Whose age did you
guess better? Why?
A
B
C
2. Did you overestimate or underestimate
C’s age?
Guess
3. How do you tell your graphing calculator to graph the best
guess line, guess = actual ?
Your partner had the following guesses.
Name
Guess Actual
Mr. Jackson
50
42
Ms. Chi
42
37
Mr. Beyer
45
40
Ms. Harris
40
28
4. Which would be a good viewing window for a scatter plot of
the above data?
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Your partner had the following guesses.
Name
Guess
Mr. King
37
Ms. Alcini
30
Mr. Golm
35
Ms. Cline
25
Actual
42
37
42
30
5. What would be a good viewing window for a scatter plot?
Explain each choice.
Xmin:
Xmax:
Ymin:
Ymax:
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Student Activity 2: Sales Goals
Overview:
Students use the context of sales goals to further refine their understanding of
the line y = x by contrasting points above and below the line.
Objective:
Algebra I TEKS
b.2.a The student identifies and sketches the general forms of linear ( y = x )
b.2.c The student interprets situations in terms of given graphs or creates
Terms:
linear parent function
Materials:
markers, 1” grid paper
Procedures:
Activity: Sales Goals
Begin by orienting students to the graphs using the following discussion
questions.
The management for a clothing store sets up weekly sales goals for their
employees. The graphs show the results for a quarter of a year (12 weeks).
Each data point represents a week’s (actual sales, sales goals) for an
employee.
• What is represented on the x-axis? [Actual sales.]
• What is represented on the y-axis? [Sales goals.]
• What is the meaning of an ordered pair in this situation? [An ordered
pair is (actual sales, sales goals).]
• Why might some employees have higher goals set for them? [Some
examples may include: more experienced employees, employees with
strong sales in the previous quarter, employees that work at peak sales
times of the day or week.]
• Why do employees have such different levels of goals in the same
quarter? [If these graphs represent the second quarter of the year, the
goals for the two weeks before Father’s Day would be higher than the
week following. The goals for the weeks previous to Easter would be
higher than the weeks after Easter.]
Pick a specific point on a graph and discuss the meaning. For example,
circle the lowest point on Amber’s graph.
• What does this point mean for Amber? [Amber set a goal to sell about
$2600 for a particular week. She actually sold about $6800, exceeding
her goal.]
• What are the meanings for the points under the line goals = actual?
[The employees exceeded their goals.]
• What are the meanings for the points over the line goals = actual?
[The employees failed to meet their goals.]
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Have students answer the questions on the following page in their groups
and sketching their graphs for Exercises 6 – 8 on 1” grid paper. Have
students view other groups’ graphs.
1.
2.
3.
4.
5.
Enrique and Amber met and exceeded their goals.
Seth failed to meet his weekly goals.
Moesha most often met her weekly goals.
Moesha and Enrique had the higher goals.
Seth and Amber had the lower goals.
6. Sample answer:
7. Sample answer:
10000
10000
8000
8000
6000
6000
4000
4000
2000
2000
Actual Sales ($)
Actual Sales ($)
8. Sample answers:
10000
10000
8000
8000
6000
6000
4000
4000
2000
2000
Actual Sales ($)
Wardrobe consultant with middle
goals who often matched the goals.
Summary:
Actual Sales ($)
Wardrobe consultant with middle
goals who often failed to meet
the goals.
Reading and interpreting scatter plots of points above and below the line
y = x helps students further refine their understanding of the linear parent
function.
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Student Activity 2: Sales Goals
A local clothing store sets weekly sales goals for their
employees. The graphs below show the quarterly results for
four employees, (actual sales, sales goals).
Moesha
Seth
10000
10000
8000
8000
6000
6000
4000
4000
2000
2000
Actual Sales ($)
Actual Sales ($)
Enrique
Amber
10000
10000
8000
8000
6000
6000
4000
4000
2000
2000
Actual Sales ($)
Actual Sales ($)
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1. Which consultant(s) most often met and exceeded their
weekly goals?
2. Which consultant(s) most often failed to meet their weekly
goals?
3. Which wardrobe consultant(s) most often matched their
weekly goals?
4. Which consultant(s) had the higher goals?
5. Which consultant(s) had the lower goals?
6. Sketch a graph of a consultant who has low goals and
consistently matched them.
7. Sketch a graph of a consultant who has high goals and did
not meet them.
8. Name a scenario that is not represented by the four original
graphs or in Exercise 6 and 7 above. Sketch a graph to match
the scenario.
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1.2 The Y-Intercept: Leaders’ Notes
1.2 The Y-Intercept
Overview:
Participants use real life experiences to build the concepts of y-intercept as the
starting point and slope as a rate of change.
Objective:
Algebra I TEKS
(c.1.C) The student translates among and uses algebraic, tabular, graphical, or
verbal descriptions of linear functions.
(c.2.A) The student develops the concept of slope as rate of change and
determines slopes from graphs, tables, and algebraic representations.
(c.2.B) The student interprets the meaning of slope and intercepts in
situations using data, symbolic representations, or graphs.
Terms:
y-intercept, slope, rate of change, increasing, decreasing, recursion
Materials:
Procedures:
Depending on the participants, briefly talk through or work through the
Student Activity, which connect recursion with graphing.
Activity 1: The Birthday Gift
Work through Activity 1 with participants, modeling good pedagogy: ask
leading questions, use appropriate wait time, have teachers present their work,
etc.
Introduce the scenario.
1. Guide participants in filling in the table, using language similar to the
following:
At time zero, Susan started with $25.
Time (weeks)
Process
Amount Saved
0
$25
$25
After the 1st week, Susan had the $25 she started with and $2.50.
Time (weeks)
Process
Amount Saved
0
$25
$25
$25 + $2.50
1
$27.50
After the 2nd week, Susan had the $25+$2.50 from week 1 and another $2.50.
In other words, Susan had the $25 she started with and two $2.50’s.
Time (weeks)
Process
Amount Saved
0
$25
$25
$25 + $2.50
1
$27.50
2
$30.00
$25 + $2.50 + $2.50 = $25 + 2($2.50)
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Note that the above step is not a natural step for many students. They are
more apt to operate recursively on the previous term , adding $2.50 to $27.50.
Teachers need to be aware that this is difficult for some students.
Time (weeks)
0
1
2
3
4
Process
$25
$25 + $2.50
$25 + $2.50 + $2.50 = $25 + 2($2.50)
25 + 2.5 + 2.5 + 2.5 = 25 + 3(2.50)
25 + 2.5 + 2.5 + 2.5 + 2.5 = 25 + 4(2.50)
Amount Saved
$25
$27.50
$30.00
$32.50
$35.00
2. Write the sentence in words and then abbreviate to variables:
After t weeks, Susan will have the $25 she started with and t ($2.50’s).
Because of convention, mathematicians write 2.5t , instead of t (2.5).
Time (weeks)
Process
Amount Saved
25 + t (2.5) = 25 + 2.5t
t
25 + 2.5t
3. Use questions to lead participants to find a suitable viewing window.
• What does x represent in this problem? [Elapsed time in weeks]
• What values make sense for x in this problem? [Answers will vary.
Sample answer. Zero weeks to 10 weeks.]
• What does y represent in this problem? [Total money saved]
• What values make sense for y in this problem? [Answers will vary.
Sample answer. No money to $60.]
4. Sample answer. The variable x stands for elapsed time in weeks so zero to
10 weeks shows a reasonable amount of time. The variable y stands for
total money saved, so $0.00 to $60.00 will show all the savings and the xaxis.
5. 25 + 2.5(7) = 42.50 . Susan will have $42.50 after 7 weeks.
6. 25 + 2.5t = 139.99 . After 46 weeks, Susan will have more than $139.99,
enough to buy the ring.
You may have to open up your window. We did as follows:
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As an extension, note that the question really asks for an inequality:
25 + 2.5t ≥ 139.99 .
7. Susan’s starting value is lower, so the line will “start” on the y-axis at 15
instead of 25. The y-intercept changed. The slope, or amount of money
she saved every week, did not change. The two lines are parallel, with the
new line translated down from the original.
8. Susan’s starting value is higher, so the line will “start” on the y-axis at 40
instead of 25. The y-intercept changed. The slope, or amount of money
she saved every week, did not change. The two lines are parallel, with the
new line translated up from the original.
9. Susan’s rate of saving has changed so the amount of money will not grow
as fast, so the line will be less steep. Susan’s rate of saving has changed.
Her starting point, or the y-intercept, did not change. The new line is not
parallel to the original line because the rate of saving has changed.
10. When the rate of saving changes, the slope of the line changes.
11. When the starting value in Susan’s saving’s plan changed, the starting
point, or y-intercept of the line, changed.
12. The point (0, y) is where a line intersects the y-axis. This point represents
the starting value of Susan’s savings plan.
Activity 2: Spending Money
Have participants work through Activity 2 in their groups. Encourage them to
practice the language they plan to use when teaching their students.
1.
Time (weeks)
Process
Amount of
Money
0
$1090
$1090
$1090 − $30
1
$1060
2
$1030
$1090 − $30 − $30 = $1090 − 2($30)
3
$1000
1090 − 30 − 30 − 30 = 1090 − 3(30)
4
$35.00
1090 − 30 − 30 − 30 − 30 = 1090 − 4(30)
2. Write the sentence in words and then abbreviate to variables:
After t weeks, Manuel will have the $1090 he started with minus t ($30’s).
Because of convention, mathematicians write 30 , instead of t (30).
Time (weeks)
Process
Amount of Money
1090 − t (30) = 1090 − 30t
t
1090 − 30t
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3. Use questions to lead participants to find a suitable viewing window.
• What does y represent in this problem? [Total amount of money]
Sample answer. $800 to $1090.]
4. Sample answer. The variable x stands for elapsed time in weeks so zero to
10 weeks shows a reasonable amount of time. The variable y stands for
total amount of money, so $800 to $1090 will show all his money.
5. 1090 − 30(11) = 760 . Manuel will have $760 after 11 weeks.
6. 1090 − 30t = 0 . After 36 weeks, Manuel will only have $10. He will
cannot spend the whole $30 the next week, only $10 and then he will be
out of money.
You may have to open up your window. An example:
7. Manuel’s starting value is higher, so the line will “start” on the y-axis at
1300 instead of 1090. The y-intercept changed. The slope, or amount of
money he spent every week, did not change. The two lines are parallel,
with the new line translated up from the original.
8. Manuel’s starting value is lower, so the line will “start” on the y-axis at
890 instead of 1090. The y-intercept changed. The slope, or amount of
money he spent every week, did not change. The two lines are parallel,
with the new line translated down from the original.
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9. Manuel’s rate of spending has changed so now the amount of money will
not deplete as fast, so the line will be less steep. Manuel’s rate of saving
has changed. His starting point, or the y-intercept, did not change. The
new line is not parallel to the original line because the rate of saving has
changed.
Activity 3: Money, Money, Money
Work through Activity 3 with participants.
1. Use questions to find a suitable viewing window.
• What does y represent in this problem? [Total amount of money]
Sample answer. No money to $1100.]
Sample answer. The variable x stands for elapsed time in weeks so zero to
38 weeks shows the time it takes Manuel to spend all of his money. The
variable y stands for total amount of money, so $0.00 to $1100 will show
both graphs.
2. 25 + 2.5(7) = 42.50 . Susan will have $42.50 after 7 weeks.
3. 25 + 2.5t = 1090 − 30t . They never do have the same amount of money
because they are saving or spending each week, not in the middle of the
week. This is shown in the table as we choose the increment to be a week
not a part of a week. After week 33, Susan has $107.50 and Manuel has
$100, which is the closest they get to each other.
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4. From earlier work, we found that Manuel had only $10 to spend after 36
weeks. So we will say that after 36 weeks, Manuel is out of money. So
the question is now, how much money does Susan have after 36 weeks?
Susan has $115 after 36 weeks.
To find this answer, we solved 1090 − 30t = 0 and used the solution to
solve 25 + 2.5(36) = 115 .
1. a. Yen started with $20.
b. Lira started with $0.00.
c. Lira is saving $30 a month.
d. Mark is saving $10 a month.
2. a. Frank started with $80.
b. Ruble started with $40.
c. Peso is spending $30 a month.
d. Ruble is spending $10 a month.
3. ii, b
4. iii, c
5. iv, a
6. i, d
Use the following questions to summarize and connect activities:
• What changes in the situation resulted in a change in the steepness of the
line? [Changing the rate of spending per week, the amount of money
spent per week. Encourage participants to use the word “rate”.]
• What changes in the situation resulted in a change in the starting point of
the line? [Changing the starting amount of money, initial amount of
money.]
• Look at your function rules. What does the constant represent in this
problem? [The initial, or starting, amount of money]
• Look at your function rules. What does the coefficient of x represent in
this problem? [Encourage the words “rate of spending”]
• Look at your function rules. If the coefficient of x is negative, what does
this represent in this problem? [Spending]
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•
Summary:
Look at your function rules. If the coefficient of x is positive, what does
this represent in this problem? [Saving]
Using real life situations, participants investigate the effects of changing the
starting point and the rate of change of a line.
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1.2 The Y-Intercept: Activity 1
Activity 1: The Birthday Gift
Susan’s grandmother gave her $25 for her birthday. Instead of
spending the money, she decided to start a savings program by
depositing the $25 in the bank. Each
week, Susan plans to save an
additional $2.50.
1. Make a table of values for the situation.
Time (Weeks)
Process
Amount Saved
$25
2. Write a function rule for the amount of money Susan will
have after t weeks.
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3. Find a viewing window for the problem situation.
Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
4. Justify your window choices.
Use your graph and table to find the following:
5. How much money will Susan have after 7 weeks? Write this
equation. Show how you found your solution.
6. Susan wants to buy a school ring. When will she have
enough money to buy the $139.99 ring? Write this equation.
Show how you found your solution.
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7. How will the line change if Susan deposits only $15 of the
$25? Graph the line. What changed? What did not
change?
8. How will the line change if Susan deposits the $25 from her
grandmother plus another $15 she already had? Graph the
line. What changed? What did not change?
9. How will the line change if Susan deposits the $25 from her
grandmother, but decides she can only save $2.00 a week?
Graph the line. What changed? What did not change?
10. What changes in the situation resulted in a change in the
steepness of the line?
11. What changes in the situation resulted in a change in the
starting point of the line?
12. Write the coordinates of the point where a line intersects the
y-axis. This point is called the y-intercept. What do these
coordinates represent in this problem?
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Activity 2: Spending Money
Manuel worked all summer and saved $1090.
He plans to spend $30 a week.
1. Make a table of values for the situation.
Time (Weeks)
Process
Amount of Money
$1090
2. Write a function rule for the amount of money Manuel will
have after t weeks.
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Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
5. How much money will Manuel have after 11 weeks? Write
this equation. Show how you found your solution.
6. When will Manuel be out of money? Write this equation.
Show how you found your solution.
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7. How will the line change if Manuel had initially earned
$1300? Graph the line. What changed? What did not
change?
8. How will the line change if Manuel spent $200 on school
clothes and started the year with only $890? Graph the line.
What changed? What did not change?
9. How will the line change if Manuel starts with the $1090, but
decides he will only spend $25 a week? Graph the line.
What changed? What did not change?
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Activity 3: Money, Money, Money
1. Manuel has $1090 and he will spend $30 a
week. Susan has $25 and will save $2.50
a week. Find a viewing window that
includes both situations.
Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
3. When will Manuel and Susan have the same amount of
money? Write this equation. Show how you found your
solution.
4. How much money will Susan have when Manuel is out of
money? Show how you found your solution.
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1.2 The Y-Intercept: Reflect and Apply
Reflect and Apply
1. The graphs represent the savings of three students.
Lira
1
2
3
4
1
Months
a.
b.
c.
d.
Yen
Mark
2
3
4
1
2
3
4
Months
Months
Which student started with the most money? Explain.
Which student started with the least money? Explain.
Which student is saving the fastest? Explain.
Which student is saving the slowest? Explain.
2. The graphs represent the spending habits of three students.
Ruble
1
2
Frank
3
4
Months
a.
b.
c.
d.
1
2
Months
Peso
3
4
1
2
3
4
Months
Which student started with the most money? Explain.
Which student started with the least money? Explain.
Which student is spending the fastest? Explain.
Which student is spending the slowest? Explain.
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1.2 The Y-Intercept: Reflect and Apply
Match the recursive routine with an equation and with a graph:
3. _____
1000 ENTER
i. y = 35 − 3 x
Ans – 20, ENTER,
ENTER, . . .
4. _____
35 ENTER,
Ans + 3, ENTER,
ENTER, . . .
a
b
ii. y = 1000 − 20 x
5. _____
1000 ENTER
iii. y = 35 + 3 x
Ans + 20, ENTER,
ENTER, . . .
6. _____
35 ENTER,
Ans – 3, ENTER,
ENTER, . . .
iv. y = 1000 + 20 x
c
d
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1.2 The Y-Intercept: Student Activity
Student Activity: Show Me the Money!
Overview:
Students connect recursive operations with graphs.
Objective:
Algebra I TEKS
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6.
Time
(weeks)
Money
Saved
0
$1090
1
2
3
4
5
6
7
$1060
$1030
$1000
$970
$940
$910
$880
$ 1,1 2 0
$ 1,0 6 0
M
o
n $ 1,0 0 0
e
y
$940
$880
0
1
2
3
4
5
6
7
Time ( weeks)
7. Subtraction, which can also be thought of as repeated addition of a negative
number.
8. linear, decreasing. Emphasize that repeated subtraction is the same as repeated
addition of a negative number.
Assessment Answers:
1. c
2. f
3. d
4. b
5. g
6. e
7. a
8. h
Summary:
Using a recursive routine, students generate points on a graph and make
generalizations. Repeated addition results in a linear graph. Repeated
addition of a positive number is an increasing line. Repeated addition of a
negative number is an decreasing line.
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Student Activity: Show Me the Money!
Susan’s grandmother gave her $25 for her birthday. Instead of
spending the money, she decided to start a savings program by
depositing the $25 in the bank. Each week, Susan plans to save
an additional $2.50.
1. Write a recursive routine to model Susan’s savings plan.
2. Fill in the table and sketch a graph to model Susan’s savings
plan:
Time
Money
(weeks)
3. What operation did you repeat in your recursive routine?
4. How does repeated addition “look” in a graphical
representation?
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Manuel worked all summer and saved $1090. He plans to spend
$30 a week.
5. Write a recursive routine to model Manuel’s spending plan.
6. Fill in the table and sketch a graph to model Manuel’s
spending plan:
Time
(weeks)
Money
7. What operation did you repeat in your recursive routine?
8. How does repeated subtraction “look” in a graphical
representation?
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Assessment
Match the recursive routines with the graphs:
1. ___
300 ENTER
Ans – 20, ENTER, ENTER, . . .
2. ___
90 ENTER
Ans + 20, ENTER, ENTER, . . .
a
3. ___
300 ENTER
b
4. ___
300 ENTER
c
5. ___
90 ENTER
6. ___
90 ENTER
d
e
f
g
h
7. ___
300 ENTER
8. ___
90 ENTER
Ans - 50, ENTER, ENTER, . . .
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1.3 Exploring Rates of Change: Leaders’ Notes
1.3 Exploring Rates of Change
Overview:
Participants use real data from a motion detector to model motion at a
constant rate over time. Participants translate among algebraic, tabular,
graphical, and verbal descriptions of linear functions.
Objective:
Algebra I TEKS
(b.1.B) The student gathers and records data, or uses data sets, to determine
functional (systematic) relationships between quantities.
(c.2.D) The student graphs and writes equations of lines given characteristics
such as two points, a point and a slope, or a slope and y-intercept.
Terms:
rate of change, constant rate
Materials:
motion detector connected to an overhead calculator, a motion detector with
graphing calculator for each group of 3-4, data collection devices, graphing
calculators
Procedures:
The room should be set up with an aisle down the middle. Set up a motion
detector pointing down the aisle, connected to an overhead calculator, so the
group can see both the participants walking down the aisle and the data
projected from the calculator on a screen in front of the room.
Work through the Student Activity, Rates of Change, with participants. Talk
through the assessment. Make sure each participant gets a chance to write an
equation for their own motion over time.
Activity 1: Wandering Around
Work through Exercise 1 with participants. This activity takes participants
from a verbal description of a situation to a graph, table, and rule representing
the situation.
1. You know that Ryan was at 9 feet at 3 seconds. Label the table and fill in
the (3, 9) as shown.
Table
Time
Distance
3
9
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Since you know that Ryan was walking at 2 feet per second, count back to
find his starting point as shown below.
Table
Graph
Time
Distance
0
3
2
1
5
2
2
7
2
3
9
A sentence to describe Ryan’s walk: Start at 3 feet and walk away from
the motion detector at 2 feet per second.
Rule: y = 3 + 2 x
Have participants do Exercise 2 and then discuss their results.
2. You know that Madeline was at 9 feet at 2 seconds. Label the table and
fill in (2, 9) as shown.
Table
Time
Distance
2
9
Since you know that Madeline was walking at 3 feet per second, count
back to find her starting point as shown below.
Table
Time
Distance
0
15
1
12
2
9
Graph
-3
-3
A sentence to describe Ryan’s walk: Start at 15 feet and walk toward the
motion detector (decrease the distance) at 3 feet per second.
Rule: y = 15 − 3 x
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Do Exercises 3 – 4 with participants.
3. You know that Robyn started at 1 foot and she was at 9 feet at 2 seconds.
Label the table and fill in (0, 1) and (2, 9) as shown.
Table
Time
Distance
0
1
2
8
2
9
From the table, you can tell that Robyn covered 8 feet in 2 seconds.
• If Robyn walked 8 feet in 2 seconds, how far did she walk in one
second? [4 feet]
• So how fast was she walking per second? [4 feet per second]
Table
Time
Distance
0
1
1
5
2
9
Graph
4
4
A sentence to describe Robyn’s walk: Start at 1 foot and walk away at 4
feet per second.
Rule: y = 1 + 4 x
4. You know that Chet was at 6 feet at 1 second and he was at 1 foot at 2
seconds. Label the table and fill in (1, 6) and (2, 1) as shown.
Table
Time
Distance
1
1
2
6
1
-5
From the table, you can tell that Chet covered 5 feet in 1 second, thus his
rate was 5 feet per second.
1
1
Time
0
1
2
Table
Distance
11
6
1
Graph
-5
-5
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A sentence to describe Chet’s walk: Start at 11 feet and walk toward the
motion detector at 5 feet per second.
Rule: y = 11 − 5 x
Activity 2: Describe the Walk
This activity takes participants from a table representing a situation to a verbal
description, graph, and rule of the situation.
Have participants work through Exercises 5 – 8 in their groups and then
discuss their results.
1.
1
1
Table
Time
Distance
0
15
1
21
2
27
Graph
6
6
Rule: y = 15 + 6 x
Sample verbal description: Start 15 feet in front of the motion detector and
walk away at 6 feet per second.
2.
Table
Graph
Time
Distance
0
24
1
1
1
23
1
1
2
22
1
1
3
21
1
1
4
20
1
1
5
19
Rule: y = 24 − x
Sample verbal description: Start 24 feet in front of the motion detector and
walk toward the motion detector (distance decreases) at 1 foot per second.
3.
3
3
Time
0
3
6
Table
Distance
1
7
13
Graph
6
6
Rule: y = 1 + 2 x
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Sample verbal description: Start 1 foot in front of the motion detector and
walk away at 2 feet per second.
4.
Table
Graph
Time
Distance
0
-1
3
1
1
2
3
1
2
5
12
4
6
17
12
4
10
29
Rule: y = −1 + 3 x
Sample verbal description: Start 1 foot behind the motion detector and walk
past it. Continue to walk away at 3 feet per second. (Think outside, or this
time, behind the box!)
The scatter plot is a finite set of points. It is based on the data collected by the
motion detector and is stored in the lists on the calculator.
The function rule is an infinite set of points.
One way to tell the difference is to trace. When tracing in the scatter plot, the
calculator will show each data point from the lists. When tracing on the
function rule, the calculator can evaluate the function at any x-value in the
domain determined by the window settings (and limited by the calculator’s
rounding off.) Often there is confusion when a participant traces on the
scatter plot and wants to know why the calculator will not let them trace to a
particular value.
Summary:
By using real data generated from their own motion to determine linear
models, participants further develop the concepts of the y-intercept as a
starting point and slope as a rate of change. They gain facility in translating
among representations: algebraic, tabular, graphical, and verbal descriptions
of linear functions.
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1.3 Exploring Rates of Change: Activity 1
Activity 1: Wandering Around
Label the table and graph. Fill in the table, sketch the graph, and
write a symbolic rule for the situation.
1. Ryan was walking away from the motion detector at 2 feet
per second. You missed where he started but you know that
he was at the 9 foot mark when the timer called out the 3rd
second.
Table
Graph
Rule:
2. Madeline was walking toward the motion detector at 3 feet
per second. You missed where she started, but you know that
she was at the 9 foot mark at the 2nd second.
Table
Graph
Rule:
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3. Robyn started 1 foot from the motion detector. You looked
up and she was at 9 feet at the 2nd second.
Table
Graph
Rule:
4. You looked up and Chet was walking! He was at the 6 foot
mark at the 1st second and the 1 foot mark at the 2nd second.
Table
Graph
Rule:
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Activity 2: Describe the Walk
Label the table and graph. Sketch the graph. Write a symbolic
rule and a description of the walk.
1.
Table
Graph
Time Distance
0
15
1
21
2
27
Verbal Description:
Rule:
2.
Table
Time Distance
3
21
4
20
5
19
Graph
Verbal Description:
Rule:
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3.
Table
Time Distance
0
1
3
7
6
13
Graph
Verbal Description:
Rule:
4.
Table
Time Distance
2
5
6
17
10
29
Graph
Verbal Description:
Rule:
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1.3 Exploring Rates of Change: Reflect and Apply
Reflect and Apply
What is the difference between the scatter plot created by the
motion detector and the graph of the function rule created by the
function grapher?
Scatter plot created by the motion detector:
Function rule:
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1.3 Exploring Rates of Change: Student Activity
Student Activity: What’s My Trend?
Overview:
Students investigate the linear model with motion data. Students also use
numeric techniques to write the equation of a line.
Objective:
Algebra I TEKS
b.1.B The student gathers and records data, or uses data sets, to determine
c.2.A The student develops the concept of slope as rate of change and
c.2.B The student interprets the meaning of slope and intercepts in situations
using data, symbolic representations, or graphs.
Terms:
Linear model, rate of change
Materials:
motion detectors, overhead calculator, data collection devices, graphing
calculators
Procedures:
The classroom should be set up with an aisle down the middle. Set up a
motion detector pointing down the aisle, connected to a viewscreen calculator,
so the class can see both the students walking down the aisle and the data
projected from the calculator on a screen if front of the room.
Activity: What’s My Trend?
1. Relate the following situation to your class: Suppose you start 2 feet from
a chair and walk away at 1.5 feet per second. Complete the table to show
where you are at each second.
0
1
2
10
t
•
•
•
2
3.5
5
?
?
How can you find the distance from the chair at 10 seconds? [Multiply
by 1.5 and add 2]
Write a sentence to describe how you can find the distance if you know
the time. [Distance is 2 plus 1.5 times the time.]
Translate the sentence to an equation. [Distance is 2 + 1.5 * time.
d = 2 + 1.5t ]
0
1
2
10
2
3.5
5
17
t
2 + 1.5 t
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Make a scatter plot of the data.
Show that your equation contains the points by graphing it.
2. Run the motion detector Ranger program. Use the following screens to set
up the experiment and then follow the instructions on the screen.
Ask a student to walk in front of the motion detector as follows. Start about
2 feet from the motion detector and then walk away from it.
When you have a satisfactory graph, press ON, quit, and graph. You should
see the graph again.
Trace to the points where time is 0 seconds, 1 second, and 2 seconds. Have
students fill in the table.
x
0
1
2
t
y
1.4
2.9
4.4
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Ask:
• How can you use the points to find how fast you were going? [Take the
difference over 1 second, about 1.5 feet per second.]
• Where did you start? [about 1.4 feet]
• How can you use your rate and where you started to figure out where you
will be in 10 seconds? [1.4 plus 10 times 1.5]
• Write a sentence to describe how you can find the distance if you know the
time. [Distance is 1.4 plus 1.5 times the time.]
• Translate the sentence in words to a sentence in symbols.
[Distance = 1.4 + 1.5 * time. d = 1.4 + 1.5t ]
Type the equation into the y= menu and graph.
Use the table to check your prediction for where you will be in 10 seconds.
Now repeat the above procedure for the rest of the walks. Find an equation to
fit the data and check your prediction with a table.
Examples of different walks follow:
Ask a student to start 11 feet away and walk toward the motion detector.
(Where were you 2 seconds before?)
Ask a student to stand approximately 4 feet from the motion detector and stand
still for the whole 4 seconds.
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Assessment: What’s My Trend?
1. b
2. a
3. c
6. y = 1.5 x + 2
7. y = 0.5 + 1x
8. y = 4 + 0.2 x
9. y = 5 − 2 x
10. y = 0.4 x
4.
11. y = 2 + x
12. You rode your bike to the park
at a rate of one mile per minute
for 3 minutes. Then you talked
to your friend at the park for 2
minutes. You both walked back
to your house at a rate of 0.6
miles per minute for 5 minutes.
5.
Note: This activity was based on the activity Rates of Change in the
TEXTEAMS Mathematical Modeling Institute for Secondary Teachers.
Summary:
Using motion detectors to gather data for their own motion, students develop
the concept of slope as a rate of change as students write functions to model
the collected data.
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Student Activity: What’s My Trend?
1. Start 2 feet away from
the motion detector and
walk away at 1.5 ft/sec.
t
d
0
1
2
10
t
2. Walk slowly away
from the motion detector
at a constant rate.
t
d
0
1
2
t
3. Walk quickly away
from the motion detector at
a constant rate.
t
d
0
1
2
t
4. Walk slowly toward
the motion detector at
constant rate.
t
d
0
1
2
t
5. Walk quickly toward
the motion detector at a
constant rate.
t
d
0
1
2
t
6. Stand still about 6 feet
in front of the motion
detector.
t
d
0
1
2
t
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Assessment: What’s My Trend?
Match the following equations with the graphs they represent.
1
___ 1. y = 5 − x
3
___ 2. y = 1 + 0.5 x
___ 3. y = 2 x + 1
a.
b.
c.
Draw a graph of each of the following:
4. Start 1 foot from the motion detector and stand still for 3 seconds.
Then walk away from the motion detector at a rate of 2 feet per
second for 2 seconds. Then walk toward it at a rate of 0.2 feet per
second for 5 seconds.
5. Start 5 feet from the motion detector and walk toward it at a rate of 2
feet per second for 1 second. Then walk toward the motion detector
at a rate of 0.3 feet per second for 3 seconds. Now walk away from
the motion detector at a rate of 0.5 feet per second for 6 seconds.
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Write an equation for each of the following:
6. Start 2 feet from the motion detector and walk away at a rate of 1.5
feet per second.
7. Start 0.5 feet from the motion detector and walk away at a rate of 1
foot per second.
Write an equation for each of the following:
8. _______________
9.
_______________
10. _______________
11.
_______________
12. Write a story for the following graph using units of hours for time
and miles for distance.
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1.4 Finite Differences: Leaders’ Notes
1.4 Finite Differences
Overview:
Participants use their cumulative concrete experiences with the linear model to
build to the abstract symbolic representations of slope. Finite differences are
used to find linear models and to discover what makes data linear.
Objective:
Algebra I TEKS
Terms:
rate of change, constant rate, constant differences, finite differences
Materials:
Procedures:
Activity 1: Rent Me!
Here we introduce rental problems, another good real world linear situation.
Have participants work through the activity and share their results.
Table
Time
3
1
2
3
4
5
6
7
Time
0
1
2
3
4
5
6
7
Cost
12.25
2.25
2.25
= 0.75
3
14.50
Cost
10.00
10.75
11.50
12.25
13.00
13.75
14.50
15.25
Function rule:
y = 10.00 + 0.75 x
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Activity 2: Guess My Function!
Play Guess My Function using an overhead graphing calculator as shown:
Put the Beach Rental equation y = 10 + 0.75 x in the graphing calculator
without showing the participants, and using a table, ask for y when x = 2
and x = 4. Give them a minute to figure out the equation.
•
How can you find the equation? [Take differences and work backwards.]
Then enter y = 14 + 1.5 x and give them 3 and 7 for x.
Discuss their methods for finding the rate and the starting point.
Have participants play Guess My Function with a partner using the tables in
Activity 2.
Answers:
1. y = 11 + 5 x
2. y = 1.3 + 1.2 x
3. y = 4 − 2 x
4. y = 25 − 7 x
5. y = 16 + 4 x
6. y = −4.9 + 3 x
7. y = 24 − x
8. y = 41 − 5 x
9. y = 11 + 6 x
10. y = 15 − x
11. y = 4 + 2 x
1
12. y = 25 − x
2
13. y = 6 + 2 x
14. y = 8.05 − .25 x
15. y = −2 x
1
16. y = 5 + x
2
Transparency: How Did You Do It?
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Now that participants have many concrete experiences finding rates of change
(slopes), we formalize the slope of a line as an algebraic rule.
You may use either Transparency 1 or 2.
Use Transparency 1 to discuss the process of finding the rate of change both
numerically and generally.
Use Transparency 2 to develop the ideas live, using Transparency 1 as a
guide.
•
•
•
What do the values in the table have to do with the points on the
graph? [Use the graph to show that the change in y , y2 − y1 , is the
vertical distance between the points. Also show that the change in x ,
x2 − x1 , is the horizontal distance between the points.]
∆y
∆y
Where do you find
in your rule? [The rate of change is
.]
∆x
∆x
How can the development of rate of change and slope through
concrete experiences enhance student understanding?
Extension: In these activities, an algebraic rule for finding the slope between
two points on a line was developed. However, up to this point we have
always found the y-intercept (starting point) by counting back in the table
using the rate of change. Later in the institute, we formalize the point-slope
form of a line, but at this point there is another way to find the equation of a
line when you have found the slope between the two points. See the example
below.
Participants are now comfortable with the form of a line, y = b + mx ,
y = y − intercept + rate( x ) . We use that to our advantage.
We will use the information given in Activity 1: Rent Me!
Find the rate of change between (3, 12.25) and (6, 14.50) which is $0.75 per
hour. Substitute this into the equation of the line, y = b + mx :
y = b + 0.75 x .
Now choose one of the given points, (3, 12.25) and substitute into the above:
12.25 = b + 0.75(3) and solve for b, b = 10 .
So now we have y = 10 + 0.75 x .
Activity 3: Finite Differences
Work through the Activity with participants. Point out that in all the linear
activities up to this point in the institute, we have assumed a constant rate of
change. Here we want to emphasize that in order for a function to be linear,
the first level of differences must be constant and conversely, if the first
differences are constant, the function is linear.
∆x
Term Number
Process Column
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Value of Term
∆y
172
0
1
2
3
4
1
1
1
1
b+a
b+a+a
b+a+a+a
b+a+a+a+a
b
b + 1a
b + 2a
b + 3a
b + 4a
a
a
a
a
b + an
n
1. If first differences are constant, then the data is linear.
2. If the data is linear, then first difference are constant.
Sample Answers for Reflect and Apply:
1.
x
0
1
2
y
11
16
21
Function rule: y = 11 + 5 x
Real world scenario: Rent a bicycle built for two at a base charge of $11.00
and $5.00 per hour.
Another possible real world scenario: Start with a base of 11 block. Add 5
blocks to each figure.
Fig 1:
Fig 2:
Fig 3:
Summary:
Building on participants’ previous experiences in the institute with rate of
change, we formalize the concept of slope and finite differences.
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1.4 Finite Differences: Transparency 1
Transparency 1: How Did You Do It?
The Slope of a Line
With a specific situation:
Table
Time Distance
2
4
10 − 4
4−2
4
10
Graph
(4, 10)
(2, 4)
}
10-4
4-2
10 − 4
x
4−2
Rule: y = starting point +
In general:
Table
x2 − x1
x
x1
x2
y
y1
y2
Graph
y2 − y1
(x 2, y 2)
(x 1, y 1)
y 2-y 1
x 2-x 1
Rule: y = starting point +
y2 − y1
x , y = starting point + rate( x )
x2 − x1
rate of change =
change in y ∆y
y −y
=
= 2 1
change in x ∆x
x2 − x1
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1.4 Finite Differences: Transparency 2
Transparency 2: How Did You Do It?
The Slope of a Line
With a specific situation:
Table
Time Distance
2
4
4
10
Graph
Rule:
In general:
Table
x
Graph
y
Rule:
rate of change =
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1.4 Finite Differences: Activity 1
Activity 1: Rent Me!
1. At the beach in Galveston, you decide to rent an umbrella
from a beach side vendor. Upon asking for the cost to rent
the umbrella for 8 hours, the vendor pulls out a worn and wet
table of prices.
“Oh, no. What will I do?” exclaimed the vendor!
Help the vendor reproduce the price list. What linear
function rule can you use?
Time
1
2
3
4
5
6
7
8
Cost
$12.25
$14.50
Rule:
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Activity 2: Guess My Function!
Write the symbolic rule below each table:
1.
x
y
2.
0
11
1
16
2
21
x
0
1
2
y
1.3
2.5
3.7
3.
x
0
1
2
y
4
2
0
4.
x
0
1
2
y
25
18
11
5.
x
1
2
3
y
20
24
28
6.
x
5
6
7
y
10.1
13.1
16.1
7.
x
3
4
5
y
21
20
19
8.
x
10
11
12
y
-9
-14
-19
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9.
x
0
2
4
y
11
23
35
10.
x
0
5
10
y
15
10
5
11.
x
0
3
6
y
4
10
16
12.
x
0
10
20
y
25
20
15
13.
x
2
5
8
y
10
16
22
14.
x
11
15
19
y
5.3
4.3
3.3
15.
x
-7
-5
-3
y
14
10
6
16.
x
-10
-4
2
y
0
3
6
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Activity 3: Finite Differences
When only one level of differences is necessary to obtain a
constant value, the algebraic rule which generates the terms of
the sequence is linear and can be written in the form b + an . The
terms of a linear sequence are in the form b, b + 1a , b + 2 a ,
b + 3a , b + 4 a , . . . , an + b .
∆x
Term #
Process Column
0
_____________
Value of Term
∆y
b
___
___
1
_____________
b + 1a
___
___
2
_____________
b + 2a
___
___
3
_____________
b + 3a
___
___
4
n
_____________
a( n ) + b
b + 4a
an + b
1. If first differences are constant, then ___________________
2. If the data is linear, then ____________________________
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1.4 Finite Differences: Reflect and Apply
Reflect and Apply
Refer to the tables in Activity 2. Choose 4 tables from the
activity. List them below. Graph each scatter plot and the
function rule. Make up a real world scenario for each. Include
one real world scenario that can be physically built, such as with
centimeter cubes, etc.
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1.4 Finite Differences: Student Activity
Student Activity: Graphs and Tables
Overview:
Students use graphing calculator programs to find the equations of lines given
graphs or tables.
Objective:
Algebra I TEKS
(b.3.B) Given situations, the student looks for patterns and represents
generalizations algebraically.
(c.1) The student understands that linear functions can be represented in
different ways and translates among their various representations.
Terms:
rate of change, y-intercept
Materials:
graphing calculators, LINEGRPH program, LINETBL program
Procedures:
Link the two programs to students, LINEGRPH and LINETBL. Briefly
demonstrate how to run each. An example of each is shown below.
The program LINEGRPH:
When you run the program, it graphs a randomly generated line in the window
[-4.7, 4.7] [-3.1, 3.1].
After examining the graph, enter the equation of the line in Y2 and change
the graph style of Y2 to the point tracer. This allows you to see graph
more easily. Below is an example of a correct equation because the
graphs are the same.
If you had entered an incorrect equation, it would graph accordingly and
you can quickly see that it is incorrect, as shown below.
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1.4 Finite Differences: Student Activity
The program LINETBL:
When you run the program, it displays a randomly generated table of a line.
After examining the table, enter the equation of the line in Y2 . Below is an
example of a correct equation because the tables return the same values.
If you had entered an incorrect equation, it would display a table where the
values generated are not the same and you can quickly see that it is
incorrect, as shown below.
Summary:
Using the power of technology to examine many examples quickly and get
instant feedback, students gain facility in finding the equations of lines given
graphs or tables.
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2.1 Out for a Stretch: Leaders’ Notes
2.1 Out for a Stretch
Overview:
Participants investigate the relationship between the “stretch” of a rubber band
attached to a container and the number of marbles in the container.
Objective:
Algebra I TEKS
(b.1.E) The student interprets and makes inferences from functional
relationships.
(c.1.A) The student determines whether or not given situations can be
represented by linear functions.
Terms:
trend line, linear model, rate of change, slope, y-intercept
Materials:
for each group of 3 –4 participants: one 8 oz. Styrofoam or paper cup, 3” long
thin rubber bands, one 8 oz. cup of marbles of the same size, large paper clips,
tape, meter sticks, graphing calculators
Procedures:
Participants should be in groups of 3 – 4.
Complete the teacher activity with all of the participants, modeling good
pedagogical practices for data collection activities. After completing the
teacher activity as a whole group, divide participants into 4 groups, each of
which will then do one of the student activities in small groups of 3 – 4.
When they have completed their respective student activities, have one small
group for each student activity present their activity to the large group. They
should include a demonstration of the experiment, a scatter plot of their data,
their trend line and discussion of its meaning, and any conclusions.
In this way, each participant will actively participate in two linear data
collection activities and see three others.
If participants have experienced the Hooke’s Law activity in the TEXTEAMS
Mathematical Modeling Institute for Secondary Teachers, you may want to
choose one of the student activities to do as the whole group activity. Then
have any teachers who are not as familiar with the activity do the “Stretch It”
teacher activity, while the other teachers complete the remaining student
activities.
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Activity: Stretch It
Briefly demonstrate the experiment, clearly showing each of the distances to
be measured.
1. Stress the importance of predicting results of the experiment before they
perform the experiment.
Have participants predict each relationship, one at a time, as you
demonstrate the distance to be measured.
• What will the values in List 2 represent? [The distance from the table
to the top of the cup.] Demonstrate this measurement and have
participants sketch their prediction of the graph of (list 1, list 2).
• What will the values in List 3 represent? [The distance from the table
to the bottom of the cup.] Demonstrate this measurement and have
participants sketch their prediction of the graph of (list 1, list 3).
• What will the values in List 4 represent? [The distance from the
bottom of the cup to the floor.] Demonstrate this measurement and
have participants sketch their prediction of the graph of (list 1, list 4).
2. Have participants create their “hanging containers” as shown and collect
the data. Have participants fill in the table accordingly. Make sure
participants measure the distances from the free hanging cup, before any
marbles are added.
• Which measure is the dependent variable and which is the independent
variable? Justify your response. [The number of marbles is the
independent variable because we are controlling this number. The
distances from the table are dependent because they rely, depend, on
how many marbles are in the cup.]
Sample data:
Number of
marbles
List 1
0
5
10
15
20
25
30
Distance from
table to top of cup
(cm)
List 2
16
16.5
16.9
17.2
17.5
17.8
18.2
Distance from table
to bottom of cup
(cm)
List 3
26.1
26.4
27
27.3
27.6
27.8
28.2
Distance from
bottom of cup to
floor (cm)
List 4
51.6
51.2
50.8
50.5
50.2
49.8
49.4
Have the recorder enter the data in a graphing calculator and link with the
other members of the group.
Have one group write their data on the transparency of Activity 1 to use with
the whole group.
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First find a trend line for List 2 (distance from the table to the top of the cup)
versus List 1 (number of marbles) together as a group.
• Is the graph linear or non-linear? [Answers will vary. It looks linear.]
• How can you decide if the data is linear? [Based on the work in 2.1.4
Finite Differences, if the first differences are relatively constant, it
makes sense to model the data with a linear model.]
4. Have participants use mental math to find first differences on the data in
List 2 as you find first differences on the transparency of Activity 1.
Number of
marbles
List 1
0
5
5
5
10
5
15
5
20
5
25
5
30
Distance from table
to top of cup (cm)
List 2
16
0.5
16.5
0.4
16.9
0.3
17.2
0.3
17.5
0.3
17.8
0.4
18.2
Distance from table to
bottom of cup (cm)
List 3
26.1
0.3
26.4
0.6
27
0.3
27.3
0.3
27.6
0.2
27.8
0.4
28.2
Distance from bottom
of cup to floor (cm)
List 4
51.6
0.4
51.2
0.4
50.8
0.3
50.5
0.3
50.2
0.4
49.8
0.4
49.4
change in y
. Take an average
change in x
of the differences in List 2 and divide by 5 (the constant difference in List 1.)
.37
For the sample data above, rate of change ≈
= 0.074 cm/marble.
5
To estimate a rate of change per marble, find
5. We do not need to estimate because we found the distance for no marbles
in the cup. For our sample data, the y-intercept is 16 cm.
6. For our sample data, y = 16 + 0.074 x
7.
8. The units of slope are centimeters per marble.
9. The y-intercept represents the distance from the table to the top of the cup
when the cup is empty.
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change in y
. Take an average of the
change in x
differences in List 3 divided by 5 (the constant difference in List 1.) For
.35
the sample data above, rate of change ≈
= 0.07 cm/marble.
5
in the cup. For our sample data, the y-intercept is 26.1 cm.
13. For our sample data, y = 26.1 + 0.07 x
11. To estimate a rate of change, find
14.
16. The y-intercept represents the distance from the table to the bottom of the
cup when the cup is empty.
change in y
. Take an average of the
change in x
differences in List 4 divided by 5 (the constant difference in List 1.) For
.37
the sample data above, rate of change ≈
= 0.074 cm/marble.
5
in the cup. For our sample data, the y-intercept is 51.6 cm.
20. For our sample data, y = 51.6 + 0.074 x
18. To estimate a rate of change, find
21.
23. The distance from the floor to the bottom of the cup when the cup is
empty.
Activity 2: Comparing Graphs
1.
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2. For the sample data:
#1: y = 16 + 0.074 x
#2: y = 26.1 + 0.07 x
#3: y = 51.6 − 0.074 x
3. The slope for trend line #3 is negative because the distance from the floor
to the bottom of the cup is decreasing. The slopes of trend lines #1 and #2
are positive because the distances from the table to the top or bottom of
the cup are increasing. The magnitudes of the slopes of the three trend
lines are very close to one another. This is because the slopes all represent
the number of centimeters the cup moved for each marble added.
4. For the sample data the difference between the y-intercept of trend line #1
and the y-intercept of trend line #2 is 26.1 − 16 = 10.1cm. This distance is
the length of the cup.
5. Sample data. One way to find when the cup will touch the floor is to find
when the distance from the bottom of the cup to the floor is zero. This
relationship is represented by trend line #3, y = 51.6 − 0.074 x . Therefore,
find x when 0 = 51.6 − 0.074 x .
Another way to find when the cup will touch the floor is to find when the
distance from the table to the bottom of the cup is equal to the height of
the table. So, the height of the table is the distance from the floor to the
table equals the distance from the table to the bottom of the cup and the
distance from the floor to the bottom of the cup which is:
26.1 + 51.6 = 77.7 cm . So the question is when does trend line #2 equal
77.7? Therefore, find x when 26.1 + 0.07 x = 77.7 .
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(The above two methods produce different solutions because they use
different trend lines that used slightly different rates of change.)
6. Sample data: 16 + 0.074( 42) = 19.108
1. Either Group 1’s marbles are much heavier than Group 2’s marbles or
Group 1’s rubberband is stretchier than Group 2’s.
2. Group 1’s table is taller than Group 2’s. Group 1’s cup is taller than
Groups 2’s by 0.5 cm.
3. The intersection point is when the distance from the table to the top of the
cup equals the distance from the floor to the bottom of the cup. It is the
number of marbles that move the cup to a point halfway between the table
and the floor.
Math Note:
Hooke’s Law: The force exerted by a spring is proportional to the distance the
spring is stretched or compressed from it’s relaxed position, that is that the
tension exerted by a stretched string is (within certain limits) proportional to
the extension, or, in other words, that the stress is proportional to the strain.
Robert Hooke, 1635- 1703, had many varied interests from physics and
astronomy, to chemistry, biology, and geology. He made many important
scientific contributions.
Note that Hooke’s law is based on a spring. In this activity, we use a rubber
band instead of a spring to achieve similar results. If desired, use a spring for
the most accurate results. Another alternative is to use a slinky to simulate a
spring.
Note: This activity was based on the activity “Linear Modeling in Science” in
the TEXTEAMS Mathematical Modeling Institute for Secondary Teachers.
Note on Student Activity Sample Assessments: The sample items are
intended to provide teachers with possible ways to assess the data collection
activities. They are not intended to show the best or only assessments
possible. Have participants suggest other possible techniques. Have
participants compare the Sample Assessments and discuss the different
objectives of each. For example, one Sample Assessment prompts students
for a graphical solution, another for a tabular solution, and another for two
solutions.
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Summary:
Participants collect and analyze data to find the relationships between the
number of marbles in the cup and selected distances from the cup. By
concluding that a linear model is a reasonable model, participants demonstrate
that a spring stretches at a constant rate, Hooke’s Law. By using differences
and estimating rates of change with real data, participants further cement the
concept of the linear model.
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2.1 Out for a Stretch: Activity 1
Activity 1: Stretch It
What is the relationship between the number of marbles in the cup
and the distances shown below?
Use large paper clips and an 8 oz. paper cup to form a “hanging
container.” Measure the distances indicated below as marbles
are added to the cup.
Table
Rubberband
List 2
List 3
Paper
clips
List 4
Floor
1. Predict a graph of the relationship between the number of
marbles in the cup and the distances shown above. You
should predict 3 different graphs.
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2. Measure the indicated distances and record in the table
below. Next, add 5 marbles to the container and measure
each distance as before. Record the new measurements.
Continue this process: add 5 marbles to the container,
measure, and record.
Number Distance from Distance from Distance from
of
table to top of table to bottom bottom of cup
marbles
cup (cm)
of cup (cm)
to floor (cm)
List 1
List 2
List 3
List 4
0
5
10
15
20
25
30
Trend line #1: Consider the relationship between the distance
from the table to the top of the cup (List 2) and the number of
marbles (List 1.)
3.
4.
Create a scatter plot using a graphing calculator.
Estimate a rate of change by finding first differences in
your data.
5. Estimate the y-intercept (starting point.)
6. Use the estimated rate and y-intercept to find a trend line for
your data.
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7. Graph the trend line over the scatter plot. Adjust the
parameters y-intercept and rate of change, if necessary, for
a better fit.
8. What are the units of slope for the trend line?
9. What is the meaning of the y-intercept in the trend line?
from the table to the bottom of the cup (List 3) and the number
of marbles (List 1.)
10. Create a scatter plot using a graphing calculator.
11. Estimate a rate of change by finding first differences in the
data.
your data.
a better fit.
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from the floor to the bottom of the cup (List 4) and the number
of marbles (List 1.)
17. Create a scatter plot using graphing calculator.
18. Estimate a rate of change by finding first differences in the
data.
your data.
a better fit.
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Activity 2: Comparing Graphs
1. Graph all three scatter plots and trend lines on your calculator
in the same window. Sketch:
2. Write the equations of the trend lines.
Trend line #1, List 2 vs List 1:
3. Compare the slopes of the trend lines. What do you find?
4. Find the difference between the y-intercept of trend line #1
and the y-intercept of trend line #2. Where is this distance in
the experiment?
5. Use your trend line to determine when the cup will touch the
floor. Describe your strategy.
6. Use your trend lines to determine how far the top of the cup
would be from the table if you added 42 marbles. Describe
your strategy.
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2.1 Out for a Stretch: Reflect and Apply
Reflect and Apply
Group 1
Group 2
1. What can you conclude about Group 1’s marbles and rubber
band compared to that of Group 2?
2. What can you conclude about Group 1’s table and cup
compared to that of Group 2?
3. What is the meaning of the intersection point shown below
from Group 1?
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2.1 Out for a Stretch: Student Activity 1
Student Activity 1: Have You Lost Your Marbles?
Overview:
Students investigate the relationship between the height of an object and the
distance the object rolls.
Objective:
Algebra I TEKS
relationships.
(b.2.D) In solving problems, the student collects and organizes data, makes and
interprets scatter plots and models, predicts, and makes decisions and critical
judgments.
Terms:
rate, slope
Materials:
each group needs 5 – 6 building blocks, 39 cm of PVC pipe, marble or steel
ball bearing that will fit inside and roll freely through the PVC pipe, metric
tape measure, graphing calculators
Procedures:
Activity: Have You Lost Your Marbles?
Briefly describe and/or demonstrate the experiment. Make sure students
measure the distance the marble rolls once it leaves the end of the pipe.
1. Stress how important it is for students to predict the results of the
experiment before they perform the experiment. Encourage students to
think about and anticipate the results of the experiment before they begin
collecting data.
2. Sample data:
Height
Distance
(blocks)
(cm)
1
19
2
53.5
3
74
4
100
5
137.5
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3. Sample data:
4. Sample data:
An estimate for the rate of change is about 30 cm/block.
5. Since the rate of change is about 30, then the y-intercept is about
19 − 30 = −11 .
6. Using our sample data, y = −11 + 30 x
7. Sample data:
8. The units of slope are centimeters per block.
9. The real world meaning of the y-intercept is that for zero blocks the
marble does not roll out of the pipe at all, it covers no distance.
10. The equation is y = −13 + 30(20) . Some solution methods:
11. For our sample data, solve: −13 + 30 x = 60 .
Some solution methods:
Table:
Other
Table:
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Trace:
Trace to the
Intersection point.
Guess and check.
Solve algebraically, if you are at a place in your curriculum where it
makes sense for students to do so: −13 + 30 x = 60
−13 + 30 x + 13 = 60 + 13
 1  30 x = 73 1 
 30 
 30 
x = 2.43
12. The more blocks placed under the pipe, the farther the marble travels.
•
•
•
Did the height of the pipe increase the distance the marble rolled?
[Yes.]
Why or why not? [The marble has more potential (stored) energy with
greater height.]
If the slope keeps increasing, will the marble roll farther each time?
Why or why not? [Up to a point, yes. When the slope is completely
vertical, the marble will not roll far because its energy is absorbed by
the ground.]
Extensions. Find the trend lines for the following and compare.
• Use different sized marbles (as long as they still travel freely through
the pipe),
• Use spheres with different masses (i.e., golf balls, ping pong balls,
steel ball bearings),
• Use different surfaces (i.e., rug, cement, dirt, table top, sheets).
Answers to Sample Assessment
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1.
2. With 7 blocks, the marble will roll approximately 163 cm.
Some solution methods are shown below.
Trace to x = 7
Solve using the trend
line and arithmetic.
Use a table.
3. x ≈ 10 . Students’ answers should be close, depending on their trend lines.
Some solution methods are shown below:
Trace to the
intersection of
y = 2 + 23 x
and y = 232
Use a table, in
two ways
Trace to y = 232
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Summary:
By collecting data and finding a trend line, students investigate the
relationship between the height of an object and the distance it rolls. Students
use real data to further their conceptualization of the linear function.
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Student Activity 1: Have You Lost Your Marbles?
What is the relationship between the height of the pipe and the
distance the marble rolls?
Measure
Roll the marble from heights of 1, 2, 3, 4, and 5 blocks. Release
the marble at the opening of the pipe. Measure the distance the
marble rolls from the end of the pipe.
1. Sketch a graph predicting the relationship between the height
of the pipe and the distance the marble rolls.
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2. Data Collection
Tasks:
• one person rolls the marble,
• one person holds the blocks and pipe,
• one person marks where the marble stops,
• one person measures the distance the marble traveled.
Height
(blocks)
Distance
(cm)
3. Make a scatter plot using a graphing calculator. Sketch below.
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4.
Use first differences to estimate a rate of change.
6. Find a trend line for the data using the estimated rate and yintercept.
7. Graph your trend line over the scatter plot. Adjust the
a better fit.
10. Use your trend line to determine how far the marble would
roll if you placed 20 blocks under the pipe. Write an
equation and solve in at least four ways.
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11. Use your trend line to determine how many blocks are
needed for the marble to roll 60 cm. Write an equation and
solve in at least four ways.
12. Make a general statement about the relationship between the
number of blocks and the distance the marble travels.
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Sample Assessment
A group collected the following data for “Have You Lost Your
Marbles?”
Height
Distance
(blocks)
(cm)
1
25
2
47.5
3
73.75
4
92
5
117
1. Create a scatter plot and find a trend line. Sketch both in an
appropriate window.
2. Use the graph to determine how far the marble would roll
with 7 blocks. Solve in two ways. Show your work.
3. Use the graph to determine how many blocks it would take
for the marble to roll 232 cm. Solve in two ways. Show
your work.
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Student Activity 2: Unidentified Circular Objects (UCO’s)
Overview:
Students investigate the relationship between the diameter of a circular light
on a surface produced by a flashlight and the distance of the flashlight from
the surface.
Objective:
Algebra I TEKS
relationships.
Terms:
diameter, trend line, linear model, rate of change, slope, y-intercept
Materials:
flashlights (one per group), rulers, yardstick or meter stick, graphing
calculators
Procedures:
Activity 1: Unidentified Circular Objects
Briefly describe and/or demonstrate the experiment. Make sure students hold
the meter stick perpendicular to the surface on which the light is shining.
collecting data.
2. Sample data:
Distance (cm)
1
2
3
4
5
6
9
10
Diameter
6.5
8.5
10.6
12.4
14.5
16.3
22.3
24.6
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3. Sample data:
4. Rate of change is approximately 2 cm/cm
5. Estimated starting point is 4.5 cm.
6. y = 4.5 + 2 x
7. Sample data:
8. The units of slope are centimeters per centimeters.
9. The real world meaning of the y-intercept is that if the flashlight was no
centimeters from the surface, the circular light pattern would have the yintercept as its diameter.
10. The equation is y = 4.5 + 2(15) = 34.5 . Some solution methods:
11. For our sample data, solve: 4.5 + 2 x = 18
Table:
Other
Table:
Trace:
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Trace to the
Intersection point.
Guess and check.
Solve algebraically, if you are at a place in your curriculum where it makes
sense for students to do so:
4.5 + 2 x = 18
4.5 + 2 x − 4.5 = 18 − 4.5
 1  2 x = 13.5 1 
 2
 2
x = 6.75
12. The further from the surface the flashlight is, the larger the circular light
pattern produced.
Answers to Sample Assessment:
1.
2. Trace to x = 3.5
3. Two graphical methods:
Trace to y = 20
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Trace to the
intersection of
y = 3.6 + 1.5 x
and y = 20
Summary:
relationship between the diameter of the circular light pattern produced by a
flashlight at varying distances from the surface. Students use real data to
further their conceptualization of the linear function.
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Activity 2: UCO’s
What is the relationship between the diameter of the circular
light pattern cast by a flashlight and the flashlight’s distance
from the circular light pattern?
Diameter
Distance
Vary the distance of the flashlight from the surface and measure
the diameter of the circular light pattern cast by the flashlight.
1. Sketch a graph predicting the relationship between the
diameter of the circular light pattern cast by a flashlight and
the flashlight’s distance from the circular light pattern.
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2. Data Collection
Hold a yardstick perpendicular to a flat surface, such as a table,
with the end starting at 0 on the flat surface. Hold a flashlight
next to the meter stick so that it will cast light on the flat surface.
Place the rim of the flashlight (light source end) at 1 cm and
measure the diameter of the distinct circular pattern formed on
the flat surface. Record in the table below. Continue to vary the
distance of the flashlight from the table and record the diameter
of the circle formed.
Distance (cm)
Diameter
1
2
3
4
5
6
9
10
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4. Use first differences to estimate a rate of change.
7. Graph your trend line over the scatter plot and adjust the
a better fit.
10. Use the trend line to determine the diameter of the circle
when the flashlight is 15 cm from the flat surface. Write the
equation and solve in at least three ways.
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11. Use the trend line to determine how far the rim of the
flashlight is from the flat surface if the diameter of the circle
is 18 cm. Write the equation and solve in at least four ways.
distance of the flashlight from the surfaces and the diameter
of the circular light pattern produced on the surface.
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Sample Assessment
A group collected the following data for Unidentified Circular
Objects.
Distance (cm)
1
2
3
4
5
Diameter (cm)
5
6.4
8
9.6
11
appropriate window.
2. Use the graph to determine what the diameter of the circular
light pattern is when the flashlight is 3.5 cm from the surface.
Show on the graph how you found the answer.
3. Use the graph to determine how far the flashlight is from the
surface when the diameter of the circular light pattern is 20
cm. Show on the graph how you found the answer.
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Student Activity 3: Going to Great Depths
Overview:
Students investigate the relationship between the height of water in a cylinder
and the number of bolts that have been added to the cylinder.
Objective:
Algebra I TEKS
relationships.
(c.2.B) The student interprets the meaning of slope and intercepts in situations
using data, symbolic representations, or graphs.
judgments.
Terms:
rate, slope
Materials:
each group needs a cylinder (obtain from a science class, use a pharmacy
medication bottle that is cylindrical, or use a cylindrical flat-bottomed
drinking glass), uniform objects that will fit in the cylinder and sink (golf
balls, marbles, centimeter cubes), water, metric ruler, graphing calculators
Procedures:
Note: for the sample data below, we used a cylinder and 4 golf balls. You
can also use marbles but instead of adding one at a time, add 5 marbles at a
time. You want the displacement to be enough to be able to measure easily.
If you use 5 marbles each time, adjust the questions accordingly.
Activity: Going to Great Depths
Briefly describe and/or demonstrate the experiment. Make sure students
measure the water level before adding any objects.
collecting data.
2. Sample data:
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Number of Objects
0
1
2
3
4
Height
(cm)
9
10.2
11.5
12.7
13.9
3. Sample data:
4. For our sample data, an estimated rate of change is 1.2 centimeters per
object.
5. For our sample data, the y-intercept is the original water level, 9 cm.
6. For our sample data, a trend line is y = 9 + 1.2 x .
7.
8. The units of slope are centimeters per object.
9. The real-world meaning of the y-intercept is that for zero objects the water
level is the original level before there were any objects added.
10. The equation is y = 9 + 1.2(9) . Some solution methods:
11. For our sample data, the highest water level recorded was 13.9
centimeters so we need to find how many objects would cause the water to
rise to 13.9 + 6 = 19.9 cm. Solve: 9 + 1.2 x = 19.9 .
Some solution methods:
Table:
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Trace:
Trace to the
Intersection point.
Guess and check.
Solve algebraically, if you are at a place in your curriculum where it
makes sense for students to do so:
9 + 1.2 x = 19.9
9 + 1.2 x − 9 = 19.9 − 9
 1  1.2 x = 10.9 1 
 1.2 
 1.2 
x = 9.083
12. The more objects placed in the cylinder, the higher the level of the water.
13. The rate of change would be higher because larger objects would displace
more water. Therefore the line would be steeper, have a higher slope.
14. If you added the same amount of water, the original water level would be
higher. Therefore the line would shift up. Also, while the objects would
still displace the same amount of water, this amount of water displaced in
a smaller container would make the rate of change increase. Therefore the
slope of the line would be steeper.
Answers to Sample Assessment
1.
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2.
Using a table.
3. x ≈ 31. Students answers should be close, depending on their trend lines.
Using a table, in
2 ways
Summary:
relationship between the height of water in a cylinder and the number of
uniform objects added to the cylinder. Students use real data to further their
conceptualization of the linear function.
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Student Activity 3: Going to Great Depths
What is the relationship between the
number of uniform objects added to the
cylinder and the height of the water in
the cylinder?
Add uniform objects to the cylinder.
Measure the height of the water with
each additional object.
1. Sketch a graph predicting the
relationship between the height of the
water and the number of objects added:
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2. Data Collection
Number of
Objects
Height
(cm)
3. Make a scatter plot using a graphing calculator. Sketch
below.
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a better fit.
10. Use your trend line to determine how high the water would
rise with 9 objects in the cylinder. Write the equation and
solve in at least three ways.
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11. Use your trend line to determine how many objects are
needed to make the water rise 6 cm higher than the highest
water level recorded (assuming the container could hold that
much water.) Write the equation and solve in at least four
ways.
number of uniform objects and the height of the water.
13. Suppose you used larger objects. Predict how the graph
would change.
14. Suppose you used a cylinder whose diameter is half that of
the original cylinder. Predict how the graph would change.
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Sample Assessment
A group collected the following data for Going to Great Depths.
Number of
Objects
0
1
2
3
4
Height (cm)
15
16.1
17
18.2
19
appropriate window.
2. Use a table on a graphing calculator to determine how high
the water level would be with 10 added objects. Show how
you found your answer.
3. Use the table on a graphing calculator to determine how
many objects were added if the water level is 45 cm,
(assuming the container is tall enough.) Show how you
found your answer.
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Student Activity 4: Height versus Arm Span
Overview:
Students investigate the relationship between the height of person and the
person’s arm span.
Objective:
Algebra I TEKS
relationships.
judgments.
(c.2.G) The student relates direct variation to linear functions and solve problems
involving proportional change.
Terms:
rate, slope, arm span, proportional relationship
Materials:
metric measuring tape or meter stick(s), graphing calculators
Procedures:
Activity: Height versus Arm Span
Briefly describe and/or demonstrate the experiment.
collecting data.
2. Sample data:
Arm Span (cm)
171.5
169
169
170
159
179
Height (cm)
170
166.5
169
175
166
184
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3. Sample data:
4. For our sample data, an estimated rate of change is 1 centimeters per 1
centimeter.
5. For our sample data, we could reason that the y-intercept is the theoretical
height of a person with no arm span, therefore the y-intercept is zero.
6. For our sample data, a trend line is y = 0 + 1x .
7.
8. The units of slope are centimeters of height per centimeters of arm span.
Since the units (centimeters) are the same,
cm
= 1 and therefore, the slope is this case is dimensionless. To illustrate,
cm
consider if the measurements would have been made in inches, feet,
cubits, or pencil lengths. The units of slope would be the same,
unit of measure
= 1.
unit of measure
9. The real world meaning of the y-intercept is that for a theoretical person
with no arm span, the person would have no height.
10. y = 1(137) = 137 . Students could solve using a table, tracing on the graph,
and evaluating on the home screen. See the other student activities for
examples of solution methods.
11. 1x = 214 . Students could solve using a table where y = x , using a table
where y = x and y = 214 , tracing on the graph, finding the intersection of
y = x and y = 214 , and using guess and check on the home screen. See
the other student activities for examples of solution methods.
12. The longer your arm span is, the taller you are.
13. Neither variable is the independent or dependent variable. There is not a
dependent relationship inherent in this situation. The two relationships
(arm span, height) and (height, arm span) are inverse relations.
14. The ratios should be somewhat constant and approximately equal to 1.
y
15. The average of the ratio should be very close to the slope of the trend
x
line, approximately 1.
y
16. If = k , then k is the constant of proportionality.
x
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Answers to Sample Assessment:
1 – 2. The points in the scatter plot for Group A all lie below the line y = x .
This means the people measured by Group A have longer arm spans than
their heights. However, the points in the scatter plot for Group B all lie
above the line y = x . This means the people measured by Group B are
taller than their arm spans.
Also, noting the different windows, the points in the scatter plot for Group
A must have generally higher coordinates than those for Group B. The
folks measured by A are taller than those measured by Group B.
3. The points should have relatively high coordinates and lie below the line
y = x.
Summary:
relationship between the height of a person and the person’s arm span.
Students use real data to further their conceptualization of the linear function,
specifically of the form y = mx , a proportional relation.
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Student Activity 4: Height versus Arm Span
What is the relationship between the height of a person and the length
of the person’s arm span?
Measure the height and arm span of students.
1. Sketch a graph predicting the relationship between the height
of the person and the person’s arm span:
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2. Data Collection
Arm Span
(cm)
Height
(cm)
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a better fit.
10. Use the trend line to determine how tall a person is with an
arm span of 137 cm. Write an equation and solve in at least
three ways.
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11 . Use your trend line to determine what arm span a 214 cm
tall person would have? Write an equation and solve in at
least four ways.
height of a person and the person’s arm span.
13. Which is the independent variable and which is the
dependent variable in this problem situation?
A linear relationship that contains the origin is called a
proportional relationship and is in the form y = mx .
14. In the table, find the average of the ratios,
y
.
x
y
above to the slope in your
x
trend line. What do you find?
15. Compare the average ratio
16. If
y
= k , what is k called?
x
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Sample Assessment
Two groups collected data for height versus arm span. Their
scatter plots in relation to the line y = x are shown below.
Group A
Group B
1. Name one difference between the people measured by Group
A and the people measured by Group B.
2. Name another difference between the people measured by
Group A and the people measured by Group B.
3. What would a graph look like of mostly tall people whose
arm spans tended to be greater than their heights? Sketch the
graph and include the line y = x .
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2.2 Linear Regression: Leaders’ Notes
2.2 Linear Regression
Overview:
Participants write a program to find a least squares linear function to model
data. They use the program and calculator regression to find linear models for
data and they discuss the correlation coefficient, r.
Objective:
Algebra I TEKS
(c.2.C) The student investigates, describes, and predicts the effects of changes
in m and b on the graph of y = mx + b .
Terms:
trend line, line of best fit, linear regression, residual, r, correlation coefficient
Materials:
graphing calculator, dynamic geometry program with a prepared
demonstration of a linear least squares fit for data, computer with a projection
device.
Procedures:
Math Note:
With the advent of graphing calculators, many have begun to use calculator
regression to find models for data. This activity is designed to help
participants understand how a linear regression model is calculated, to discuss
when and how to use calculator linear regression, to think about the
pedagogical issues associated with calculator linear regression, and to
correctly understand how to use (or not to use) the correlation coefficient, r.
The least squares method of finding a line of best fit is accessible for teachers,
especially if looked at geometrically, using a dynamic geometry program and
also if looked at numerically, using a graphing calculator.
Begin by showing a geometric demonstration of the least squares method for
finding a line of best fit and discuss as follows.
Help orient participants by pointing out the data set (points) and the trend line.
Make sure the trend line is not close to the line of best fit so that the “squares”
can be seen.
When finding a line of best fit, we desire to minimize the distance between the
y-values of the data and the function values of the line. (A statistician might
say that we want to minimize the difference between an observed value of the
response variable and the value predicted by the regression line.) The closer
the line is to the data, the smaller the differences will be. Because the
differences may be positive (if the data point is above the line) or negative
(the data point is below the line), we then look at the squares of the
differences. Hence we are really looking to minimize the squares of the
differences. This can be shown on the dynamic geometry program as shown
below.
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The following graphs come from a demonstration sketch from the Geometer’s
Sketchpad. The original sketch:
2
P6
P5
P4
P2
yint
-2
P3
P1
Slope
2
4
Points P1 through P6 represent data points. A
line is drawn through the points and from each
data point to the line a square is constructed.
Drag the y-intercept and slope of the line so
that the sum of the areas of the squares is
minimized. That line is the least squares
regression line for the data.
Bill Finzer, 3/95
Total Area = 0.82 inches 2
-2
Show participants the data set and the trend line. Point out the differences
between the data and the trend line, and the visuals representing the square of
the differences (shaded squares) as shown below. Use the Transparency to
illustrate.
This square represents
2
( y1 − f ( x1 )) for
point P 1 (x1, y1 )
2
P6
}
P2
}
y int
}
P3
P1
Slope
-2
2
-2
}
P5
P4
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The differences between
the y-values of the points and the
function values of the line.
233
Use the dynamic geometry program to raise the y-intercept of the trend line
and watch the differences get larger, and hence, the squares get bigger. Note
the total area gets bigger as the squares get bigger.
2
P6
P5
P4
P2
y int
P3
P1
Slope
-2
2
4
-2
Use the dynamic geometry program to lower the y-intercept of the trend line
and watch the differences get smaller, and hence, the squares get smaller.
Note the total area gets smaller as the squares get smaller.
2
P6
P5
P4
P2
P3
P1
y int
-2
2
4
Slope
-2
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Use the dynamic geometry program to change the slope of the trend line to
make it steeper and watch the squares change. Note the total area.
2
P6
P5
P4
P2
y int
-2
P3
P1
Slope
2
4
-2
Use the dynamic geometry program to change the slope of the trend line to
make it less steep and watch the squares change. Note the total area.
2
P6
P5
P4
P2
Slope
y int
P3
P1
-2
2
4
-2
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Change both the y-intercept and the slope to minimize the total area. When
the area is as small as you can get it, you have a line of best fit using a least
squares method. If you desire, you can display the equation of your line of
best fit as shown.
2
y = 0.39x + 0.33
P6
P5
P4
P2
P3
y int
P1
Slope
-2
2
4
-2
•
•
Using the method of least squares, why do we square the differences
and then sum them? Why do we not just sum the differences? [Some
of the differences might be positive, while others might be negative.
Adding these together, they may cancel each other out. You would get
a small sum but you might not have a very good fit. Also the points
that are furthest from the line adds much more weight than points close
to the line, when their differences are squared.]
How has this geometrical approach added to your understanding of
linear regression?
Math Note: The difference between the y-value of the data point and the
function value of the trend line is called the residual. Therefore, we want to
minimize the sum of the squares of each of the residuals to achieve a better
fit. A method of discussing the appropriateness of a model is to look at a
residual plot, (x-value, residual). This leads more into statistics and modeling
and will not be discussed more in the institute.
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Activity 1: Sums of Squares
Now participants use a numeric approach to finding a model using least
squares. To do this, write a program that does the following with participants.
The program should:
Find the differences between the y-values of the data points and the associated
y-values on the line.
Square the differences.
Sum the squares.
Display the sum.
To do this, the participant should first put the data in List 1 (x-values) and List
2 (y-values) and the first guess of a trend line in the function grapher (y=).
Help participants write the program, using questions like the following:
• How do you denote the difference between a point’s y-value and the
function’s y-value? [Given the point (L1, L2) and the function y1, the
difference is denoted L2 − y1( L1) .]
• Where should we put those differences? [Store the difference in List
3.]
• What do we need to do with the differences? [Square them, (List 3)2]
• Where should we put those squares? [Store the squares in List 4.]
• What do we need to do with the squares? [Sum them, sum(List 4)]
• What do we want to do with the sum of the squares once it is
calculated? [See it on the calculator screen. Display the sum.]
Using the following data, here is a sample of how you can demonstrate the
program:
Enter the data.
Graph the data.
Propose a trend line and find the sum of squares.
Adjust the trend line and try for a smaller sum.
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Adjust the trend line again and try for a smaller sum.
Adjust the trend line again and try for a smaller sum.
When you are satisfied with your accuracy, when you have minimized the
sum of squares, then you have a reasonable fit for your data. A line of best fit
is the line that minimizes the sum of squares. Try to minimize the sum of
squares to get the best fit you can.
Activity 2: Line of Best Fit
Have participants work through the activity in groups, comparing their trend
lines and sums of squares with each other in their group.
1. Enter the data
2. Enter a trend
line.
3. Use the program to find
the sum of squares.
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Repeat to find better and better trend lines as the sum of squares
decreases.
Equation
y = 111 + 0.9 x
y = 100 + 1.6 x
y = 99.86 + 1.63 x
Sum of Squares
3881
88
76.0252
Note that the person whose equation most closely matches the linear
regression line of best fit shown below will have the least sum of squares.
We will return back to this after the next activity.
5. Answers will vary.
Discuss with participants the different processes used to find a trend line.
In this activity, we looked for the smallest sum of squares. In the data
collection activities originally we took first differences, estimated a rate of
change and a starting point, and adjusted the function based on the graph.
• How do these methods compare?
• What different things do you learn from each method?
• Which method do you think you will use in the rest of the institute?
Why?
Math Note: The least sum of squares that you find with the above program
will vary dramatically from data set to data set. The more linear the data, the
smaller the least sum of squares can be. The less linear the data is, the bigger
the residuals will be, the larger the least sum of squares will be. Also, the
nature of the data can affect the magnitude of the least sum of squares. For
instance, if the data deals with relatively large numbers such as distances
between planets, the least sum of squares will probably be a similarly large
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number (unless, of course, the data is completely linear, in which case the sum
of squares is zero.) Likewise, if the data deals with relatively small numbers
such as the size of insect eyes, the least sum of squares will probably be a
similarly small number. This is all about scale. The least square sums are not
scale insensitive. So if you are measuring in centimeters versus meters, the
sums of squares may be very different in magnitude.
Activity 3: The Correlation Coefficient
A big part of thinking algebraically is doing and undoing. Here we start with
first differences and ask participants to make data sets based upon the given
first differences. Here we again return to the idea that if a data set has
constant first differences, it is linear.
a.
x
0
1
2
3
4
5
y
100
129
89
124
137
119
1. Sample data sets. Note: for each different first value, there will be a
different data set.
b.
c.
d.
x
y
x
y
x
y
0
0.2
0
-10
0
2000
29
-15
20
1
-14.8
1
10
1
2004
-40
21
8
2
6.2
2
18
2
2008
35
15
5
3
21.2
3
23
3
2012
13
-4
17
4
17.2
4
40
4
2016
-18
8
32
5
25.2
5
72
5
2020
4
4
4
4
4
2.
4.
Not linear data, low r
value
Slightly more linear
data, slightly higher
but still low r value
More linear, higher r
value
Exactly linear, r = 1
5. As the data is more linear, the r gets closer to 1. See the following note
about data that correlates negatively.
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Note that all of the r values shown above are positive. This is because the
data correlated positively. If the data correlates negatively, the r values are
negative.
You can quickly demonstrate this by looking at data that correlates negatively.
Note that r is close to –1. Data that is linear with a negative slope has an r
close to –1. The same pattern holds true. That is that the more linear the data,
the better a line will model the data. This is indicated by r close to 1, this
time by r being close to –1.. In other words, if a linear function models the
data well, you will have an r value close to 1. The next part of the activity
examines the converse of this statement.
6.
7. Note that the second differences below are constant. This means that the
data is quadratic, not linear.
x
y
First
Second
Differences Differences
0
0
1
2
1
1
3
2
2
4
5
2
3
9
7
4
16
8. Just because you have r close to 1, you may not have found a good (an
appropriate) model. The data above is quadratic, yet a linear model yields
r close to 1.
If you have a good model, the r will be close to 1. If you have a model
that yields r close to 1, then you may or you may not have a good model.
The correlation coefficient, r, measures the strength and direction of the
linear associate between two variables. An underlying key to the
discussion is that one must look at the data to see if a linear model is
appropriate, and then interpret the r value in that context.
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Math note: Statisticians often find it useful to square the correlation
coefficient, r. This statistic, called the coefficient of determination, is a
measure of the proportion of total variation in the observed values of y (the
dependent variable) that is explained by the observed values of x (the
independent variable). The value of the coefficient of determination may vary
from zero to one. A coefficient of determination of zero indicates that none of
variation in the dependent variable is explained by the independent variable.
On the other hand, a coefficient of determination of one indicates that 100%
of the variation in y has been explained by the regression equation. Thus, if a
researcher finds that there is a correlation coefficient of +0.5 between IQ and
reading speed, then the r-squared value of 0.25 tells us that 25% of the
variation in reading speed of the subjects is related to the individual IQ's.
Note that this also means that 75% is related to other factors, so that much of
reading speed is not accounted for by IQ.
If r 2 is large (98%, 89%, etc.), the model is providing a good fit to the data
and we can have confidence in its ability to predict. If r 2 is small (10%, 18%,
25%, etc.), the model is not providing a good fit. If the data fall perfectly all
along a straight line, then the model is a perfect fit, and r 2 is 1.0. In general,
the extent to which the data points are lined up along the line or scattered
away from it determines the strength of the correlation r 2 . Keep in mind that
while r 2 indicates the strength of correlation, one still requires r to indicate
the direction of the correlation (+ or -). So, one needs both statistics to tell the
whole story.
Math Note: The value of r can vary, depending on the size of the data set.
Larger data sets yield more confidence in the trend and therefore the model.
A linear model can often fit reasonably well over a small set of data, but does
not represent the trend over the long run behavior of the data. Teachers
should use caution when using models to make predictions far from the data
set, especially when the data set is small.
Caution: It is suggested that you not become involved in a discussion of
statistics. The intention in this activity is not to teach a statistics course. It is
to caution teachers from making inferences about r that are not true and about
lines of best fit when the model may not be appropriate.
Discuss with participants the difference between trend lines and lines of best
fit.
• When in the institute did we find trend lines?
• When in the institute did we find lines of best fit?
• How do trend lines and lines of best fit compare?
Note: Throughout the institute, we have found trend lines. We have used real
life problem situations to develop the concepts of slope as a rate of change and
y-intercept as a starting point. We have also used first differences to find rates
of change and y-intercepts. Also, in the data collection activities, we used first
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differences to estimate rates of change and y-intercepts for trend lines. We
then adjusted the parameters to visually obtain a better fit.
Our objectives in all of these activities finding and using trend lines was to
build understanding of the linear function. Our objective was not to obtain a
line of best fit in the easiest way. Had we entered table values into lists and
found regression lines as “end-all” answers, we would have missed
opportunities to learn about the attributes of linear functions.
When and how to use regression models is a topic for discussion. Certainly
calculator regression can be used effectively in the midst of a larger problem,
where the objective is to use the regression model to learn about a concept.
Teachers may have to wrestle with the fact that some students may know how
to find regression models with technology before the teacher wants students to
have that knowledge. Effective assessment is essential in provoking students
to really think, using technology as a tool for understanding and not as a
crutch that hinders further progress.
Sample Answers to Reflect and Apply:
1. A trend line is an estimate for a linear function to model a situation. It
may or may not be a line of best fit. A line of best fit is the best linear
function to model a situation, usually found by linear regression. In the
case of perfectly linear data, it is simply the line that contains those points.
4. Enter the two points into lists as shown below and find the line that
contains them by using linear regression. Note that the correlation
coefficient is 1 or –1 because through any two points there is exactly one
line.
For example: find the equation of the line that contains (0, 0) and (1, 10).
Find the equation of the line that contains (4.3, 20.4) and (5.1, -10.5)
Summary:
By looking at a geometric and numeric approach to finding lines of best fit by
the method of least squares, participants gain added understanding of
calculator linear regression models. Calculator linear regression should be
used to further understanding of the linear model or as an intermediate step in
a bigger problem.
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2.2 Linear Regression: Transparency
Transparency: Least Squares
This square represents
( y1 − f ( x1 ))2 for 2
point P1 (x1, y1)
P6
}
P2
}
y int
}
P3
}
P1
Slope
-2
2
-2
P5
P4
4
The differences between
the y-values of the points
and the function values
of the line .
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2.2 Linear Regression: Activity 1
Activity 1: Sums of Squares
Write a program that will, step by step, find the sum of squares
between a linear function and a data set.
The program will:
• Find the differences between the line and the
y-values of the data points.
• Square the differences.
• Sum the squares.
• Display the sum.
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Activity 2: Lines of Best Fit
Using your program and the following data, find a line of best
fit.
1. Enter the data into lists in your calculator.
x
y
10
120
20
129
30
146
40
163
50
187
60
199
70
212
2. Enter a guess for a trend line into the function grapher in
your calculator.
3. Use your program to find a line of best fit. Record your trend
lines and the corresponding sums of squares:
Equation
Sum of Squares
4. Compare your equation and your least sum of squares with
your group.
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5. In your group, refer to the data collection activities in 2.1
Out For a Stretch. Enter a data set from one of the activities.
Use your program to find a line of best fit. Compare that line
with those from your group members. Also compare that line
with the trend line you found when you first completed the
activity.
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Activity 3: The Correlation Coefficient
1. Construct a data set given the first differences shown.
a.
b.
c.
d.
x
y
x
y
x
y
x
y
0
0
0
0
29
-15
20
1
1
1
1
-40
21
8
2
2
2
2
35
15
5
3
3
3
3
13
-4
17
4
4
4
4
-18
8
32
5
5
5
5
2. Sketch a scatter plot of each data set above.
a.
b.
c.
4
4
4
4
4
d.
3. What do you notice about the above scatter plots? What is
the big visual picture in each of the above graphs?
4. Using linear regression on your calculator, find a line of best
fit for each data set above. Record the equation of the line
and the value of the correlation coefficient, r.
a.
b.
c.
d.
5. What is true about the value of r as the data becomes more
linear?
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Is the converse true?
6. Using the linear regression on your calculator, find a line of
best fit for the data below. Note the value of the correlation
coefficient, r.
x
0
1
2
3
4
5
y
0
1
4
9
16
25
7. Find first differences and then second differences for the data
above. What do you find?
8. If r is close to 1, have you necessarily found the most
appropriate model? Why or why not?
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2.2 Linear Regression: Reflect and Apply
Reflect and Apply
1. What is the difference between a trend line and a line of best
fit?
2. When do you believe students should find trend lines and
when should they find lines of best fit?
3. How can you use technology to enhance student
understanding, without allowing the student to rely on the
technology as a crutch with little understanding of what the
technology is doing?
4. How can you use linear regression on your calculator to find
the equation of the line between two points?
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3.1 Solving Linear Equations: Leaders’ Notes
3.1 Solving Linear Equations
Overview:
Participants solve linear equations with concrete models and make
connections between the concrete model, abstract, and symbolic
representations.
Objective:
Algebra I TEKS
(c.3.B) The student investigates methods for solving linear equations and
inequalities using concrete models, graphs, and the properties of equality,
selects a method, and solves the equations and inequalities.
Terms:
concrete model, addition property of equality, subtraction property of equality
Materials:
algebra tiles, overhead algebra tiles
Procedures:
Participants should be seated at tables with plenty of elbow room in groups of
3—4.
“Why use algebra tiles? Manipulating algebra tiles combines an algebraic and
a geometric approach to algebraic concepts using an array-multiplication
model similar to that employed in many elementary school classrooms. Our
experience leads us to believe that students benefit from seeing algebra
concepts developed from such a geometric perspective.
Furthermore, we believe that we reach a broader group of students by
sequencing instruction from the concrete level, through the pictorial level, and
finally to the abstract—or symbolic—level. Such sequencing gives students
several modes, in addition to just abstract manipulations, that help them
understand and solve algebraic problems. The algebra tiles give a frame of
reference to students who are not abstract thinkers.” Leitze, Annette Ricks
and Kitt, Nancy A., “Using Homemade Algebra Tiles to Develop Algebra and
Prealgebra Concepts,” Mathematics Teacher, September, 2000, 462.
Before learning to solve linear equations with algebra tiles, students should be
familiar with the tiles, what the tiles represent, and the relationships that exist
among the tile pieces. Students should recognize that color represents positive
or negative quantities and that the shape of the tile determines the value it
represents (unit, x, x 2 .) Students should also have experiences adding and
subtracting integers and should understand the concept of “zero pairs.” A
zero pair consists of a negative tile and a positive tile pair. Together their sum
is zero. For example, a negative unit tile and a positive unit tile form a zero
pair, and a positive x tile and a negative x tile form a zero pair, etc.
Begin by explaining zero pairs. Lead participants through the following
examples, emphasizing that to maintain equality, manipulations made on one
side must also be made on the other side.
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Have participants solve x + 5 = 7 on their own and then discuss the two
different algebra tile solution methods on Transparency 1.
•
Which method looks easier? Do you think that it always will be?
[Participants may suggest that using the subtraction property of
equality is easier. They may change their minds on the next example.]
Have participants solve x + ( −4) = 3 on their own and then discuss the two
different algebra tile solution methods on Transparency 1.
Discuss the differences between using the addition property of equality and
using the subtraction property of equality. The technique must match the
symbolic representation of solutions. The goal is to help participants gain
facility in seeing the connection between the two methods, choosing the most
expedient method, and being able to perform either whenever called for.
Math note: The addition property of equality: if a = b , then a + c = b + c .
The subtraction property of equality: if a = b , then a − c = b − c .
Activity 1: Concrete Models
Work through one Exercise with participants, demonstrating the process of
solving using tiles, drawing a sketch showing the manipulations, and writing
the symbolic version. Sketches should not be tedious, but should be viewed
as a bridge from the concrete to the symbolic. Eventually sketches will be
used as mental representations to assist students in understanding the symbolic
manipulation.
Make sure that participants write the symbolic representation that matches
their concrete manipulation. For example, if they remove 2 negative unit tiles
then they should write −( −2) . If they add two positive unit tiles, they should
write +2 .
Have participants work in pairs to complete the activity. One participant
manipulates the tiles and the other participant records the actions on paper.
Participants should switch roles periodically.
One possible sequence is shown for each Exercise. Note that in the example
answers, the addition property of equality is used. This is primarily for
consistency and is not the only or necessarily the most expedient way.
Discuss both ways with participants.
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1. Sample solution:
4 x + 8 + ( −4 x ) = 5 x + 4 + ( −4 x )
4 x + 8 = 5x + 4
8= x+4
8 + ( −4) = x + 4 + ( −4)
4=x
2.
Sample solution:
11 = 3 x + 2
x + 20 − x − 9 = 7 x + 2 − 4 x
11 + ( −2) = 3 x + 2 + ( −2)
9 = 3x
3= x
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3.
Sample solution:
x = −x + 1
x + 7 + ( −7) = 4 x + 3 − 5 x − 2
x + x = −x + 1 + x
2x = 1
x=
1
2
Activity 2: Using Concrete Models
Have the participants work through the activity. Have a participant work
through an exercise using overhead tiles. Ask another participant to quickly
demonstrate a different way to manipulate the tiles to solve the same problem.
Continue with the other exercises.
1. Sample solution:
8 x − 12 + ( −3 x ) = 3 x + 13 + ( −3 x )
8 x − 12 = 3 x + 13
5 x − 12 + 12 = 13 + 12
5 x = 25
x=5
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2.
Sample solution:
x − 6 = −x − 6
x − 6 = 2 x + 4 − 3 x − 10
x = −x
x − 6 + 6 = −x − 6 + 6
At this point in the problem, x = − x , you can ask,
• What number equals its opposite? [Zero, 0 = −0 , therefore x = 0 .]
An alternative method is to add a positive x tile to both sides, resulting in
2 x = 0 . Therefore each x tile is equal to zero.
3.
Sample solution:
3 x + 7 − (3 − x ) = ( x + 2 ) + x
4x + 4 = 2x + 2
2x + 4 = 2
4 x + 4 + ( −2 x ) = 2 x + 2 + ( −2 x )
2 x + 4 + ( −4) = 2 + ( −4)
2 x = −2
x = −1
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4.
Sample solution:
− x − 10 = x + 2
2 x − 9 − (3 x + 1) = 5 x − ( 4 x − 2)
−10 = 2 x + 2
− x − 10 + x = x + 2 + x
−10 + ( −2) = 2 x + 2 + ( −2)
−12 = 2x
−6 = x
You may need some additional examples, depending on the level of your
participants.
•
What are some ways that you used to record the algebra tiles? [Some
teachers use dots and lines, some use circles and ovals, etc. Have
participants share their recording strategies. Suggest that participants
help students use the drawings as a bridge from the concrete tiles to
the abstract algebraic notation.]
1. See Transparency 2 for an example.
2. See Transparency 1 for an example.
Summary:
Students come to algebra classes with varied backgrounds and learning styles.
Using concrete models to introduce and support algebraic solution strategies
bridges the gap between student informal understanding to abstract
understanding. Sequencing instruction from the concrete, through the
pictorial, to the abstract gives students several ways to understand algebraic
problems.
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3.1 Solving Linear Equations: Transparency 1
Transparency 1: x + 5 = 7
Algebra tile solution of
x + 5 = 7, using the addition
property of equality
x + 5 = 7, using the
subtraction property of
equality
x+5= 7
x+5= 7
x + 5 + ( −5) = 7 + ( −5)
x +5−5 = 7−5
x=2
x=2
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3.1 Solving Linear Equations: Transparency 2
Transparency 2: x + ( −4) = 3
x + ( −4) = 3, using the
addition property of equality
x + ( −4) = 3, using the
subtraction property of
equality
x + ( −4) = 3
x + ( −4) = 3
You do not have –4 to
subtract from both sides, so
add 4 zero pairs.
x + ( −4) + 4 = 3 + 4
x + ( −4) = 3 + 4 + ( −4)
x=7
x + ( −4) − ( −4) = 7 + ( −4) − ( −4)
x=7
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3.1 Solving Linear Equations: Activity 1
Activity 1: Concrete Models
Use algebra tiles to solve each equation. Sketch each step and
record the symbolic representation for each step.
1.
____________________
original equation
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2.
____________________
original equation
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3.
____________________
original equation
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Activity 2: Using Concrete Models
Build each equation and solve with algebra tiles. Record the
intermediate steps and the solution.
1. 8 x − 12 = 3 x + 13
2. −6 + x = 2 x + 4 − 3 x − 10
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3. 3 x + 7 − (3 − x ) = ( x + 2) + x
4. 2 x − 9 − (3 x + 1) = 5 x − ( 4 x − 2)
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3.1 Solving Linear Equations: Reflect and Apply
Reflect and Apply
Write an equation that is readily solved with an algebra tile set,
and:
1. is more easily solved using the addition property of equality.
2. is more easily solved using the subtraction property of
equality.
3. What are your classroom goals for using tiles?
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3.2 Stays the Same: Leaders’ Notes
3.2 Stays the Same
Overview:
Participants solve linear equations in one variable, making connections
between algebraic solution steps, algebra tile solution steps, and graphical
solution steps.
Objective:
Algebra I TEKS
Terms:
algebraic solution method, algebra tile solution method
Materials:
algebra tiles, overhead algebra tiles, graphing calculators, 1” grid paper,
markers
Procedures:
The big idea in this activity is that just as each algebraic step in solving a
linear equation can be modeled with algebra tiles, each algebraic step can be
modeled with a graphic representation. As you graph each side of a resulting
equation in the solving process, the x-value of the intersection point remains
the same. This is because the solution to each resulting equation in the
solving process has the same solution. In other words, equivalent equations
have the same solution. This is an important connection.
Work through the following example with participants on the overhead
projector with the algebra tiles and on the overhead calculator.
Example: 5 + x = 3 x + 1
5 + x = 3x + 1
[0, 4.7] [0, 10]
WINDOW
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5 + x + (− x ) = 3x + 1 + (− x )
•
What remains the same? [The xcoordinate of the intersection points.
This is the solution, x = 2 , to both
equations: the original and the
equivalent equation.]
•
What remains the same? [The xcoordinate of the intersection points,
which is the solution to each equivalent
equation, x = 2 .]
5 = 2x + 1
5 + ( −1) = 2 x + 1 + ( −1)
4 = 2x
2=x
(You can use a
vertical line to show
that each intersection
point has the same xvalue.)
•
What remained the same throughout the entire solving process? [The xcoordinate of the intersection points, which is the solution to each
equivalent equation, x = 2 .]
• Could we have solved the equation in a different way? [Yes, in a few
different ways. For example, you could have added –5 to both sides first.]
• Would solving the equation differently change the solution to the
equation? [No.]
• Would it change the resulting graphs? [Yes.]
Work through the following example to demonstrate how solving the problem
in a different way changes the way the graph looks, but the x-coordinate of the
resulting intersection points remain the same, x = 2 .]
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5 + x = 3x + 1
[0, 4.7] [-8, 10]
WINDOW
5 + x + ( −5) = 3 x + 1 + ( −5)
x = 3x − 4
x + ( −3 x ) = 3 x − 4 + ( −3 x )
−2 x = −4
•
− x = −2
If the opposite of
x is –2, − x = −2 ,
then what is x?
x = 2.
x=2
Activity: Stays the Same
Have participants work on the activity in their small group. Use the model
just completed by everyone.
Assign one of the Exercises to each group, and have each group make a poster
size presentation on 1” grid paper of the graphical results of their Exercise.
They should list the algebraic steps and sketch the corresponding lines in one
appropriate window. You might suggest that they use a different color for
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each step and sketch the corresponding lines in the same color. Then have
them make a second poster size presentation, showing the same Exercise, but
with the equation solved in a different order, thus obtaining different graphs,
all with the same x-coordinate for the relevant intersection points.
Have participants present their work to the rest of the group.
•
•
•
•
•
Did anyone solve this Exercise different from the two shown by this
group? [If so, briefly describe the results.]
What connections does this activity build?
What are some of the important concepts or ideas that you want students
to understand as a result of this activity?
How might students view the algebraic solution method differently after
completing this activity?
How do you think this activity might impact how you teach solving one
variable linear equations?
One way of completing each exercise follows. There are many other correct
ways.
1. A sample solution:
7 + 3 x = −8 − 2 x
[-4.7, 4.7] [-20, 3.1]
(−7) + 7 + 3x = ( −7) − 8 − 2 x
3 x = −15 − 2 x
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3 x + 2 x = −15 − 2 x + 2 x
5 x = −15
x = −3
x + 5 =1− x
[-4.7, 4.7] [-7, 5]
x + 5 + x =1− x + x
2x + 5 = 1
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2 x + 5 + ( −5) = 1 + ( −5)
2 x = −4
x = −2
3 − 2x = x − 6
[-4.7, 4.7] [-6, 10]
3 − 2x + 6 = x − 6 + 6
−2 x + 9 = x
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2x − 2x + 9 = 2x + x
9 = 3x
3= x
−2 x + 4 = −5 + 4 x
[-4.7, 4.7] [-3.1, 11]
−2 x + 4 + 2 x = −5 + 4 x + 2 x
4 = −5 + 6 x
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4 + 5 = −5 + 5 + 6 x
9 = 6x
3
=x
2
1. The original equation is −2 − 6 x = 3 − x .
−2 − 6 x = 3 − x
−2 − 6 x + 2 = 3 − x + 2
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−6 x = 5 − x
−6 x + x = 5 − x + x
−5 x = 5
x = −1
Summary:
Building on the work for solving one variable equations with concrete models
(algebra tiles), participants make connections between the concrete model, the
algebraic solution method, and a graphical look at the steps involved in each.
Just as the solution to equivalent equations is the same, so too is the xcoordinate of the intersection points when the equivalent equations are
graphed. This gives teachers one more way to meet all of the learning styles
present in their classrooms.
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3.2 Stays the Same: Activity
Activity: Stays the Same
Solve each of the following problems, showing each step, in the
three ways below. Sketch the algebra tile solution and the
graphical solution.
1. 7 + 3 x = −8 − 2 x
Algebra Tile
Solution:
Algebraic
Solution:
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Solution:
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2. x + 5 = 1 − x
Algebra Tile
Solution:
Algebraic
Solution:
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Solution:
275
3. 3 − 2 x = x − 6
Algebra Tile
Solution:
Algebraic
Solution:
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Solution:
276
4. −2 x + 4 = −5 + 4 x
Algebra Tile
Solution:
Algebraic
Solution:
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Solution:
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3.2 Stays the Same: Reflect and Apply
Reflect and Apply
1. Fill in the missing steps below:
Algebra Tile
Algebraic
Solution:
Solution:
Graphic
Solution:
2. Write a linear equation that:
• has an integer solution,
• can be solved with one algebra tile set, and
• whose solution can be found graphically in the
window [-10,10] [-10, 10].
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3.3 Solving Linear Inequalities: Leaders’ Notes
3.3 Solving Linear Inequalities
Overview:
Participants use problem situations and technology to explore linear
inequalities.
Objective:
Algebra I TEKS
(c.3.A) The student analyzes situations involving linear functions and
formulates linear equations or inequalities to solve problems.
(c.3.C) For given contexts, the student interprets and determines the
reasonableness of solutions to linear equations and inequalities.
Terms:
linear inequality, strict inequality
Materials:
transparencies of the Student Activity: Age Estimates from 2.1.1 The Linear
Parent Function, graphing calculators
Procedures:
Activity 1: Linear Inequalities in One Variable
Many of the problem situations explored throughout the institute have focused
on situations in which linear equations were solved. Many of those linear
equations, however, could very well have been considered linear inequalities.
Ask participants to recall some of these problem situations that could have
been considered as linear inequalities.
Have participants work through the activity. Circulate and ask guiding
questions. Discuss their results, using the following answers as a guide.
1. 5 + 4 x
2. 5 + 4 x ≤ 50
3. A tabular approach:
the y-values are
less than or equal
to 50.
For these x-values
and less,
so, x ≤ 11.25
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4. Algebraic: 5 + 4 x ≤ 50
Graphic:
When is this line, y = 5 + 4 x ,
less than , or under, this
line, y = 50 ?
In other words, for
what x-values is the
diagonal line under
the horizontal line?
Answer:
for these x-values
4 x ≤ 45
x ≤ 11.25
11.25
Note that the graphing calculator makes no distinction between inequalities
and strict inequalities. Participants will use the same graph on the graphing
calculator to find or confirm solutions for both inequalities and strict
inequalities.
5. 15 + 3 x
6. 5 + 4 x ≤ 15 + 3 x
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7. A sample solution.
Algebraic:
5 + 4 x ≤ 15 + 3 x
Graphic:
5 + x ≤ 15
4 x ≤ 10 + 3 x
x ≤ 10
x ≤ 10
10
Discuss how using multiple representations for solving linear inequalities
makes connections and develops understanding.
• How does the solution to a linear equation differ from the solution to a
linear inequality? [The solution to a linear equation is one value, one
location on the number line. The solution to a linear inequality is an
infinite set of values.]
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Extension: Boolean Algebra and Inequalities
Many graphing calculators have the capability to use Boolean algebra on the
home screen and/or with graphing to support solution sets of linear
inequalities.
On the home screen, inequalities produce either a True = 1 or a False = 0 as
follows.
You can demonstrate what happens when you add to or multiply both sides of
an inequality as follows:
Using the Boolean and graphing features, have participant investigate
4 x + 5 ≤ 50 , first on the home screen as follows:
You can also investigate 4 x + 5 ≤ 50 by graphing. The calculator graphs
y = 1 when 4 x + 5 ≤ 50 and it graphs y = 0 when 4 x + 5 ≤ 50 .
•
What are the implications for assessment when students understand
how to graph solution sets as shown above?
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Activity 2: Linear Inequalities in Two Variables
Show participants the transparencies of the Student Activity: Age Estimates
from 2.1.1 The Linear Parent Function.
• What were the big ideas in the activity? [Guessed famous people’s
ages, graphed the scatter plot of (guess, actual age), discussed the line
y = x , over-guessing, and under-guessing.]
Have participants work on Exercises 1 – 4. Circulate and ask guiding
questions. Discuss results.
1. All ordered pairs representing over-guesses are under and not including
the line y = x and do not include the line itself.
2. Represent the shaded set with y < x .
• How can you graph y < x on your graphing calculator? [Help
participants with individual calculators.]
3. All of the ordered pairs representing under-guesses are above, not
including, the line y = x .
4. Represent the shaded set with y > x .
• How can you graph y > x on your graphing calculator? [Help
participants with individual calculators.]
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Work through the rest of the activity with participants. Discuss the scenario,
demonstrating free-throws, 2-point field goals, and 3-point field goals, if
necessary. Emphasize that the situation will only consider 2- and 3-point
shots and the total combined score for Friday’s game.
5. Ask participants to fill in the table. After participants have filled in most
of their table, ask for some suggestions from the group and write them on
the Activity transparency. Make sure to include some examples that give
a combined score of more than 63.
Sample entries
Score for 2-point Score for 3-point
Total Score
Shots
Shots
24(2)
5(3)
63
20(2)
10(3)
70
0(2)
21(3)
63
31(2)
1(3)
63
40(2)
15(3)
125
•
•
How did you get the total score? [Multiply 2 times the number of 2point shots and add to the product of 3 times the number of 3-point
shots, 2 x + 3 y .]
How did you know if your choices for numbers of shots fit the
problem situation ? [If the total score was greater than or equal to 63.]
6. 2 x + 3 y ≥ 63
2
7. a. y ≥ 21 − x
3
2
x
3
c. Use a friendly window for your calculator so that cursor has integer
values.
b. Graph y = 21 −
Each ordered pair listed in the table lies above the line y = 21 −
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x.
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284
d. Randomly choose some non-integer points above and below the line.
Use mental math to decide if the point satisfies the inequality or check
on the home screen as shown:
This ordered pair works because 82.5 > 63.
This is a non-example because 39.5 < 63.
This ordered pair works because 71.5 > 63.
Again, each ordered pair that works lies above the line
2
y = 21 − x .
3
e. To satisfy both the problem situation and the inequality, the points
2
must lie above the line y = 21 − x .
3
f. Help participants with individual calculators.
g. The solution set for y ≥ 21 −
2
x is the set of ordered pairs that lie
3
2
x.
3
h. Use the free-floating cursor to demonstrate that for that specific xvalue, the y-values of the ordered pairs in the solution set are all
greater than the y-value of the line (Use trace to get the y-value of the
line.)
above the line y = 21 −
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8.
a. The solution set for y ≥ 21 −
2
x is the set of ordered pairs that lie
3
2
x.
3
b. Then use the free-floating cursor to demonstrate that for that specific
x-value, the y-values of the ordered pairs in the solution set are all
greater than the y-value of the line (Use trace to get the y-value of the
line.)
above the line y = 21 −
Have participants look for general descriptions of the solution(s) for the
problems, not specific solution(s). See the answers for below for examples.
1. 13 = 3 x − 5 : The solution to a linear equation in one variable is x = a ,
which is represented graphically by one location on the number line.
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2. 13 > 3 x − 5 : The solution to this strict linear inequality in one variable is
x < a , which is represented graphically by a subset of the number line.
The solution is all x-values where the line y = 13 is above the line
y = 3x − 5 .
3. 13 < 3 x − 5 : The solution to this strict linear inequality in one variable is
x > a , which is represented graphically by a subset of the number line.
The solution is all x-values where the line y = 13 is under the line
y = 3x − 5 .
4. y > 3 x − 5 : The solution to this strict linear inequality in two variables is
the infinite set of ordered pairs that comprise the section of the plane
above the line y = 3 x − 5 .
5. y < 3 x − 5 : The solution to this strict linear inequality in two variables is
the infinite set of ordered pairs that comprise the section of the plane
below the line y = 3 x − 5 .
Summary:
Just as linear functions and linear equations can be developed using problem
situations from life, so can linear inequalities. Linear inequalities in one
variable have solutions in one dimension, which are sets of numbers, and
linear inequalities in two variables have solutions in two dimensions, which
are sets of ordered pairs.
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3.3 Solving Linear Inequalities: Activity 1
Activity 1: Linear Inequalities in One Variable
Bianca and Joe are starting their own pet grooming
business called Bianca and Joe’s. They have
figured that they can spend no more than $50 a
month on flea shampoo.
Joe has found a local dealer of pet shampoo, The Pet
Pantry, who sells quart bottles for $4.00 a bottle plus
a $5.00 handling fee per order.
1. Write an expression that represents the amount of money
charged by The Pet Pantry for an order of shampoo.
2. Write an inequality that represents the amount Bianca and
Joe’s is willing to pay per month for The Pet Pantry’s
shampoo.
3. Solve the inequality using the table on your calculator.
4. Solve the inequality algebraically step by step. Next to each
step, show the graphical solution.
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Bianca is wondering if they can save money by shopping
around. She found another local dealer, The Canine Corner,
who sells the shampoo for $3.00 a quart bottle plus a $15.00
handling charge per order.
5. Write an expression that represents the amount of money
charged by The Canine Corner for an order of shampoo.
6. Write an inequality that represents when it is less expensive
to buy an order with The Pet Pantry compared to The Canine
Corner.
7. Solve the inequality algebraically step by step. Next to each
step, show the graphical solution.
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Activity 2: Linear Inequalities in Two Variables
Recall the Student Activity “Age Estimates” from 2.1.1 The
Linear Parent Function. You estimated the age of famous
people and graphed the scatter plot (my guess, actual age).
Actual
1. On the graph below, shade the region that contains the
ordered pairs representing the over-guesses possible for the
activity.
2. How can you represent the shaded set with an inequality?
Guess
Actual
3. On the graph below, shade the region that contains the
ordered pairs representing the under-guesses possible for the
activity.
4. How can you represent the shaded set with an inequality?
Guess
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The Stars basketball team never makes a free-throw.
Considering only 2-point and 3-point shots, what
possible combination of shots could they score in
Friday’s basketball game to meet or beat their season
average score of 63 points?
5. Investigate numerically some combinations of shots that
would meet or beat their season average of 63.
Score for 2-point
Shots
Score for 3-point
Shots
Total Score
6. Write an inequality in two variables that represents the
situation.
7. Solve the inequality graphically:
a. Solve for y.
b. On your graphing calculator graph the line
y = ____________ .
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c. Use the free-floating cursor to find the ordered pairs listed
in your table above. Where do these ordered pairs lie on
the graph?
d. Use the free floating cursor to find points that satisfy
2
y ≥ 21 − x and the problem situation. Where are these
3
points?
e. Use the free floating cursor to find other points that satisfy
2
y ≥ 21 − x , but not the problem situation. Where are
3
these points?
2
f. Shade on your graphing calculator y ≥ 21 − x . Sketch.
3
g. Use the word “over” or “under” to describe the solution
2
2
set for y ≥ 21 − x in terms of the line y = 21 − x .
3
3
h. For a specific x-value, how do the y-values of the ordered
pairs in the solution set compare to the y-value of the line?
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8. Suppose the problem situation had called for the inequality,
2
y ≤ 21 − x . Predict the graph and produce it on your
3
graphing calculator. Sketch.
a. Use the word “over” or “under” to describe the solution
2
2
set for y ≤ 21 − x in terms of the line y = 21 − x .
3
3
b. For a specific x-value, how do the y-values of the ordered
pairs in the solution set compare to the y-value on the
line?
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3.3 Solving Linear Inequalities: Reflect and Apply
Reflect and Apply
Describe the general solution(s) to the following. Do not solve
for specific solutions.
1. 13 = 3 x − 5
2. 13 > 3 x − 5
3. 13 < 3 x − 5
4. y > 3 x − 5
5. y < 3 x − 5
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3.4 Systems of Linear Equations and Inequalities: Leaders’ Notes
3.4 Systems of Linear Equations and Inequalities
Overview:
Participants use a table to develop a system of linear inequalities. They solve
the system using various methods and make connections between a system of
inequalities and a system of equations.
Objective:
Algebra I TEKS
(c.4.A) The student analyzes situations and formulates systems of linear
equations to solve problems.
(c.4.B) The student solves systems of linear equations using concrete models,
graphs, tables, and algebraic methods.
(c.4.C) For given contexts, the student interprets and determines the
reasonableness of solutions to systems of linear equations.
Terms:
system of linear equations, system of linear inequalities
Materials:
Procedures:
Note: Systems of linear inequalities is not an Algebra I TEKS , but is an
Algebra II TEKS. We use a system of inequalities in this activity to stretch
teacher understanding. Using a situation that describes a system of linear
inequalities lends itself well to developing the system in a table.
Activity 1: Using a Table
Describe the scenario of the scout group going to the movies. Have
participants fill in a few rows of the table. Some participants may interpret
the problem to mean that exactly 10 people will go to the show or that they
must spend exactly $45. Clarify that they can take any number of people up
to and including 10 as long as they spend $45 or less.
•
What factors might the troop consider when deciding how many adults
and children will go to the movies? [Sample Answers. The troop may
want to consider the best adult-child ratio, or they may want the most
number of children to go to the movie. Perhaps the adults want to get
out of the heat and see the movie.]
Ask for some examples and fill in the table on the transparency. Ask for at
least one example that someone tried that did not work. Record it and draw a
line through it as shown below.
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Number
of Adults
2
5
1
2
3
x
Number of
children
4
5
9
8
7
y
Total number
of people
6
10
10
10
10
x + y ≤ 10
Cost for
Adults
2(6)
5(6)
1(6)
2(6)
3(6)
6x
Cost for
Children
4(3.50)
5(3.50)
9(3.50)
8(3.50)
7(3.50)
3.5y
Total Cost
26
47.50
37.50
40
42.50
6 x + 3.5 y ≤ 45
Ask participants to look for patterns in the table. Help them use their patterns
to develop the inequalities in the last row of the table, using the questions
below.
• How did you find the total number of people? [Add the number of
adults and the number of children.]
• How can we write this symbolically? [If x is the number of adults (fill
in the first box with x) and y is the number of children (fill in the
second box with y) then the total number of people is x + y ≤ 10 .]
• How did you find the cost for the adults? [Multiply the number of
adults times 6, or 6x .]
• How did you find the cost for the children? [Multiply the number of
children times 3.5, or 3.5 y .]
• How did you find the total cost? [Add the cost for the adults, 6x, and
the cost for the children , 3.5y.]
• How did you know if that met the requirement? [It had to be less than
$45.]
• How can you write this as an inequality? [The sum of the cost for the
adults, 6x, and the cost for the children, 3.5y, must be less than 45,
6 x + 3.5 y ≤ 45 .]
Have participants write the system of inequalities below the table.
x + y ≤ 10
6 x + 3.5 y ≤ 45
Activity 2: Solve the System Graphically
Lead the participants with the overhead calculator through the activity using
the following suggestions.
1. To graph x + y ≤ 10 , solve for y and graph y = 10 − x .
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Use the free floating cursor to find points that satisfy y ≤ 10 − x . Look for
points under the line y = 10 − x . Use mental math to check that sum of
the x-coordinate and the y-coordinate is less than or equal to 10.
• Where do the solutions to the equality, y = 10 − x , exist? [The
solutions are the ordered pairs, the set of which is the line y = 10 − x .
In other words, the solutions lie on the line y = 10 − x .]
• How can you verify your answer? [Trace to points on the line to
confirm that the sum of the x-coordinate and the y-coordinate is equal
to 10.]
Move the cursor around and discuss.
•
Which of the above satisfy the problem situation? [Only the first
ordered pair found above. The second screen, which shows the
ordered pair (1, 5.2) satisfies the inequality but not the problem
situation as you cannot take 5.2 people to the movie. The third screen,
which shows the ordered pair (-2.2, 2.2) again satisfies the inequality,
but not the problem situation as you cannot have negative people.]
Shade y ≤ 10 − x
•
What do the shaded points represent? [Each ordered pair represents
(number of adults, number of children) such that no more than 10
people go to the movie.]
2. To graph 6 x + 3.5 y = 45 , solve for y and graph y =
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6
−
x.
3.5 3.5
297
Use the free floating cursor to find points that satisfy y ≤
Look for points under the line y =
•
45
6
−
x.
3.5 3.5
Which of the above satisfy the problem situation? [Only the first
ordered pair found above because 5 adults and 2 children satisfy both
conditions of having less than 10 people and costing less than $45.
The second screen, which shows the ordered pair (4, 0.6) satisfies the
inequality but not the problem situation as you cannot take 0.6 people
to the movie. The third screen, which shows the ordered pair 4, -1)
does satisfy the inequality, but not the problem situation as you cannot
have negative people.]
Shade for y ≤
•
45
6
−
x.
3.5 3.5
45
6
−
x
3.5 3.5
What do the shaded points represent? [Each ordered pair represents
(number of adults, number of children) such that the total cost is no
more than $45.]
3. Graph the
system of
equations and
the solution.
•
What does this solution represent? [The solution (4, 6) represents the
number of adults, 4, and the number of children, 6, to go to the movie
such that the price will be exactly $45.]
Graph the system of inequalities. The solution is the double shaded
section.
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•
•
Where is the solution to the system of inequalities? [The
double shaded region.]
What does the solution to the system of inequalities, the points
in the double shaded section, represent? [The solution (x, y)
represents the number of adults, x, and the number of children,
y, to go to the movie such that no more than 10 people will go
and the price will be no more than $45.]
Use the free floating cursor to discuss various points as follows:
• Do the points satisfy the system and/ or the problem situation and
why?
This point (2, 5) means that
2 adults and 5 children go. It
satisfies the system and the
problem situation because
2 + 5 < 10 and
2(6) + 5(3.5) < 45 .
This point (1, 10) means that
1 adult and 10 children go. It
does not satisfy the problem
situation because you cannot
take 11 people. It does not
satisfy the system because
1 + 10 < 10 .
This point (-3, 2) does not
satisfy the problem situation
because you cannot have a
negative number of adults. It
does satisfy the system
because −3 + 2 < 10 and
−3(6) + 2(3.5) < 45 .
This point (2, -1) does not
satisfy the problem situation
because you cannot have a
negative number of scouts. It
does satisfy the system
because 2 + ( −1) < 10 and
2(6) + ( −1)(3.5) < 45 .
•
What is the relationship between the sets, the set that satisfies the
problem situation and the set that is the solution to the system of
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inequalities? [All of the points that satisfy the problem situation
are in the set that is the solution to the system of inequalities. The
solution to the system of inequalities contains many other points.
The solution to the system includes points in the second and third
quadrants, none of which satisfy the problem situation. Also the
solution to the system includes all of the non-integer ordered pairs
bounded by the two inequalities. The problem situation includes
only natural number values in the ordered pairs.]
Activity 3: Solve the System Symbolically
Have participants solve the system using any algebraic methods they choose.
Ask participant to consider the following question while they are solving.
• How does the algebraic solution to the system of equations relate to
the graphical solution of the system of inequalities?
1.
Solve using substitution
x + y = 10
6 x + 3.5 y = 45
Solve for y: y = 10 − x
Substitute: 6 x + 3.5(10 − x ) = 45
Solve for x: 6 x + 35 − 3.5 x = 45
2.5 x = 10
x=4
Substitute to find y: y = 10 − 4
y=6
The solution is (4, 6)
Solve using linear combination
(elimination)
x + y = 10
6 x + 3.5 y = 45
Multiply −6( x + y = 10)
Add to 6 x + 3.5 y = 45
− 6 x − 6 y = −60
+ 6 x + 3.5 y = 45
− 2.5 y = −15
y=6
Substitute to find x: x = 10 − 6
x=4
The solution is (4, 6)
2. The solution to the system of linear equations is an ordered pair, the
intersection of the two lines. The solution is written as an ordered pair
(x, y).
The solution to the system of linear inequalities is a set of ordered pairs.
The solution is drawn as a graph that represents all of the ordered pairs
contained in the intersection of the shaded regions.
Explain to participants that we are looking for general descriptions of the
solution(s) for the problems, not specific solution(s). See the answers for
below for examples.
1. 13 = 3 x − 5 : The solution to a linear equation in one variable is x = a ,
which is represented graphically by one location on the number line.
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2. 13 > 3 x − 5 : The solution to this strict linear inequality in one variable is
x < a , which is represented graphically by a subset of the number line.
The solution is all of the x-values where the line y = 13 is above the line
y = 3x − 5 .
3. y > 3 x − 5 : The solution to this strict linear inequality in two variables is
the set of ordered pairs that comprise the section of the plane above the
line y = 3 x − 5 .
4. 3 x − y = 13
x + y = 11 :
The solution to a system of linear equations can be the intersection point if
the lines intersect, a line if the equations of the lines represent the same
line, or there can be no solution if the lines are parallel.
5. 3 x − y > 13
x + y < 11:
The solution to a system of linear inequalities is the intersection of the two
inequalities. It is a set of ordered pairs if the intersection set is not empty.
It is represented by the double shaded area.
6. Answers will vary. An example:
Here is a question from a recent End of Course Algebra I test:
A truck is carrying 1500 pounds of cargo that occupies 138 cubic feet of
space. A television weighs 50 pounds and occupies a space of 4 cubic
feet. A microwave oven weighs 30 pounds and occupies a space of 3 cubic
feet. Which system of equations can be used to find the total number of
televisions, t, and microwaves, m, that are in the truck?
Changed to inequality problem:
A truck is carrying at most 1500 pounds of cargo that occupies at least
138 cubic feet of space. A television weighs 50 pounds and occupies a
space of 4 cubic feet. A microwave oven weighs 30 pounds and occupies a
space of 3 cubic feet. Which system of equations can be used to find the
total number of televisions, t, and microwaves, m, that are in the truck?
59t + 30 m ≤ 1500
4t + 3m ≥ 138
A good window: [0, 60] [0, 40]
Summary:
Building on the work of solving equations and inequalities, we end the section
with solving systems of equations and systems of inequalities. A table is used
to build the linear inequalities from the written situation. Solving the system
step by step graphically builds understanding of systems of equations and
inequalities. We make the distinction between the solution to a system of
linear equations as being an ordered pair, the intersection point, and the
solution to a system of linear inequalities as being a set of ordered pairs.
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3.4 Systems of Linear Equations and Inequalities: Activity 1
Activity 1: Using a Table
A local scout troop and leaders are going to a
movie.
• A maximum of 10 people can go.
• They can spend $45 or less for the admission price.
• The movie theater charges $6.00 per adult and $3.50 per
child.
Use the table to investigate possible combinations of people that
satisfy the conditions.
Number Number
Total
Cost for
of Adults
of
number
Adults
children of people
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Cost for
Children
Total
Cost
302
Activity 2: Solve the System Graphically
x + y ≤ 10
6 x + 3.5 y ≤ 45
1. Solve x + y ≤ 10 graphically.
2. Solve 6 x + 3.5 y ≤ 45 graphically.
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3. Consider the system
x + y ≤ 10
6 x + 3.5 y ≤ 45
• Graph the system of equations:
x + y = 10
6 x + 3.5 y = 45
• Find the solution to the system of equations graphically.
• Graph the solution to the system of inequalities.
• What part of the solution to the system of inequalities
makes sense in the problem situation?
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Activity 3: Solve the System Symbolically
1. Solve the system of equations symbolically using at least 2
different methods.
x + y = 10
6 x + 3.5 y = 45
2. How does the solution for the system of equations compare
to the solution for the system of inequalities?
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3.4 Systems of Linear Equations and Inequalities: Reflect and Apply
Reflect and Apply
Describe in general the solution(s) to the following.
1. 13 = 3 x − 5
2. 13 > 3 x − 5
3. y > 3 x − 5
4. 3 x − y = 13
x + y = 11
5. 3 x − y > 13
x + y < 11
6. Create a problem that results in a system of inequalities.
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3.4 Systems of Linear Equations and Inequalities: Student Activity
Student Activity: Concrete Models and Systems of Linear
Equations
Overview:
Students use concrete models to solve a system of linear equations.
Objective:
Algebra I TEKS
(c.4.B) The student solves systems of linear equations using concrete
models, graphs, tables, and algebraic methods.
Terms:
system of linear equations
Materials:
algebra tiles
Procedures:
Students should be seated with plenty of elbow room to work with the algebra
tiles.
Work through the following example with students.
Explain that you need to choose one tile shape to represent the variable x, a
different tile shape to represent the variable y, and a different tile shape to
represent one unit.
x=
y=
unit =
Example:
3x + 4 y = 2
x − 4y = 6
3x + 4 y = 2
x − 4y = 6
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Add the two models together to eliminate the y terms.
3x + 4 y = 2
+( x − 4 y = 6)
4x = 8
x=2
Replace x with 2 in one of the original equations.
3x + 4 y = 2 ,
3(2) + 4 y = 2
Solve for y.
6 + 4y = 2
(−6) + 6 + 4 y
= 2 + ( −6)
4 y = −4
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y = −1
So the
solution is
x = 2, y = −1
Now solve the system by substitution.
3x + 4 y = 2
x − 4y = 6
Solve for x.
x − 4y + 4y =
6 + 4y
x = 6 + 4y
Substitute 6 + 4 y for x in 3 x + 4 y = 2 .
3(6 + 4 y) + 4 y
=2
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16 y + 18 + ( −18) =
2 + ( −18)
16 y = −16
y = −1
Substitute y = −1 in 3 x + 4 y = 2 .
3 x + 4( −1) = 2
3x − 4 = 2
3x − 4 + 4 =
2+4
3x = 6
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x=2
So the
solution is
x = 2, y = −1
Have students complete the student activity and discuss.
Activity: Concrete Models and Systems
1. An example of solving the system:
x + 2y = 1
− x + 3y = 9
Add the two models together to eliminate the x terms.
x + 2y = 1
+ ( − x + 3 y = 9)
5 y = 10
y=2
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Replace y with 2 units in one of the original equations.
x + 2 y = 1,
x + 2( 2 ) = 1
Solve for x.
x + 4 =1
x + 4 + ( −4) =
1 + ( −4)
x = −3
So the
solution is
x = −3, y = 2
2 x + 3y = 1
− x + 2 y = −4
In this case, the student solves by linear combination.
2.
2 x + 3y = 1
− x + 2 y = −4
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Multiply − x + 2 y = −4 by 2.
2( − x + 2 y = −4 )
−2 x + 4 y = −8
Add the two models together to eliminate the x terms.
2 x + 3y = 1
+( −2 x + 4 y = −8)
7 y = −7
y = −1
Replace y with -1 in one of the original equations.
2 x + 3 y = 1,
2 x + 3( −1) = 1
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Solve for x.
2x − 3 = 1
2x − 3 + 3 =
1+ 3
2x = 4
x=2
So the
solution is
x = 2, y = −1
2 x + 3y = 1
− x + 2 y = −4
In this case, the student solves by substitution.
2.
2 x + 3y = 1
− x + 2 y = −4
Solve − x + 2 y = −4 for x.
−x + x + 2y + 4 =
−4 + x + 4
2y + 4 = x
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Substitute 2 y + 4 for x in 2 x + 3 y = 1.
2(2 y + 4) +
3y = 1
7y + 8 = 1
7 y + 8 + ( −8)
= 1 + ( −8)
7 y = −7
so
y = −1
Replace y with –1 in 2 x + 3 y = 1
2 x + 3( −1) = 1
2x − 3 = 1
2x − 3 + 3 = 1 + 3
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2x = 4
x=2
So the
solution is
x = 2, y = −1
•
Summary:
How does solving with tiles help you understand the algebraic steps?
Using concrete models to solve systems of linear equations helps students
understand and make connections when using the linear combination
(elimination) algebraic solution method.
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Student Activity: Concrete Models and Systems of Linear
Equations
1. Solve the following system using algebra. Sketch each step
and write the algebraic representation for each step.
3x + 4 y = 2
x − 4y = 6
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2. Solve the following system using algebra tiles in two ways,
by substitution and by linear combination. Sketch each step
and write the algebraic representation for each step.
2 x + 3y = 1
− x + 2 y = −4
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1.1 Quadratic Relationships: Leaders’ Notes
1.1 Quadratic Relationships
Overview:
Participants use lists to develop a quadratic function representing the volume
of a sandbox with a fixed depth. Using the quadratic function, participants
solve quadratic equations numerically and graphically.
Objective:
Algebra I TEKS
(d.2) The student understands there is more than one way to solve a quadratic
equation and solves them using appropriate methods.
(d.2.B) The student relates the solutions of quadratic equations to the roots of
their functions.
Terms:
quadratic function, zero of a function, root of a function, solution of an
equation
Materials:
graphing calculators, pieces of lumber or cardboard to simulate lumber
Procedures:
Activity 1: Building a Sandbox
Work through the activity with participants, using the overhead graphing
calculator to demonstrate. Begin by discussing the situation of building a
rectangular sandbox. Use the 1 foot wide lumber or cardboard to simulate a
sandbox.
1. Have participants roughly sketch some possible sandboxes from a bird’s
eye view. Examples:
14
8
1
7
11
4
•
•
•
•
•
What is fixed in this situation? [Two things are fixed. The depth of
the sandbox is 1 foot deep and the perimeter of the sandbox is 30.]
How will you fill in the depth column? [The depth is fixed. It will
always be 1 foot.}
If the perimeter is 30, what kind of widths make sense for the
situation? [Widths ranging from more than 0 feet to less than 15 feet.]
How does the length relate to the width? [The length is always
15 − width .]
How does the volume relate to the width and length? [The volume is
the product of the width, length, and depth, v = w ∗ l ∗ d . In this case
since the depth is always 1, v = w ∗ l ∗ 1 = w ∗ l ]
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2. Sample dimensions:
Width
Length
2
13
4
11
6
9
8
7
10
5
12
3
14
1
Depth
1
1
1
1
1
1
1
Volume
26
44
54
56
50
36
14
3. Encourage participants to predict the general shape of the graph.
4. Use the following to help participants enter the data into lists.
Ask participants to put the table values for width into a list in the
calculator.
• How is the length related to the width? [The sum of the width and the
length is always 15.]
• How can you use an expression for the length in terms of the width to
fill in the lengths into the list in your calculator. [ 15 − list 1.]
• What is the depth of each sandbox? [Depth is always fixed at 1 foot.]
• What expression can you use for volume? [ (list 1)(list 2) • 1.]
Sample scatter plot:
Note: Some calculators allow you to name lists. For this situation, you
could name lists WIDTH, LENTH, and VOLUM.
5.
(list 1)(list 2) • 1
6. Sample:
Extension: Ask participants to predict how the situation and the graph would
change if the depth of the sandbox is 1.5 feet instead of 1 foot. Change the
volume function to be V = x(15 − x ) * 1.5 and graph. How would the situation
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and the graph change if the depth of the sandbox is 0.75 feet. Again change
the volume function to be V = x(15 − x ) * 0.75 and graph. This is an
introduction to transformations, which will be explored in depth in 1.2
Transformations.
Note: The purpose of the following questions, solving quadratic equations, is
to familiarize participants with the different types of equations that arise from
quadratic functions. One of the common struggles that students have is
differentiating between questions that ask for an input value and questions that
ask for an output value. We use some non-algebraic solution methods here to
introduce quadratic equations. We want to build confidence in reading
questions and solving equations with power of technology and students will
then be able to solve symbolically with more understanding.
Briefly discuss each of the solution methods shown below, emphasizing the
power of multiple representations in promoting understanding.
Graph:
Table:
Home screen:
Using function notation on
the home screen:
7. Solution:
V = 38.1875 ft 3
•
Does this question give an input value and ask for an output value or
does the question give an output value and ask for an input value?
[The question gives an input value, 3.25 feet, and asks for an output
value, the volume. You could also use the terms domain and range in
this question and answer.]
8. Solution: width = 7.5 ft
You can do some work on the home screen to get a feel for the answer:
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Graph, using trace to get an Graph, using the calculator
approximation:
to find the maximum value:
Table, looking for the x-value that yields the highest yvalue:
•
•
[The question gives an output value, the maximum volume, and asks
for an input value, the width.]
Where do we usually see a maximum question like this?
[Traditionally, maximum and minimum problems have usually been
reserved for calculus, but can readily be examined using technology in
earlier courses. This kind of question naturally arises when studying
quadratic functions.]
9. Solution:
width = 5 ft,10 ft
Graph, using trace to get an Graph, using trace to get an
approximation for one
approximation for the other
solution:
solution:
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Table:
Graph, using the calculator
to get an exact answer for
one solution, by finding the
intersection of
y = x(15 − x ) • 1 and
y = 50 :
to find the other solution,
by finding the intersection
of y = x(15 − x ) • 1 and
y = 50 :
to get an exact answer for
one solution, by finding the
zero (root) of
y = x(15 − x ) • 1 − 50 :
to find one solution, by
finding the (root) zero of
y = x(15 − x ) • 1 − 50 :
“Trace” to the x-value where the volume is 50. Two solutions.
Find the intersection where y1 = y2 . Two solutions.
Find the zero (root) of y1 − y2 = 0 . Two solutions.
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•
[The question gives an output value, 50 ft3, and asks for an input value,
the width.]
An important discussion to have with participants is to compare the 3
different table methods and 3 different graph methods – trace, intersection,
zero.
• How does solving quadratic equations in many ways add to your
understanding?
• Why might one be more inclined to use the zero method when solving
a quadratic equation and not a linear equation. [Linear equations are
solved algebraically by getting all of the variables on one side and the
numbers on the other side and solving for the variable. Quadratic
equations are often solved algebraically by getting everything on one
side equal to zero and then solving using factoring, completing the
square, or with the quadratic equation.]
Note: When solving linear equations, there is one solution. Often students
mistakenly find only one solution to a quadratic equation when solving
symbolically. Now with a picture in their heads of a quadratic equation being
a parabola intersecting a line, they will be more apt to consider how many
solutions they are looking for.
• If you think of the solution to a quadratic equation as the intersection
between a parabola and a line, how many solutions are possible?
Make a sketch of each to justify your answer. [Two, one, or no real
solutions.]
Two solutions
*
•
One solution
No real solutions
*
*
If you use the “zero” method, that is setting everything equal to zero
and solving, how will your sketch of the possible solutions change?
[The sketch is essentially the same except the line is now the x-axis,
y = 0 .]
Note: In this problem, we used a quadratic function to represent the volume
of a sandbox as the width and length varied with a fixed perimeter and a fixed
depth. The units of measure of volume are cubic, in this case, cubic feet.
Some discussion may arise that area can be modeled with a quadratic function
and the unit of measure is square units, while volume can be modeled with a
cubic function, the unit of measure is cubic units. In the particular scenario of
the sandbox, the volume can be modeled by a quadratic function because one
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of the three dimensions, depth, is a fixed quantity. Hence,
V = l ⋅ w ⋅ d = x(15 − x ) ⋅ 1, which is a quadratic function.
Activity 2: Projectile Motion
Have participants work through the activity, using the following to discuss.
1.
•
What are a reasonable domain and range for the situation? [See the
above window.]
Graph:
Table:
Home screen:
Using function notation on
the home screen:
2. Solution:
h = 96 ft
•
[The question gives an input value, 2 feet. and asks for an output
value, the height.]
3. Solution:
width = 1 ft, 4 ft
Graph, using trace to get an Graph, using trace to get an
approximation for one
approximation for the other
solution:
solution:
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Table:
finding the intersection of
y = −16 x 2 + 80 x and
y = 64 :
by finding the intersection
of y = −16 x 2 + 80 x and
y = 64 :
finding the zero (root) of
y = −16 x 2 + 80 x − 64 :
by finding the zero (root)
of y = −16 x 2 + 80 x − 64 :
“Trace” to the x-value where the volume is 64. Two solutions.
Find the intersection where y1 = y2 . Two solutions.
Find the zero (root) of y1 − y2 = 0 . Two solutions.
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•
[The question gives an output value, 64 ft, and asks for an input value,
the time.]
•
How does solving quadratic equations in many ways add to your
understanding?
Graph:
Table:
4. Solution:
t = 5 sec
•
[The question gives an output value, 0 feet. and asks for an input
value, the time.]
Graph, using trace to get an Graph, using the calculator
approximation:
to find the maximum value:
5. Solution:
height = 100 ft
Table, looking for the x-value that yields the highest yvalue:
6. The big idea of this question is to use the symmetry of a parabola to find
the vertex. Once you know the roots of a parabola, you can find the xcoordinate of the vertex by finding the average of the roots. Then you can
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find the y-coordinate of the vertex by substituting the x-coordinate into the
equation.
2
1. a. h(2.5) = −16(2.5) + 64(2.5) , h = 60. Participants may also answer,
−16 x 2 + 64 x = 60 , x = 2.5. If they do, discuss that there is another
solution to the equation in addition to the one that is shown in the graph.
b. −16 x 2 + 64 x = 48 , x = 1, 3
c. −16 x 2 + 64 x = 60 , x = 1.5, 2.5
ft
2. v0 = 96
sec
Summary:
Using the natural quadratic relationship of volume where the depth is fixed,
participants build a quadratic function. They solve arising quadratic equations
in several non-algebraic ways, making connections and building
generalizations. Participants further their study by solving equations that arise
from a projectile motion situation.
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1.1 Quadratic Relationships: Activity 1
Activity 1: Building a Sandbox
The Cano family is building a rectangular sandbox one foot
deep. Diana has decided to use lumber that is one foot wide.
She collected 30 feet of lumber to enclose the sandbox.
1. Sketch a few possible sandboxes.
2. Fill in the table with some possible dimensions:
Width
Length
Depth
Volume
3. Predict: what do you think a scatter plot of (width, volume)
will look like?
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4. Enter the table values into lists in your calculator, using
expressions where appropriate. Create a scatter plot of
(width, volume) in your calculator in an appropriate window
and sketch:
5. Write the expression in the lists that you used for volume.
6. Enter the expression for volume into the function grapher.
Sketch the graph over the scatter plot above.
Using your function for volume:
7. If the width of the sandbox is 3.25 feet, find the volume of
sand necessary to fill the box. Solve, using your calculator:
Graphically
With a Table
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8. What dimensions would allow for the greatest volume of
sand? Solve, using your calculator:
Graphically
With a Table
9. The family decides they can afford to buy 50 ft3 of sand.
What dimensions should they build the sandbox? Solve,
using your calculator:
Graphically
With a Table
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Activity 2: Projectile Motion
It can be shown that after being thrown straight up into the air
with a velocity of 80 ft/sec, a ball’s height t seconds after being
thrown can be represented by h = −16t 2 + 80t (ignoring air
resistance).
1. Find an appropriate viewing window for h = −16t 2 + 80t for
this problem situation. Sketch the graph. Justify your
window choice.
2. How high is the ball after 2 seconds?
Graphically
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332
3. When was the ball 64 feet above the ground?
Graphically
With a Table
4. When did the ball hit the ground?
Graphically
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5. What is the maximum height that the ball reached?
Graphically
With a Table
6. The ball was thrown from a height of 0 ft. In Exercise 4, you
found that the ball hit the ground, height = 0, at ______ sec.
a. Based on this information, how can you find the time at
which the ball reached its maximum height? Explain your
strategy.
b. Evaluate the function to find the maximum height.
c. What is this point (time, maximum height) called on the
parabola, h = −16t 2 + 80t ?
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1.1 Quadratic Relationships: Reflect and Apply
Reflect and Apply
1. Write an equation and its solution for the following screens.
a. __________________
equation
__________________
solution
b. __________________
equation
__________________
solution
c. __________________
equation
__________________
solution
2. The following equation represents the height of an object
after t seconds when thrown straight up from the ground:
h = −16t 2 + v0 t . At what initial velocity, v0 , would you have
to throw the ball to get it to a maximum height of 144 feet?
(Hint: Use your graphing calculator to find v0 such that
maximum height the ball reaches is about 144 feet. Graph
h = −16t 2 + v0 t in the window [0, 6] [0, 150], guessing and
checking values for v0 until the maximum height is 144 feet.)
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1.2 Transformations: Leaders’ Notes
1.2 Transformations
Overview:
Participants investigate the effects of changing the parameters of quadratic
function of the form y = ax 2 + c . They apply this understanding by fitting a
quadratic to real data. Participants extend their understanding and investigate
the effects of changing the parameter h in quadratic functions of the form
2
y = ( x − h) .
Objective:
Algebra I TEKS
(d.1.B) The student investigates, describes, and predicts the effects of
changes in a on the graph of y = ax 2 .
(d.1.C) The student investigates, describes, and predicts the effects of
changes in c on the graph of y = x 2 + c .
Terms:
parameter, transformation, scale factor, translation
Materials:
graphing calculators, patty paper
Procedures:
Transformations of functions is an important concept to aid students in
graphing various functions and understanding the behavior of various
functions. In these activities, participants explore the effects of changing
parameters of quadratic functions. They use the power of graphing calculators
to find many examples quickly, make and check conjectures, and apply what
they have learned. Exploring transformations with parabolas is a natural
starting place as participants can watch the vertex “travel” around the
coordinate system. In later courses, students will apply the lessons learned to
other parent functions, and they will add other transformations to their
graphing toolkit.
Have participants work together in groups, comparing observations on
Activities 1 – 3. Discuss their answers to Exercise 5 for each activity. Also,
look at table values. See the notes for each activity for an example.
Activity 1: Investigating the Role of a
2.
1.
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3.
4.
5. The value a is a vertical scale factor. For a > 1, the parabola is vertically
stretched. As x increases, the y-values increase faster than for y = x 2 .
For 0 < a < 1 , the parabola is vertically compressed. As x increases,
the y-values decrease faster than for y = x 2 . For a < 0, the graph is a
reflection over the x-axis.
•
Does it change the shape of the graph? [For a = −1, the shape of the
graph does not change. It is a reflection over the x-axis. For a > 1, the
shape does change because the parabola is vertically stretched. For
0 < a < 1 , the shape also changes because the parabola is vertically
compressed.
Choose an Exercise and look at table values, both on the graphs and in the
table as shown. Use the questions below to discuss.
•
•
•
How do the y-values (function values) of y = 2 x 2 compare with those
of the parent function y = x 2 ? [The y-values are twice as much.]
How do the y-values (function values) of y = 0.5 x 2 compare with
those of the parent function y = x 2 ? [The y-values are half as much.]
Why did the vertex remain the same? [Any number times zero is still
zero, x ⋅ 0 = 0 .]
This process of looking at y-values to compare functions may seem
unnecessary because it is so obvious, but it lays important groundwork for
students. In later courses, students will be expected to discern when a
question is referring to function values (y-values) and when a question is
referring to x-values.
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Activity 2: Investigating the Role of c
1.
2.
3.
4.
[4.7, 4.7] [-250, 50] 15
WINDOW
5. For c > 0 , the graph is vertically translated (shifted) up c units. For c < 0 ,
the graph is vertically translated (shifted) down c units.
• Does the shape change? [No, vertical translations are shape preserving
transformations.]
Put a piece of patty paper over the graph for Exercise 1 and trace the parent
function y = x 2 . Slide or shift the patty paper up and down until the parent
function is directly over the translated functions to show that indeed the shape
does not change.
• Why did we not use patty paper to look at the transformed functions
y = ax 2 ? [Dilations are shape changing transformations. If you take
the parent function traced on the patty paper and try to make it “fit”
one of the stretched or compressed functions from Activity 1, it will
not work. The shapes are different.]
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•
•
•
How do the y-values (function values) of y = x 2 + 2 compare with
those of the parent function y = x 2 ? [The y-values are 2 more.]
How do the y-values (function values) of y = x 2 − 3 compare with
those of the parent function y = x 2 ? [The y-values are 3 less.]
Why did the vertex change? [ 0 + 2 = 2 , 0 − 3 = −3 .]
Activity 3: Investigating the Role of h
Horizontal translations are not listed in the Algebra I TEKS. This activity is
intended to enhance teachers understanding of transformations and is not
intended for an average algebra I class.
1.
2.
3.
4.
[-47, 47] [-31, 31]
WINDOW
5. For h > 0, the graph is horizontally translated (shifted) left h units. For
h < 0, the graph is horizontally translated (shifted) right h units.
• Does the shape change? [No, horizontal translations are shape
preserving transformations.]
function y = x 2 . Slide or shift the patty paper left and right until the parent
function is directly over the translated functions to show that indeed the shape
does not change.
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•
•
•
How do the y-values (function values) of y = x 2 + 2 compare with
those of the parent function y = x 2 ? [The y-values have all been
shifted up two in the table, left two on the graph.]
How do the y-values (function values) of y = x 2 − 1 compare with
those of the parent function y = x 2 ? [The y-values have all been
shifted down one in the table, right one on the graph.]
Why did the vertex change? [The y-value of zero has been shifted
too.]
Activity 4: Transformations
Change
Add 3 to the function
Multiply by 1/3
Multiply by 3
Replace x with ( x − 2)
Multiply by – 1
Subtract 2 from the
function
Replace x with ( x + 1)
Multiply by 2
New Equation
y = x2 + 3
1
y = x2
3
y = 3x 2
2
y = ( x − 2)
y = −x2
y = x2 − 2
Change in Graph
Vertical translation up 3
Scale change of 1/3
y = ( x + 1)
y = 2x2
2
y = ( x − 3)
Horizontal translation left 1
Scale change of 2
Horizontal translation right 3
2
Scale change of 3
Reflection over the x-axis
Vertical translation down 2
2. a. Horizontal translation left 5, vertical translation down 1
b. Scale change of 3, vertical translation up 2
c. Reflection across the x-axis, scale change of 1/3, horizontal translation
left 1.
2
3. a. y = 2( x − 1)
2
b. y = −( x + 2)
2
c. y = ( x − 3) − 2
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d. y = ( x + 3) + 1
2
4.
5.
6.
7.
8.
a.
b. As the elapsed time increases, the distance from the motion detector
increases.
c.
d. The choice for c is the minimum data value.
e.
•
What is the significance of a = 16 in this problem? [The formula for
the distance of an object dropped from an initial height of d0 at an
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1 2
at + d0 . The acceleration due to the
2
ft
1
force of gravity, a, is −32
, so d = ( −32)t 2 + d0 = −16t 2 + d0 .
2
sec
2
(The proof of this statement is dependent on an understanding of
derivatives that are studied in calculus.) The 16 in the problem
situation is positive because the motion detector is measuring the
distance from the motion detector to the book instead of the distance
from the book to the ground.
initial velocity of zero is d =
Extension:
Ask participants to find first and second differences for the data.
• What do the second differences imply about the choice of a quadratic
function for a model for the data? [Since second differences are
constant, the data can be modeled with a quadratic function.]
• How are the second differences and your value for a related? [The
second differences as shown below are 0.32 feet per 0.1 sec per 0.1
ft
ft
. The acceleration due to the force of gravity,
sec, 0.32 0.1sec = 32
sec 2
0.1sec
ft
the force pulling the book down to the ground, is −32
. The
sec 2
second difference, 32, is positive because the motion detector is
measuring the distance from the motion detector to the book instead of
the distance from the book to the ground.
Discuss with participants the term “appropriate viewing windows,” especially
with respect to the graph of quadratic functions.
• What should an appropriate viewing window show about a quadratic
function? [A complete graph]
• What is a complete graph of a quadratic function? [The window
would include x-intercepts, if any, the direction the parabola opens,
and the vertex. Some participants may want to show the y-intercept.]
Sample answers. Window may vary, but should show similar graphs.
1.
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2. A window
without the yintercept.
2. A window that
shows the yintercept.
3.
4. A window that
shows the yintercept.
4. A window
without the yintercept.
5.
6.
7.
8.
Summary:
y4, A
y3, D
y2, C
y1, B
9. y2, D
10. y1, A
11. y4, B
12. y3, C
Using technology to see many examples quickly, participants connect
transformations of quadratic functions with the vertex form of the equation of
2
a parabola y = a( x − h) + b . Participants use transformations to fit a
quadratic function to data, i.e. the distance of a dropped object from a motion
detector.
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1.2 Transformations: Activity 1
Activity 1: Investigating the Role of a
Sketch a graph of the following using a graphing calculator.
Your observations should include some table values.
1 2
2
2
2.
,
y
=
2
x
y
=
x , y = 0.2 x 2
1. Function: y = x
2
Observations:
Observations:
3. y = 5 x , y = − x
2
2
Observations:
1 2
x , y = −25 x 2
10
Observations:
4. y = −
5. In general, what effects do different values of a have on the
graph of y = ax 2 ?
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Activity 2: Investigating the Role of c
2. y = x 2 − 0.5 , y = x 2 − 1
1. y = x 2 , y = x 2 + 2 , y = x 2 + 3
Observations:
Observations:
3. y = x 2 + 1.5, y = x 2 − 2.5
Observations:
4. y = x 2 + 15 , y = x 2 − 200
Observations:
5. In general, what effects do different values of c have on the
graph of y = x 2 + c ?
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Activity 3: Investigating the Role of h
1 2
2
2
2
2

1. y = x , y = ( x + 2) , y = ( x + 1) 2. y = ( x − 1) , y = x −

2
Observations:
Observations:
3. y = ( x − 3)2 , y = ( x + 2)2
Observations:
4. y = ( x + 22)2 , y = ( x − 15)2
Observations:
5. In general, what is the effect on the graph of y = x 2 , when
you replace x with ( x + h) ?
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Activity 4: Transformations
1. Fill in the blanks.
Change from the
parent function,
y = x2
Add 3 to the function
New Equation
y = x2 + 3
Change in Graph
Vertical translation up 3
Scale change of 1/3
y = 3x 2
y = − x2
Vertical translation down 2
Horizontal translation left 1
Multiply by 2
2. Describe the transformations on y = x 2 that will produce the
graph for each function below. Verify on your calculator.
a. y = ( x + 5)2 − 1
b. y = 3 x 2 + 2
1
c. y = − ( x + 1)2
3
3. Write an equation for each graph below. Each graph is a
relative of the parent function y = x 2 (shown in bold).
a.
b.
c.
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Using your knowledge of transformations on the parent function
y = x 2 , graph the following relatives. Verify with your
calculator.
1
4. Function: y = 2( x + 3)2
5. Function: y = ( x − 2)2
2
Describe transformations:
6. Function: y = − x 2 − 1
7. Function: y = ( x + 2)2 + 1
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8. A book was dropped under a motion detector. The following
data was collected (elapsed time, distance from the motion
detector).
Elapsed Time
(sec)
0
0.1
0.2
0.3
0.4
0.5
Distance from the
Motion Detector
(feet)
1.38
1.54
2.02
2.82
3.94
5.38
a.
b.
c.
d.
Set up a scatter plot in an appropriate window.
Why is the graph increasing?
Graph y = x 2 over the scatter plot.
Choose a value for c and graph y = x 2 + c over the scatter
plot. Explain your choice for c.
e. Guess and check a value for a and graph y = ax 2 + c over
the scatter plot.
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1.2 Transformations: Reflect and Apply
Reflect and Apply
Using your knowledge of transformations on the parent function
y = x 2 , find an appropriate viewing window for the following
functions. Sketch each function in the window and note the
window.
1. y = − x 2 − 50
2. y = ( x − 2000)2 + 5000
3. y = 0.001x 2
4. y = −500( x + 5)2
Match the function with a table:
x
y1
y2
y3
y4 ___ 5.
-2
4
0.8
0.2
20
___ 6.
-1
1
0.2
0.05
5
___ 7.
0
0
0
0
0
___ 8.
1
1
0.2
0.05
5
2
4
0.8
0.2
20
x
-2
-1
0
1
2
y1
1
-2
-3
-2
1
y2
5
2
1
2
5
y3
4.5
1.5
0.5
1.5
4.5
y4
2
-1
-2
-1
2
Match with a graph:
___
y = 5x
A
y = 0.05 x 2 ___
B
2
C
y = 0.2 x ___
D
2
___
y=x
2
____ 9. y = x 2 + 1 ___
____ 10. y = x 2 − 3 ___
____ 11. y = x 2 − 2 ___
1
____ 12. y = x 2 + ___
2
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C
B
A
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1.3 Lines Do It Too: Leaders’ Notes
1.3 Lines Do It Too
Overview:
Participants connect their knowledge of transformations with quadratic
functions with the equations of lines. The point-slope form of a line is looked
at from a transformational perspective.
Objective:
Algebra I TEKS
(c.2.C) The student investigates, describes, and predicts the effects of changes
in m and b on the graph of y = mx + b .
Terms:
transformation, translation, dilation, increasing function, decreasing function,
rate of change, slope, y-intercept
Materials:
graphing calculators, patty paper or blank transparencies
Procedures:
In these activities, participants explore the effects of changing parameters of
linear functions. We have done similar work in previous activities in the
institute, but always in context at more concrete level. Here participants move
out of context to a more abstract level, using the power of graphing calculators
to find many examples quickly, to make and check conjectures, and to apply
what they have learned. After their work with transformations of parabolas,
participants can connect those lessons with quadratic functions now to linear
functions. In later courses, students will apply transformations to other parent
functions, and they will add other transformations to their graphing toolkit.
Activity 1: Exploring Slope
Have participants work through the Exercises. Then discuss Exercise 5.
5. Some generalizations about rate of change, a, in y = ax : The larger a is,
steeper the line is and the higher the rate is. The smaller a is, the more
shallow the line is and the lower the rate is. If a is positive, then the
function is increasing (rising from left to right). If a is negative, the
function is decreasing (falling from left to right).
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On the transparency of Exercise 1, plot the point (1, _) for each of the graphed
lines as shown.
(1, 5)
(1, 3)
•
(1, 2)
(1, 1)
Compare the y-coordinates of these points with the a in y = ax . [The ycoordinates of these points are the same as the a in y = ax .]
Tell participants that we can think of y = ax as a transformation of y = x .
• What kind of transformation? [Dilation, stretch or compression by a scale
factor of a.]
Take the point (1,1) and stretch it to (1, a). Some participants may mistakenly
think that the line is rotated by a. Looking at table values may help to show
that all of the y-values of the line y = ax are a times the y-values of the line
y = x . On a blank transparency, trace the line y = x and the point (1, 1) and
place it on the transparency of Activity 1, Exercise 1. Rotate the transparency
about the origin to show that the point (1, 1) does not rotate to the point (1, 2).
The big idea of this activity is to help participants look at y = ax as a
transformation of y = x , as a dilation. For a > 1, the line is vertically
stretched by a scale factor of a. As x increases, the y-values increase faster
than for y = x . For 0 < a < 1 , the line is vertically compressed by a scale
factor of a. As x increases, the y-values decrease faster than for y = x . For
a < 0, the graph is a reflection over the x-axis.
Activity 2: Exploring Vertical Shifts
Have participants work through the Exercises. Then discuss Exercise 3.
3. Some generalizations about the y-intercept, b, in y = x + b : [The larger b
is, the higher the y-intercept. The smaller b is, the lower the y-intercept.
For b > 0, the y-intercept is above x-axis. . For b < 0, the y-intercept is
below the x-axis.
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In each of the graphs, plot the point (0, _) as shown.
(0, 3)
(0, 1)
(0, 0.5)
(0, 0)
•
Compare the y-coordinates with the point (0,0) from y = x and with the b
in y = x + b . [The y-coordinates of these points are the same as the b in
y = x + b .]
Tell participants that we can think of y = x + b as a transformation of y = x .
• What kind of transformation? [Translation, vertical shift by b.]
function y = x and the point (0, 0). Slide or shift the patty paper up and down
so that the point (0,0) shifts to (0,b) in all of the rest of the graphs.
The big idea of this activity is to help participants look at y = x + b as a
transformation of y = x , as a translation. For b > 0 , the graph is vertically
translated (shifted) up b units. For b < 0, the graph is vertically translated
(shifted) down b units.
Before moving to the next Activity, tell participants that now we are going to
look at these translations in yet another way.
On the transparency, put parentheses in Exercise 1 as follows.
1
d. y =  x + 
1a. y = x
b. y = ( x + 1)
c. y = ( x + 3)

2
Ask participants to graph the following points along with you.
For Exercise 1b, graph the point (-1, 0) on the line y = ( x + 1) .
For Exercise 1c, graph the point (-3, 0) on the line y = ( x + 3) .
1
For Exercise 1d, graph the point (-0.5, 0) on the line y =  x +  .

2
(-3, 0) (-1, 0)
(0.5, 0) (0, 0)
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•
•
•
•
•
What do these points suggest? [That the lines have been translated
horizontally.]
Where else have you seen horizontal translations? [In the vertex form
2
of parabola, y = a( x − h) + k , the parent function y = x 2 is
horizontally translated left h units if h < 0 and right h units if h > 0.]
Looking at Exercise 2b, what does writing the function as y = ( x − 2)
suggest? [It suggests that y = ( x − 2) can be graphed as a horizontal
translation right 2 units.]
If you horizontally translate the point (0, 0) on the parent function
y = x right 2 units, what point to you get? [(2, 0)]
Is (2, 0) on the line y = x − 2 ? [Yes.]
With the functions written y = x + b , you can think of the graph of y = x + b
as a vertical translation of y = x by b units. With the functions written
y = ( x − b) , you can think of the graph of y = ( x − b) as a horizontal
translation of y = x by b units .
Activity 3: Exploring Horizontal Shifts
Work through a few of the Exercises with participants, using the notes below.
Then have participants work through the rest of the Exercises. Ask groups to
present their strategies for a few of the Exercises, demonstrating how they had
graphed the lines and what connections they had made between the methods.
You can also use patty paper to show that both methods graph the same line.
1. y = x + 4 is a vertical translation of y = x up 4 units and y = ( x + 4) is a
horizontal translation of y = x left 4 units.
2. y = ( x + 2) + 1
First graph the line y = ( x + 2) + 1 by
translating the line y = x left 2 units
and up 1 unit.
y = x+3
Simplify y = ( x + 2) + 1 = x + 3 and
graph y = x + 3 as a line with yintercept 3 and slope 1.
(0, 3)
(-2, 1)
3. y = ( x − 5) + 3
y= x−2
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First graph the line y = ( x − 5) + 3 by Simplify y = ( x − 5) + 3 = x − 2 and
translating the line y = x right 5
graph y = x − 2 as a line with yunits and up 3 units.
intercept -2 and slope 1.
(5, 3)
(0, -2)
4. y = ( x + 1) − 4
First graph the line y = ( x + 1) − 4 by
translating the line y = x left 1 unit
and down 4 units.
y = x−3
Simplify y = ( x + 1) − 4 = x − 3 and
graph y = x − 3 as a line with yintercept -3 and slope 1.
(0, -3)
(-1, -4)
5. y = 2( x + 2) + 1
First graph the line y = 2( x + 2) + 1
by translating the line y = x left 2
units, up 1 unit, and then vertically
stretching the line by a scale factor
of 2.
y = 2x + 5
Simplify y = 2( x + 2) + 1 = 2 x + 5
and graph y = 2 x + 5 as a line with
y-intercept 5 and slope 2.
(0, 5)
(-2, 1)
6. y =
1
( x − 4) + 3
2
y=
1
( x − 4) + 3
2
by translating the line y = x right 4
units, up 3 units, and then vertically
compressing the line by a scale
1
factor of .
2
(4, 3)
First graph the line y =
1
x +1
2
1
1
( x − 4) + 3 = x + 1
2
2
1
and graph y = x + 1 as a line with
2
1
y-intercept 1 and slope .
2
Simplify y =
(0, 1)
7. y = 3( x + 1) − 4
y = 3x − 1
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First graph the line y = 3( x + 1) − 4
by translating the line y = x left 1
unit, down 4 units, and then
vertically stretching the line by a
scale factor of 3.
Simplify y = 3( x + 1) − 4 = 3 x − 1
and graph y = 3 x − 1 as a line with
y-intercept -1 and slope3.
(0, -1)
(-1, -4)
8. y = −2( x + 1) + 2
First graph the line y = −2( x + 1) + 2
by reflecting the line y = x across
the x-axis , translating the line y = x
left 1 unit, up 2 units, and then
vertically stretching the line by a
scale factor of 2.
y = −2 x
Simplify y = −2( x + 1) + 2 = −2 x
and graph y = −2 x as a line with yintercept 0 and slope -2.
(0, 0)
(-1, 2)
9. y = −( x − 2) + 1
First graph the line y = −( x − 2) + 1
by reflecting the line y = x across
the x-axis , translating the line y = x
right 2 units, and then up 1 unit.
(2, 1)
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y = −x + 3
Simplify y = −( x − 2) + 1 = − x + 3
and graph y = − x + 3 as a line with
y-intercept 3 and slope -1.
(0, 3)
356
1
y = − x −1
3
1
10. y = − ( x + 3)
3
1
1
1
First graph the line y = − ( x + 3) by Simplify y = − ( x + 3) = − x − 1
3
3
3
1
reflecting the line y = x across the
and graph y = − x − 1 as a line
x-axis , translating the line y = x
3
1
left 3 units and then vertically
with y-intercept -1 and slope − .
3
compressing the line by a scale
1
factor of .
3
(-3, 0)
•
(0, -1)
Which method of graphing is better? [“Which is better?” is the wrong
question to ask. They are both valuable methods. The emphasis is on
understanding linear functions in different ways, with different
representations. We want to connect different representations to build
understanding. Some say that either method is valid, but to say this
may imply that it is sufficient to use either exclusively, therefore
limiting students’ understanding. However, the big idea is not to let
everyone use what ever way they prefer, but to teach many ways of
looking at something, to make connections between methods, and thus
to build understanding of the concept from a wider, broader, more
inclusive prospective.]
Activity 4: A Different Perspective
1. Prompt participants to ask: What transformations do you need to
transform the line y = x to contain the points (2, 10) and (5, 4)?
You still need to find the slope.
Table
Time
Distance
3
2
10
5
4
-6
4 − 10
ft
. Now sketch the line y = x and reflect it over
= −2
5−2
sec
the x-axis to get the line y = − x .
So the slope is
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y = −x
y=x
Now sketch one of the given points, say (5, 4). The goal is to shift the line
y = x so that the transformed line contains (5, 4). So, one way to do this is to
first shift y = x right 5. This is written y = −( x − 5) .
y = −x
y = −( x − 5)
right 5
Next, translate y = x up 4 units. This is written y = −( x − 5) + 4 .
y = −( x − 5) + 4
up 4
y = −( x − 5)
Next, vertically stretch the line by a scale factor of 2, written
y = −2( x − 5) + 4 .
y = −( x − 5) + 4
y = −2( x − 5) + 4
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Of course, you could have chosen to do all of the above so that the line
contained the other point (2, 10).
The equation of the line above, y = −2( x − 5) + 4 , simplifies to y = −2 x + 14 ,
which is the result using the previous method of counting back in the table
after finding the rate of change.
2. Find the rate of change, which is about 23 cm/block. Choose a point, say
(1, 25). So the line y = x will be translated right 1, y = ( x − 1) , translated
up 25, y = ( x − 1) + 25 , and vertically stretched by a scale factor of 23,
y = 23( x − 1) + 25 .
1. The point-slope form of the equation of a line from a transformational
perspective can be seen as transforming the line y = x by translating the
line y = x horizontally x1 units, translating the line y = x vertically y1
units, and vertically stretching or compressing the line y = x by a scale
factor of a.
The vertex form of the equation of a parabola from a transformational
perspective can be seen as transforming the quadratic function y = x 2 by
translating y = x 2 horizontally h units, translating y = x 2 vertically k
units, and vertically stretching or compressing y = x 2 by a scale factor of
a.
2. The equation y = af ( x − b) + c is a general way of describing
transformations of any function, f ( x ) . The variable a is the same vertical
scale factor as in the two specific equations above. The variable b is
horizontal translation as is the x1 in the point-slope form of the equation of
a line and the h in the vertex form of the equation of a parabola.
Replacing x with ( x − b) in the function f ( x ) has the effect of translating
the graph of f ( x ) horizontally by b units just as the x1 and the h did for
the line and the parabola respectively. The variable c a vertical
translation, up if c > 0 and down if c < 0 .
3. Participants should recognize that y = 2 ⋅ f ( x ) − 1 means to vertically
stretch the original function by a scale factor of 2 and vertically translate
the original function down 1 unit. Therefore the graph of y = 2 sin x − 1 is
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4. Participants should recognize that y = − f ( x ) + 3 means to reflect the
original function over the x-axis and vertically translate the function up 3
units. Therefore the graph of y = − x + 3 is
Summary:
Using technology to explore changing the parameters of the equation of a line,
participants make connections between transformations of parabolas
transformations of lines. Understanding transformations of the line y = x
builds deeper understanding of the point-slope form of the equation of a line.
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1.3 Lines Do It Too: Activity 1
Activity 1: Exploring Slope
Sketch a graph of the following in the same viewing window.
1. y = x
y = 3x
y = 2x
y = 5x
2. y = x
y = −x
y = −2 x
y = −5 x
3. y = x
y = 0.5 x
y = 0.25 x
1
y= x
5
4. y = x
y = −0.5 x
1
y=− x
3
y = 0.2 x
5. Summarize the effects of a on the graph of y = ax .
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Activity 2: Exploring Vertical Shifts
Sketch a graph of the following in the same viewing window.
1
d. y = x +
b. y = x + 1
c. y = x + 3
1a. y = x
2
2a. y = x
b. y = x − 2
c. y = x −
1
2
d. y = x −1
3. Summarize the effects of b on the graph of y = x + b .
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Activity 3: Exploring Horizontal Shifts
1. Describe the following in two ways: y = x + 4 .
Graph the following lines in two ways. First, as transformations
of y = x . Then simplify each linear function to y = mx + b or
y = b + mx and graph.
2. y = ( x + 2) + 1
3. y = ( x − 5) + 3
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363
5. y = 2( x + 2) + 1
1
6. y = ( x − 4) + 3
2
8. y = −2( x + 1) + 2 9. y = −( x − 2) + 1
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7. y = 3( x + 1) − 4
1
10. y = − ( x + 3)
3
364
Activity 4: A Different Perspective
The following problems are found previously in the institute.
Approach them this time from a transformational perspective.
1. You looked up twice during Juan’s walk. You noted that he
was at the 10 foot mark at the 2nd second and that he was at
the 4 foot mark at the 5th second. Complete the table, graph,
and find a rule for his walk. (Assume he was walking at a
constant rate.)
Table
Graph
Rule:
2. You collected the following data. Find a trend line.
You were investigating the
Height
Distance
relationship between the
(blocks)
(cm)
height of the pipe and the
1
25
distance the marble rolls when
2
47.5
released in the pipe at that
3
73.75
height
4
92
5
117
Measure
Trend Line:
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1.3 Lines Do It Too: Reflect and Apply
Reflect and Apply
1. Discuss each of the following from a transformational
perspective. How are the two equations similar?
Point-Slope form of the
equation of a line
y = m( x − x1 ) + y1
Vertex form of the equation of
a parabola
y = a( x − h ) 2 + k
2. For any function, f ( x ), how do the above two equations
relate to y = a ⋅ f ( x − b) + c ?
Given the graphs of the following functions, sketch the indicated
transformations.
4. Given f ( x ) = x
3. Given f ( x ) = sin x
Sketch
2 ⋅ f ( x ) − 1 = 2 sin x − 1
Sketch
− f ( x) + 3 = − x + 3
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2.1 Connections: Leaders’ Notes
2.1 Connections
Overview:
Participants make connections between the roots of quadratic functions and
the solutions to quadratic equations and the factors of quadratic polynomials
and the x-intercepts of a parabola. They connect this understanding to the
vertex, polynomial, and factored form of the equation of a parabola. Using
this understanding, participants model a vertical jump, finding the height of
the jump.
Objective:
Algebra I TEKS
(d.2.B) The student relates the solutions of quadratic equations to the roots of
their functions.
Terms:
root, zero, x-intercept, factor, solution, vertex, polynomial
Materials:
graphing calculators, data collection devices, light sensors, laser pointers or
flashlights
Procedures:
The connections between a function’s, y = f ( x ) , roots, the zeros of the graph
of f ( x ) , the solution(s) to the equation f ( x ) = 0 , and the linear factors of the
polynomial f ( x ) (if f ( x ) is a polynomial) are all very important connections
for students to make about functions. In this activity, participants make these
connections for quadratic functions. In later courses, students will apply these
lessons to higher order polynomials and other functions.
Activity 1: Roots, Factors, x-intercepts, Solutions
Have participants complete the activity. Give a transparency of the table in
Exercise 8 to a group and have them present their results.
Note: The graphs are meant to be sketches using the roots. The important
criteria to look for are correct roots and direction of the parabola. We are not
concerned with the maximums and minimums of the parabolas in this activity.
1.
2. The x-intercepts are –3 and 2.
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3.
4. The x-intercepts are the same.
5. ( x + 3)( x − 2) = x 2 + x − 6
Show the algebra tile (area) method for simplifying the expression.
6. The x-intercepts are the same.
7. Participants can solve the equation y4 = 0 with the graph, table,
multiplication property of zero, etc. Solutions are x = −3, x = 2 .
8. If your participants do not have much experience with algebra tiles to
expand expressions like ( x + 3)( x − 2) = x 2 + x − 6 , then add a column to
the table that includes an area model to obtain the polynomial form.
Graph y4
y1
y2
y3
(factored form)
Roots
x+3
x−2
( x + 3)( x − 2)
-3 and 2
y4
(polynomial
form)
2
x + x−6
Solutions
y4 = 0
x = −3, 2
x 2 + 3x + 2
x +1
x+2
( x + 1)( x + 2)
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-2 and -1
x = −2, − 1
368
x 2 − 5x + 4
x−4
x −1
( x − 4)( x − 1)
x = 1, 4
1 and 4
x 2 + 7 x + 12
x+3
x+4
( x + 3)( x + 4)
x = −4, − 3
-4 and -3
x 2 − 3x
x
x−3
x( x − 3)
x = 0, 3
0 and 3
x 2 − x − 12
x−4
x+3
( x − 4)( x + 3)
x = −3, 4
-3 and 4
x2 + 6x + 8
x+2
x+4
( x + 2)( x + 4)
x = −4, − 2
-4 and -2
x2 − x − 6
x+2
x−3
( x + 2)( x − 3)
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-2 and 3
x = −2, 3
369
Transparency: Which Form?
Ask participants to recall the situations and activities where the three different
forms of a quadratic equation were explored.
Sandbox problem, V = x(15 − x ) ⋅ 1, is the factored form.
Projectile problem, h = −16t 2 + 80t , is in the polynomial form.
2
Transformations, y = a( x − h) + k , are in the vertex form.
• What different kinds of information can you readily see from each
form? [In the sandbox volume problem, V = x(15 − x ) ⋅ 1, you can
quickly see the roots of the function, 0 and –15, where the volume of
the sandbox is zero.
In the projectile motion problem, h = −16t 2 + 80t , you can easily see
the acceleration due to the force of gravity because –16 ft/sec2 is half
of –32 ft/sec2. You can see the initial velocity of the ball, 80 ft/sec,
and the initial height of the ball at zero feet.
2
In 1.2 Transformations, y = a( x − h) + k , you can see the scale factor
and the vertex.]
In Activity 2, we will work more with these different forms of a quadratic equation.
Activity 2: Which Form?
Work through Exercise 1 with participants. Then have them complete the rest
of the activity, circulating and asking guiding questions.
1. Since the x-intercepts of the graph are –3 and 5, factors are ( x + 3)( x − 5) .
−3 + 5
Average the x-intercepts
= 1 so the x-coordinate of the vertex is 1.
2
Evaluate (1 + 3)(1 − 5) = ( 4)( −4) = −16 . But the y-coordinate of the vertex
1
shown is –8, so there must be a vertical scale factor of . Thus the
2
1
factored form of the equation is y = ( x + 3)( x − 5) .
2
2. Using the distributive property or an algebra tile area model, the
1
15
polynomial form is y = x 2 − x − .
2
2
1
2
3. The vertex form is y = ( x − 1) − 8
2
4. y = ( x − 6)( x − 9)
5.
a. The x-intercepts are 0 and 2.
b. Since the x-intercepts are 0 and 2, factors are x and ( x − 2) and the xcoordinate of the vertex is 1. As the maximum height of the ball is 4.9
meters, the y-coordinate of the vertex is 4.9. Therefore the vertex is
(1, 4.9).
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c. Evaluate y = x( x − 2) at the x-coordinate of the vertex, x = 1,
y = 1(1 − 2) = −1 . But the y-coordinate of the vertex is 4.9, so there
must be a vertical scale factor of –4.9. Therefore, y = −4.9 x( x − 2) is
the factored form.
2
d. The vertex form is y = −4.9( x − 1) + 4.9 .
e. The polynomial form is y = −4.9 x 2 + 9.8 x .
• What information can be readily seen by looking at each form of the
equation? [The factored form shows the x-intercepts, in this situation,
the times when the ball was on the ground. The vertex form shows the
vertex, in this case the maximum height of the ball. The polynomial
form shows the y-intercept, in this case, the height of the ball at
time=0.]
6. One way is to work backwards.
5
3
and
x=−
x=
2
2
5
3
(2 ) x = − (2 )
(2 ) x = (2 )
2
2
2 x = −5
2x = 3
2x + 5 = 0
2x − 3 = 0
Therefore y = (2 x + 5)(2 x − 3) .
Activity 3: Jump!
Demonstrate the activity for participants, by having a participant jump when
prompted by the program. Repeat if necessary to get an appropriate graph, as
the example below. Then have participants complete the activity.
To add a note of competition, have the three participants with the highest
jump found in the experiment come to the front of the group and jump. If
they leave the ground at the same time, the last one to land is the winner. See
if the result matches the predicted result.
The program for this activity should record intensity of light. When the
jumper is standing in between the flashlight and the sensor, the sensor should
record the ambient light level in the room. When the jumper jumps, the
sensor should record the higher intensity level of the flashlight. When the
jumper returns to the ground, the sensor should again record the ambient light
level in the room. It is obviously important that the jumper takes off and lands
in relatively the same place on the floor, in order to effectively block the
flashlight’s light.
1. Sample data:
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2.
3. The quadratic function to model the jump (time, height) is
y = −192( x − root1 )( x − root2 ) . Recall that the position equation is
1
h = − at 2 + v0 t + h0 . The force of acceleration due to gravity is
2
ft
in
−32 ft 12in
1
in
−32
=
⋅
= −384
and   ( −384) = −192
.
2
2
2


sec
sec
sec
1 ft
2
sec 2
cm
Participants could also use −490
to find the maximum jump in
sec 2
centimeters.
For our sample data, the model is y = −192( x − 0.06)( x − 0.54).
4.
Jump heights from a few inches to around 15 inches are reasonable.
1a. c < 4 . The equation f ( x ) = 0 has two real solutions.
b. c = 4. The equation f ( x ) = 0 has one real solution.
c. c > 4 . The equation f ( x ) = 0 has no real solutions.
2. For y = Ax 2 + Bx + C , you can see the y-intercept, C. You can also see A
and B which are useful for certain application problems.
For y = a( x − x1 )( x − x2 ) , you can see the value of a and the roots, which
are also the x-intercepts and the solutions to y = 0 .
2
For y = a( x − h) + k , you can see the value of a and the vertex (h, k),
where k is the maximum or minimum value of the function.
Summary:
Roots, solutions, x-intercepts, and factors are often taught as isolated concepts.
Bringing them all together helps participants make connections and builds
understanding about the different forms of the equation of a quadratic
function.
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2.1 Connections: Transparency
Transparency: Which Form?
Match the quadratic equations with the correct form:
Sandbox problem:
V = x (15 − x ) ⋅ 1
Vertex form
Projectile problem:
h = −16t 2 + 80t
Factored form
Transformations:
y = a( x − h)2 + k
Polynomial form
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2.1 Connections: Activity 1
Activity 1: Roots, Factors, x-intercepts, Solutions
1. Graph the two functions in the same viewing window and
sketch.
y1 = x + 3
y2 = x − 2
2. What are the x-intercepts of the above equations?
3. Add to your sketch the graph of y3 = y1 ⋅ y2 = ( x + 3)( x − 2) .
4. How do the x-intercepts of y3 = ( x + 3)( x − 2) compare to the
x-intercept of y1 = x + 3 and of y2 = x − 2 ?
5. Using algebra tiles, simplify y3 = ( x + 3)( x − 2) to rewrite in
polynomial form and graph this expression in y4 .
6. How do the x-intercepts of y4 compare to those above?
7. Solve y4 = 0 .
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8. Complete the table.
y1
y2
Graph y4
x+ 3 x− 2
x+ 1 x+ 2
y3
(factored form)
( x + 3)( x − 2)
( x − 4)( x − 1)
Roots
- 3 and 2
-3 and -4
Spring 2001
Solution(s)
to y4 = 0
y4
(polynomial form)
x2 + x − 6
x = −3, 2
375
Graph y4
y1
y2
y3
(factored form)
Roots
Spring 2001
Solution(s)
to y4 = 0
y4
(polynomial form)
y = x2 − x − 12
x = −2, 3
376
Activity 2: Which Form?
Write the equation of the graph in three forms:
1. Factored form
2. Polynomial form
3. Vertex form
4. Write an equation for a quadratic function that has xintercepts 6 and 9 and has a vertical scale factor of 1.
5. A soccer goalie kicks the ball from the ground. It
hits the ground after 2 seconds, reaching a
maximum height of 4.9 meters.
a. Find the x-intercepts for the quadratic function that
models the relationship (time, height).
b. Find the vertex of the quadratic function.
Write the quadratic function in
c. factored form,
d. vertex form,
e. polynomial form.
6. Explain how you can find the factored form of the equation
for this quadratic function, given it has a vertical scale factor
of 1.
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Activity 3: Jump!
How high can you jump? You can
use the time that you are in the air to
find your vertical jump.
Set up the experiment as shown,
creating a photo-gate with the
flashlight and the light sensor.
Interrupt the signal by standing
between the laser pointer and
the sensor. Run the program,
jumping when prompted
1. Sketch the results.
2. Trace and record the time when you left the floor and the
time when you landed.
3. Use the two times to create a quadratic function that models
your jump (time, height).
4. Use your function to find your maximum jump.
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2.1 Connections: Reflect and Apply
Reflect and Apply
1. Name values for c such that f ( x ) = x 2 − 4 x + c satisfies each.
a. The graph of f has two x-intercepts. What does this imply
about the solution(s) to f ( x ) = 0 ?
b. The graph of f has exactly one x-intercept. What does this
imply about the solution(s) to f ( x ) = 0 ?
c. The graph of f has no x-intercepts. What does this imply
about the solution(s) to f ( x ) = 0 ?
2. What information can you readily see from each form of a
quadratic equation?
y = Ax 2 + Bx + C
y = a( x − x1 )( x − x2 )
y = a( x − h ) 2 + k
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2.2 The Quadratic Formula: Leaders’ Notes
2.2 The Quadratic Formula
Overview:
Participants program the quadratic formula into the graphing calculator and
use the program to solve quadratic equations at appropriate times.
Objective:
Algebra I TEKS
Terms:
quadratic equation, quadratic formula
Materials:
graphing calculators, 1” graph paper, markers, meter sticks
Procedures:
Activity 1: Programming the Quadratic Formula
Work with participants to write a program for their calculators.
A sample program:
Have participants check their programs by trying a simple quadratic equation
like 0 = x 2 − x − 6 , x = −2, 3
• What result do you get for 0 = x 2 + x + 1 ? [Since the roots are
imaginary, the calculator may return an error message or “non-real
answer” or if the calculator has an imaginary mode, it may return the
imaginary roots. Discuss the results with participants.]
• When do you think it is appropriate for students to use the program?
Extension: You could include in the quadratic formula program a conditional
statement that tests whether the discriminant is negative. If b 2 − 4 ac < 0 , then
the program would return a line “No real solutions.”
Activity 2: Hang Time
Have participants work together on the activity in groups. They should make
a poster size presentation on 1” grid paper with their graphs color coded and
clearly labeled.
Participants will have to deal with the “Pluto problem,” that Pluto’s g is so
small that it is difficult to clearly show the graphs representing the rest of the
planets and include the graph representing Pluto. One way of dealing with it
is the following example.
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2.
Planet
Gravity at the
Surface, g
 ft 
 sec 2 
11.84
28.16
Vertical Height Model
Hang
Time
(sec)
h(t ) = −5.92t 2 + 10t + 3
h(t ) = −14.08t 2 + 10t + 3
1.95
0.94
4. Earth
6. Mars
1. Jupiter
32
12.16
84.48
h(t ) = −16t 2 + 10t + 3
h(t ) = −6.08t 2 + 10t + 3
h(t ) = −42.24t 2 + 10t + 3
0.85
1.90
0.41
2. Saturn
2. Uranus
3. Neptune
36.80
36.80
35.84
h(t ) = −18.4t 2 + 10t + 3
h(t ) = −18.4t 2 + 10t + 3
h(t ) = −17.92t 2 + 10t + 3
0.76
0.76
0.77
8. Pluto
1.28
h(t ) = −0.64t 2 + 10t + 3
15.92
7. Mercury
5. Venus
3.
8
7
6
4
1
0.41 sec
2,3
5
0.94 sec
1.90 sec
1.95 sec
0.76 sec
0.77 sec
0.85 sec
WINDOW: [0, 2] .1 [-.5, 8] 1
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381
7.22 ft
8
7.11 ft
4.78 ft
4.40 ft
4.36 ft
4.56 ft
3.59 ft
7
6
4
5
2,3
1
WINDOW: [0, 2] .1 [-.5, 8] 1
42.06 ft
8
15.92 sec
[0, 16] 1 [-5, 45] 10
•
Is this the kind of problem where it is appropriate to use technology?
It makes sense to choose to solve Exercise 3 by completing the square because
the value of B in Ax 2 + Bx + C is an even number, therefore making it easy to
1
find B . It makes sense to solve Exercise 2 by factoring because you can.
2
1. Solved with the quadratic formula calculator program, x ≈ −0.134, 3.195
2. Solved by factoring, x 2 + x − 12 = 0 = ( x + 4)( x − 3) , x = −4, 3.
3. Solved by completing the square, 0 = x 2 − 4 x + 1.
2
0 = x 2 − 4 x + ( −2) + 1 − 4
2
0 = ( x − 2) − 3
2
3 = ( x − 2)
± 3 = x−2
x =2± 3
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4. Solved with the quadratic formula calculator program, x ≈ −0.25, 4.625 .
5. For Exercise 1, a ball was thrown up from a height of 2.1 meters at an
initial velocity of 15 meters/sec. The 4.9 is one-half the force of gravity
measured in meters per second per second. The solution to the equation
answers the question, “When does the ball hit the ground?”
For Exercise 4, a ball was thrown up from a rooftop 18.5 feet high at an
initial velocity of 70 feet/sec. The 16 is one-half the force of gravity
measured in feet per second per second The solution to the equation
answers the question, “When does the ball hit the ground?”
6.
2
2
2
a. ( x + 4)
b. ( x + 9)
c. ( x − 5)
x 2 + 8 x + 16
x 2 + 18 x + 81
x 2 − 10 x + 25
x
4
x
x2
4x
4
4x
16
d. ( x + 3)( x − 2)
x2 + x − 6
x
3
x
x
2
-2 −2x
Summary:
x
9
x
x2
9x
9
9x
81
e. ( x − 3)( x − 4)
x 2 − 7 x + 12
-3
x
x2
3x
x
-6
-4 −4x
−3x
12
x
-5
x
x2
−5x
-5
−5x
25
f. ( x + 6)( x + 5)
x 2 + 11x + 30
x
6
x2
6x
5 5x
30
x
The quadratic formula programmed into a graphing calculator can provide
students with a powerful tool to solve quadratic equations. Care should be
exercised about when students should use the program. In the midst of a big
problem where the program is used as a tool to quickly obtain otherwise
cumbersome solutions and students must ascertain at which point to use the
program, certainly this is an appropriate time to use technology.
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2.2 The Quadratic Formula: Activity 1
Activity 1: Programming the Quadratic Formula
1. Write a program that will find the roots of a quadratic
equation using the quadratic formula.
The program should:
• Ask the user to input A, B, C from y = Ax 2 + Bx + C
• Find the root(s)
• Display the root(s)
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384
Activity 2: Hang Time
If you were to jump around on different
planets, your motion would differ
because the acceleration due to the force
of gravity is different. Imagine that you
jump from a 3 foot high platform with an
initial velocity of 10 ft/sec. How would
your hang time compare on each
different planet?
1. Based on the values of g below, make some predictions. On
which planet would you land first or last? On which planet
would you jump the highest or lowest?
2. Complete the table.
Planet
Gravity at the
Surface, g
 ft 
 sec 2 
Mercury
11.84
Venus
28.16
Earth
32
Mars
12.16
Jupiter
84.48
Saturn
36.80
Uranus
36.80
Neptune
35.84
Pluto
1.28
Vertical Height
Model
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Hang
Time
385
3. Make a graph, showing the graphs of the jumps (time,
height), each labeled with the maximum height of the jump
and the time you would land on that planet.
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2.2 The Quadratic Formula: Reflect and Apply
Reflect and Apply
Solve the equations. Choose one to solve by completing the
square. Choose one to solve by factoring. Explain your choices.
1.
2.
3.
4.
0 = −4.9 x 2 + 15 x + 2.1
x 2 + x − 12 = 0
0 = x2 − 4x + 1
−16 x 2 + 70 x + 18.5 = 0
5. Choose two of the above equations and give them a realworld context, including what the various numbers represent
and what question the equation answers.
6. Fill in the missing values. Then write two expressions for the
total area for each figure.
a.
b.
c.
?
4
x
x2
?
?
?
16
d.
x
?
?
?
9x
9
?
81
e.
?
3
x
x2
?
?
?
-6
x
?
?
x2
−5x
-5
?
?
f.
x
?
?
x2
−3x
-4
?
?
Spring 2001
x
?
?
?
6x
5
?
30
387
2.2 The Quadratic Formula: Student Activity
Student Activity: Investigate Completing the Square
Overview:
Students investigate completing the square with algebra tiles.
Objective:
Algebra I TEKS
Terms:
complete the square
Materials:
algebra tiles, graphing calculator
Procedures:
Have students work through Exercises 1 – 4 in groups. As a whole group,
discuss their answers using the following.
Note: This activity assumes that students have prior experience with
representing, adding and subtracting polynomials with algebra tiles (area
model), with using algebra tiles to model monomial and binomial
multiplication and with modeling factoring trinomials with algebra tiles.
Note: This activity uses a concrete model to lay the foundation for the
algebraic work of completing the square that students will do in Algebra II.
1. a. You need 9 unit tiles to complete the square.
b. The dimensions of the completed square are ( x + 3) by ( x + 3) .
2
c. x 2 + 6 x + 9 = ( x + 3)
2. a. You need 16 unit tiles to complete the square.
b. The dimensions of the completed square are ( x − 4) by ( x − 4) .
2
c. x 2 − 8 x + 16 = ( x − 4)
3. The number of unit tiles needed to complete the square is the square of
half of the coefficient of x. For x 2 + Bx , the number of unit tiles needed is
2
 B .
 2
2
9
3
4. a. x 2 − 3 x + =  x − 
4 
2
2
2
b
b
b. x 2 + bx +   =  x − 
 2

2
Do the following 2 examples with students.
The first example is to write the equation, y = x 2 + 4 x + 5 in vertex form.
Complete the square.
y = x2 + 4x + 4 − 4 + 5
y = ( x 2 + 4 x + 4) − 4 + 5
y = ( x + 2) + 1
2
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Have students quickly sketch a graph.
The second example is to solve a quadratic equation, 0 = x 2 − 10 x + 19 , by
completing the square.
Complete the square:
0 = x 2 − 10 x + 19
0 = x 2 − 10 x + 25 − 25 + 19
0 = ( x 2 − 10 x + 25) − 25 + 19
0 = ( x − 5) − 6
2
6 = ( x − 5)
± 6 = x−5
x = 5± 6
2
Have students complete the rest of the Activity.
5. y = x 2 + 6 x + 4
y = x2 + 6x + 9 − 9 + 4
y = ( x 2 + 6 x + 9) − 9 + 4
y = ( x + 3) − 5
2
6. 3( x + 1) − 6 = 0
2
3( x + 1) = 6
( x + 1)2 = 2
x +1 = ± 2
x = −1 ± 2
2
7. x 2 − 4 x − 8 = 0
x2 − 4x + 4 − 4 − 8 = 0
( x 2 − 4 x + 4) − 4 − 8 = 0
( x − 2)2 − 12 = 0
( x − 2)2 = 12
x − 2 = ± 12
x − 2 = ±2 3
x =2±2 3
Summary:
Using algebra tiles to complete the square based on the area model of
multiplication gives students a geometric approach to understanding the
algebraic steps to complete the square.
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Student Activity: Investigate Completing the Square
1. Create a partial square with algebra tiles to
represent x 2 + 6 x as shown.
a. How many unit tiles do you need to
complete the square?
b. What are the dimensions of the
completed square?
c. x 2 + 6 x + ? = ( x + ?)2
2. Create a partial square with algebra tiles
to represent x 2 − 8 x as shown.
a. How many unit tiles do you need to
complete the square?
b. What are the dimensions of the
completed square?
c. x 2 − 8 x + ? = ( x − ?)2
3. How does the number of unit tiles to complete the square
compare to each respective coefficient of x?
4. Based on the above, complete the two square diagrams.
b. x 2 + bx + ? = ( x − ?)2
a. x 2 − 3 x + ? = ( x − ?)2
x2
−1.5 x
−1.5 x
x2
b
x
2
?
b
x
2
?
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5. Write the function in vertex form by completing the square
y = x 2 + 6 x + 4 . Then sketch a graph.
6. Solve the quadratic equation written in vertex form.
3( x + 1)2 − 6 = 0
7. Solve the equation by completing the square.
x2 − 4x − 8 = 0
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3.1 Exponential Relationships: Leaders’ Notes
3.1 Exponential Relationships
Overview:
Participants explore exponential growth and decay situations. Participants
develop the ideas of the common multiplier or ratio as the base of an
exponential function and the starting point as the y-intercept of an exponential
function.
Objective:
Algebra I TEKS
(b.3.B) Given situations, the student looks for patterns and represents
generalizations algebraically.
algebraic methods.
Terms:
growth, decay, recursion,
Materials:
graphing calculators, sheets of blank paper
Procedures:
In this section, we explore exponential relationships similar to the way we
explored linear relationships in II. Linear Functions. The connection between
the linear starting point and y-intercept is analogous to the connection
between the exponential starting point and y-intercept. The connection
between the added constant and the slope of linear functions is analogous to
the constant multiplier and the base of exponential functions. Participants will
learn to write exponential functions using a starting point and a common
multiplier or ratio just as they learned to write linear functions using a starting
point and a common difference. Encourage participants to make connections
between what is happening in the problem situation and the parameters in the
exponential functions.
Do the Student Activity with participants, depending on the level of your
participants.
Activity 1: Paper Folding
Work through Activity 1 with participants.
Introduce the scenario and demonstrate a couple of folds. Have participants
fold a piece of paper in half as many times as they can.
1. Lead participants in filling in the table, using language similar to the
following:
For no folds, you have 1 layer of the piece of paper
Number of folds
Process
Number of Layers
0
1
1
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After the 1st fold, you have 2 layers.
Number of folds
Process
0
1
1
1⋅ 2 = 2
Number of Layers
1
2
After the 2nd fold, you have 4 layers. In other words, you have twice as many
as before.
• How can you write 1 ⋅ 2 ⋅ 2 with exponents? [ 1 ⋅ 2 2 ]
Number of
Process
Number of Layers
folds
0
1
1
1
2
1⋅ 2 = 2
2
4
1 ⋅ 2 ⋅ 2 = 1 ⋅ 22
When completing the table, one is more apt to operate recursively on the
previous term, continuing to multiply by 2. The emphasis here is on
expressing the number of layers in terms of the number of folds in order to
develop the function rule.
Continue to fill in the table, establishing the pattern.
Process
Number of
Number of
folds
Layers
0
1
1
1
2
1⋅ 2 = 2
2
2
4
1⋅ 2 ⋅ 2 = 1⋅ 2
3
3
8
1⋅ 2 ⋅ 2 ⋅ 2 = 1⋅ 2
4
4
16
1⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1⋅ 2
5
5
32
1⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1⋅ 2
6
6
64
1⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 1⋅ 2
n
n
1⋅ 2 ⋅ 2 ⋅ . . . ⋅ 2 ⋅ 2 = 1⋅ 2
1 ⋅ 2n
1 44 2 4 43
n factors
2. Write the function as follows.
• After n folds, how many 2’s will be multiplied by each other? In other
words, how many factors of 2 are there is the expression? [n]
• How can you write 2 ⋅ 2 ⋅ . . . ⋅ 2 ⋅ 2 ? [ 2 n ]
1 44 2 4 43
n factors
Number of
folds
n
Process
1 ⋅ 2 ⋅ 2 ⋅ . . . ⋅ 2 ⋅ 2 = 1 ⋅ 2n
1 44 2 4 43
Number of
Layers
1 ⋅ 2n
n factors
3. Use questions to lead participants to finding a suitable viewing window.
• What does x stand for in this problem? [Number of folds]
Sample answer. Zero folds to 10 folds.]
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•
•
What does y stand for in this problem? [Total number of layers of
paper]
What values make sense for y in this problem? [Answers will vary.
Sample answer. No layers to 2^10, which is 1024.]
4. Sample answer. The variable x stands for number of folds so zero to 10
folds shows a few more folds than I could actually fold with my piece of
paper. The variable y stands for number of layers, so zero to 60 will show
about all of the layers and the x-axis.
5. 1 ⋅ 2(18) = 262,144 . You will have 262,144 layers after 18 folds.
6. Because there are 5 reams of paper, you want to know when the value of y
is 5 times 500, or 2500. 1 ⋅ 2 x = 2500 . You need to fold at least 12 times
to get as thick as a box of paper. At the eleventh fold, the paper would
only be as thick as 2048 sheets of paper, so you would need to fold it one
more time, which gets you as thick as 4096 sheets of paper.
A numerically powerful exercise is have participants guess and check a
more exact answer to 1 ⋅ 2 x = 2500 , using the home screen.
•
•
Explain how this answer applies to this problem situation? [It makes
no sense as the domain values must be whole numbers.]
Usually we solve this kind of an equation how and when? [In a typical
curriculum this type of exponential equation is solved using logarithms
in Algebra II and Precalculus. With the power of technology, using
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graphs, tables, and the home screen, we can get good approximations
of the solution and build understanding about exponential
relationships.]
Discuss the Activity:
• What operation is being repeated in this problem? [multiplication]
• What function results from repeated multiplication? [exponential
functions]
• Earlier in the institute, we worked with another repeated operation,
repeated addition. What function results from repeated addition?
[linear functions]
• What kind of graphs result from repeated addition? [linear]
• What kind of graphs result from repeated multiplication?
[exponential]
• Do you think this paper folding activity is an example of exponential
growth or decay? [growth]
Activity 2: Measure with Paper
Work through the Activity with participants, discussing using the following.
1. Each sheet of paper measures approximately 0.004 inches. The
measurement can be converted using dimensional analysis.
2 inches 1 ream
2
⋅
=
inches per sheet = 0.004 inches per sheet .
1 ream 500 sheets 500
2. Develop the table in the same way as Activity 1.
Number of folds
0
1
2
3
4
5
6
n
Process
0.004
0.004 ⋅ 2
0.004 ⋅ 2 ⋅ 2 = 0.004 ⋅ 2 2
0.004 ⋅ 2 ⋅ 2 ⋅ 2 = 0.004 ⋅ 2 3
0.004 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 0.004 ⋅ 2 4
0.004 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 0.004 ⋅ 2 5
0.004 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 0.004 ⋅ 2 6
0.004 ⋅ 2 ⋅ 2 ⋅ . . . ⋅ 2 ⋅ 2 = 0.004 ⋅ 2 n
1 44 2 4 43
Thickness
(inches)
0.004
0.008
0.016
0.032
0.064
0.128
0.256
0.004 ⋅ 2 n
n factors
3. y = 0.004 ⋅ 2 n
• How does this pattern and table compare with the table in Activity 1?
[In Activity 1, we were counting layers of paper. Here we are
measuring how thick the folded paper is. The output (range) values in
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both tables are doubling each time. The starting values for each table
are different.]
• How might the table help you find a viewing window?
paper. The variable y stands for height of the folded paper in inches, so
zero to 4.1 inches ( 0.004 ⋅ 210 ≈ 4.096 ) will show most of the graph and
the x-axis.
6. 0.004 ⋅ 2(15) = 131.072 . 131 ÷ 12 ≈ 10.92 so you have approximately 11
feet of paper after 15 folds. This is one foot higher than a 10 foot ceiling.
It is as high as double a 5.5 foot person.
7. If the Eiffel Tower is 1050 feet tall, its height in inches is
1050 ⋅ 12 = 12, 600 . The equation you are solving here is
0.004 ⋅ 2 x = 12, 600 . When you fold the paper 21 times, the value is
8388.6. When you fold the paper 22 times, the height is 16,777, so it must
be folded 22 times to reach the top of the tower.
Again, a numerically powerful exercise is to have participants guess and
check a more exact answer to 0.004 ⋅ 2 x = 12, 600 , using the home screen.
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•
•
Does this approach have meaning in the problem situation? [No, the
domain values must be whole numbers.]
How does the table and graph compare with the table and graph in
Activity 1? [In Activity 1, we were counting layers of paper. Here we
are measuring how thick the folded paper is. The output (range)
values in both tables are doubling each time. The starting values for
each table are different. Look at both tables.]
•
Find a viewing window to compare the graphs of y = 1 ⋅ 2 x and
y = 0.004 ⋅ 2 x .
•
Compare the two rules, y = 1 ⋅ 2 x and y = 0.004 ⋅ 2 x . [They both have
the same common multiplier. Each function value doubles with each
increase by 1 of x.]
Look at the table values again. How do you know that this data is not
linear? [Look at differences. There is no common difference.
Remember, if there is not a common difference, the data is not linear.
We did not obtain these table values from repeated addition.]
What operation is being repeated in this problem? [Multiplication]
What kind of function results from repeated multiplication?
[Exponential function]
What is the common multiplier for Activities 1 and 2? [The common
multiplier is 2 because the values double each time.]
How can you find the common multiplier of 2 just by looking at the
values in the table? [Divide each y-value by the previous y-value to
obtain the common multiplier.]
What is the relationship between the two functions? [ y2 = 0.004 y1.
This is a transformation, a vertical compression.]
•
•
•
•
•
•
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Activity 3: Regions
Fold a piece of paper again and show participants the regions formed. Before
you fold the paper, you have one region equal to the entire piece of paper.
When you fold the paper once, you have two regions, each of which is onehalf the entire sheet of paper. When you fold the paper again, you have 4
regions, each of which is one-fourth the entire sheet of paper.
Have participants complete the Activity. Then discuss using the following.
Number of
folds
0
1
2
3
4
5
n
Process
1
1
2
1 1
12
1⋅ ⋅ = 1⋅
2 2
2
1 1 1
13
1⋅ ⋅ ⋅ = 1⋅
2 2 2
2
1 1 1 1
14
1⋅ ⋅ ⋅ ⋅ = 1⋅
2 2 2 2
2
1 1 1 1 1
15
1⋅ ⋅ ⋅ ⋅ ⋅ = 1⋅
2 2 2 2 2
2
1 1
1 1
1n
1⋅ ⋅ ⋅ . . . ⋅ ⋅ = 1⋅
2 42 4 2 4 24 32
2
1
1⋅
Fraction of the
Piece of Paper
1
1
2
1
4
1
8
1
16
1
32
1n
1⋅
2
n factors
1n
2
How does this pattern and table compare with the table in Activity 1?
[In Activity 1, we were counting layers of paper. Here we are finding
the fraction of the piece of paper for each region formed by the fold.
The output (range) values in the first table are being doubled with each
new fold. In the second table, the fractions are being multiplied by
one-half with each new fold. The starting values for each table are
different.]
2. y = 1 ⋅
•
Sample:
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paper. The variable y stands for fraction of the paper for each region, so
–0.1 to 1 inches will show most of the graph and the x-axis.
9
1
1
5. 1 ⋅   =
= 0.001953125 . After the ninth fold, the region is one-five
 2
512
hundred twelfth of the original paper. This is comparable to a sheet of
paper from an entire ream of paper.
Note: Some calculators have the capability to change a decimal to a
fraction, which may be useful here.
x
1
1
6. The equation you are solving here is 1 ⋅   =
= 0.0025 . When you
 2
400
1
fold the paper 8 times, you get regions that are
of the paper, which is
256
1
not enough. So you must fold it 9 times to get regions that are
of the
512
1
paper, smaller than
of the paper.
400
Again, a numerically powerful exercise is have participants guess and
1x
1
check a more exact answer to 1 ⋅
=
= 0.0025 , using the home
2
400
screen.
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•
•
•
What meaning does this approach have in this problem situation?
{None, the number of folds must be a whole number.]
How does the table and graph compare with the table and graph in
Activity 1? [In Activity 1, we looked at exponential growth, y = 1 ⋅ 2 x .
x
1

Here we are looking at exponential decay, y = 1 ⋅
. The output
 2
(range) values in Activity 1 are doubling each time. The output
(range) values in Activity 2 are halving each time The starting values
for each table are the same, 1.]
Find a viewing window to compare the graphs of y = 1 ⋅ 2 x and
x
1
y = 1 ⋅   . [The graph of exponential growth, y = 1 ⋅ 2 x , is
 2
x
1
increasing. The graph of exponential decay, y = 1 ⋅   is
 2
decreasing.]
Activity 4: How Big is a Region?
Work through the Activity with participants, discussing using the following.
1. A sheet of typing paper is 93.5 in2.
Number of folds
0
1
Process
93.5
93.5 ⋅
1
2
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Area of a Region
93.5
46.75
400
1 1
12
93.5 ⋅ ⋅ = 93.5 ⋅
2 2
2
1 1 1
13
93.5 ⋅ ⋅ ⋅ = 93.5 ⋅
2 2 2
2
1 1 1 1
14
93.5 ⋅ ⋅ ⋅ ⋅ = 93.5 ⋅
2 2 2 2
2
1 1 1 1 1
15
93.5 ⋅ ⋅ ⋅ ⋅ ⋅ = 93.5 ⋅
2 2 2 2 2
2
1 1
1 1
1n
93.5 ⋅ ⋅ ⋅ . . . ⋅ ⋅ = 93.5 ⋅
2 42 4 2 4 24 32
2
1
2
3
4
5
n
23.375
11.6875
5.84375
2.921875
93.5 ⋅
1n
2
n factors
n
1
3. y = 93.5 ⋅  
 2
• How does this pattern and table compare with the table in Activity 3?
[In Activity 3, we were looking at the fraction of the piece of paper for
each region. Here we are measuring the area of each region. The
output (range) values in both tables are halving each time. The
starting values for each table are different.]
Sample:
paper. The variable y stands for the area of a region in inches2, so –10 to
100 inches2 will show most of the graph and the x-axis.
(10 )
1
6. 93.5 ⋅   ≈ 0.0913 . So you have approximately a tenth of a square
 2
inch of paper after 10 folds. A small pill might measure 0.1 in2.
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x
1
7. The equation you are solving here is 93.5 ⋅   = 1.55 × 10 −5 . When you
 2
fold the paper 23 times you get an area smaller than a plant cell.
Have participants guess and check a more exact answer to
x
1
93.5 ⋅   = 1.55 × 10 −5 , using the home screen.
 2
Discuss the Activity.
• How does the table compare with the table in Activity 3? [The output
(range) values in both tables are halving each time. The starting
values for each table are different. Look at both tables.]
•
How does the graph compare with the graph in Activity 3? Find a
x
1

and
viewing window to compare the graphs of y = 1 ⋅
 2
x
1

.
y = 93.5 ⋅
 2
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x
x
•
1
1
Compare the two rules, y = 1 ⋅   and y = 93.5 ⋅   . [They both
 2
 2
1
have the same common multiplier, . Each function value halves
2
with each increase by 1 of x. The y-intercepts, starting value, are
different.]
•
How can you find the common multiplier of
•
1
by looking at the
2
values in the table? [Divide each y-value by the previous y-value to
obtain the common multiplier.]
What is the relationship between the two functions? [ y2 = 93.5 y1 .
This is a transformation, a vertical stretch.]
Discuss all 4 Activities.
Compare each table with the function that represents the table values. Note
x
that each function follows: y = starting point ⋅ (common multiplier ) or
x
y = starting point ⋅ (common ratio) .
• Which functions are increasing? [The functions where the common
multiplier or ratio is greater than 1 are increasing.]
• What do we call increasing exponential functions? [exponential
growth]
• Which functions are decreasing? [The functions where the common
multiplier or ratio is between 0 and 1.]
• What do we call decreasing exponential functions? [exponential
decay]
• How can you determine the base of the exponential function b in
y = a ⋅ b x from the table values? [Divide each y-value by the previous
y-value.]
• How can you determine the y-intercept (starting point in the problem
situations), a, in y = a ⋅ b x from the table values? [The value of a is
the y-value when x = 0 .]
1. y4, b
2. y3, d
3. y1, c
4. y2, a
5. ii, b
6. i, a
7. iii, c
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Summary:
Building on the work with repeated addition and linear functions, participants
look at repeated multiplication and exponential functions.
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3.1 Exponential Relationships: Activity 1
Activity 1: Paper Folding
Fold a piece of paper in half. Fold it in half again. Continue
folding, filling in the table below.
1.
Number
Process
Number of
of folds
Layers of
Paper
0
1
2
3
4
5
6
n
2. Write a function for the number of layers of paper you will
have if you fold the paper n times.
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Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
Use the home screen, graph, and table to find the following:
5. If you fold the paper 18 times, how many layers of paper will
you have? Write the equation. Show how you got your
solution.
6. A box of paper is 5 reams of paper deep. A ream of paper
has 500 sheets of paper. About how many folds would you
need to be at least as thick as a box of paper? Show how you
found your solution.
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Activity 2: Measure with Paper
A ream of paper measures approximately 2 inches thick.
1. If a ream is 500 sheets of paper, approximately how thick is a
piece of paper?
2. Folding paper again, build the table and find a model.
Number
of folds
0
Process
Thickness
(inches)
1
2
3
4
5
6
n
3. Write a function for how thick the stack will be, in inches, if
you fold the paper n times.
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Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
6. If you fold the paper 15 times, how many inches of paper will
you have? Compare this measurement to something in the
room that has approximately the same measurement.
7. The Eiffel Tower is approximately 1050 feet tall. If you had
a big enough piece of paper, how many folds would you need
to match or exceed that height?
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Activity 3: Regions
When you fold the piece of paper, you split the paper into
regions, bounded by the fold lines. What fraction of the piece of
paper is each region formed? Complete the table below.
Number
of folds
0
Process
Fraction of the
Piece of Paper
1
1
2
3
4
5
6
n
2. Write a function for the fraction of a piece of paper for each
region, if you fold the paper n times.
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Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
5. If you fold the paper 9 times, what fraction of the piece of
paper is each region? Write your answer as a fraction. Give
an example of a different situation where that fraction might
appear.
6. Your school has a paper confetti machine that cuts 8.5” by
11” sheets of paper into about 400 pieces. What is the least
number of times you need to fold the paper to get regions that
1
of the piece of paper?
are no larger than
400
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Activity 4: How Big is a Region?
A piece of paper typing paper measures 8.5” by 11” inches.
1. What is the area in inches2 of a piece of typing paper?
Number
of folds
0
Process
Area of a
Region
1
2
3
4
5
n
3. Write a function for the area of a region, in inches2, if you
fold the paper n times.
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Sketch your graph:
Note your window:
Xmin:
Xmax:
Xscl:
Ymin:
Ymax:
Yscl:
6. If you fold the paper 10 times, what is the area of a region?
Compare this measurement to something in real life that has
approximately the same measurement.
7. Some plant cells have an area of approximately 1.55 × 10 −5
in2. How many folds do you need to have a region with at
least that small of an area?
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3.1 Exponential Relationships: Reflect and Apply
Reflect and Apply
Match each function with a column in the table and with a
graph:
x
y1
y2
y3
y4
-2
9
0.56
45
0.11
-1
3
1.67
15
0.33
0
1
5
5
1
1
0.33
15
1.67
3
2
0.11
45
0.56
9
1. y = 1 ⋅ (3)
a
x
1 x

2. y = 5 ⋅
 3
1 x

3. y = 1 ⋅
 3
b
c
d
4. y = 5 ⋅ (3) x
Match each function with a recursive routine and with a graph:
i. 1000, ENTER
ANS*(1+0.08),
ENTER . . .
ii. 1024, ENTER
ANS*(0.25),
ENTER . . .
iii. 27, ENTER
ANS*(4/3),
ENTER . . .
1
5. y = 1024 ⋅  
 4
x
a
b
c
6. y = 1000 ⋅ (1.08) x
4
7. y = 27 ⋅  
 3
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413
3.1 Exponential Relationships: Student Activity
Student Activity: Recursion Again
Overview:
Students use recursive routines on the home screen to explore three
exponential relationships.
Objective:
Algebra I TEKS
algebraic methods.
Terms:
interest, recursion, fractal
Materials:
Procedures:
Work together with students on Exercise 1.
1. Discuss the problem. Find the interest for one year numerically.
•
•
•
•
•
•
How do we find the interest for one year? [multiply 1000 by 0.08 to
get 80.]
How do we find the total amount after one year? [add 1000 and 80,
which can be written 1000 + 1000 ⋅ 0.08 ]
How much money would you have at the end of the first year?
[$1080]
How do we find the interest for year 2? [multiply 1080 by 0.08 to get
86.40]
How do we find the amount after two years? [add 1080 and 86.40,
which can be written 1080 + 1080 ⋅ 0.08 ]]
How much money would you have at the end of the second year?
[$1166.40]
•
Using the distributive property, how can you rewrite 1000 + 1000 ⋅ 80
and 1080 + 1080 ⋅ 0.08 ? [ 1000 + 1000 ⋅ 0.08 = 1000(1 + 0.08) and
1080 + 1080 ⋅ 0.08 = 1080(1 + 0.08) .]
•
What pattern do you see? [The amount for each year is equal to the
amount from the year before times (1 + 0.08) .]
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•
Do you think this pattern will continue? Find the amount at the end of
year three in two ways. [At the end of the second year, you had
$1166.40 so the interest for year three is 1166.4 ⋅ 0.08 = 93.312 .
Therefore, the total amount at the end of year three is
1166.4 + 93.31 = 1259.712 , so you have $1259.71.. Using our pattern,
the amount at the end of year three is 1166.4(1 + 0.08) = 1259.712 , so
you have $1259.71.]
We can use the pattern, that the amount for each year is equal to the
amount from the year before times (1 + 1.08) , to investigate the scenario
recursively on the home screen.
Start students on the recursive routine and have them solve the remaining
problems.
a.
b.
c.
d.
e.
f.
g.
$1080.00
$1469.33
$2158.90
10 years
15 years
$46,901.61
{0, 1000}, {ANS(1)+1, ANS(2)*(1+0.08)}
2.
a.
b.
c.
d.
e.
Have students complete the Exercise and then discuss.
256 units2
64 units2
0.00390625 units2
stage 6
{0, 1024}, {ANS(1)+1, ANS(2)*(0.25)}
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Note: The fractal produced by this recursive process is called the Sierpinski
Triangle.
3.
Have students complete the Exercise.
In stage 1, you now have 4 segments that are
segment. In other words, you have
1
the length of the previous
3
4
of 27, which is 36.
3
4
of each of the previous segments. In other words, you
3
4
4
4
have of of 27, which is of 36, which is 48.
3
3
3
a. 36 units
b. 48 units
c. 359.5939643 units
d. stage 15
e. {0, 27}, {ANS(1)+1, ANS(2)*(4/3)}
In stage 2, you have
Note: The fractal produced by this recursive process is called the Koch
Snowflake.
Summary:
Using repeated multiplication in recursive routines, students gain intuition for
the exponential growth and decay patterns in interest and two fractals.
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Student Activity: Recursion Again
1. Suppose you have $1000 in a savings account
earning 8% interest compounded annually. Solve each
problem using recursion on the home screen of your
calculator.
a. How much money would you have at the end of the first
year?
b. How much money would you have at the end of five years?
c. How much money would you have at the end of ten years?
d. How long would it take you to double your money?
e. If you have about $3,172, how long has your money been
invested?
f. If you wanted to retire in 50 years, how much money would
you have then?
g. Write the recursive routine that you used.
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2. Given the following sequence of figures, solve each problem
using recursion on the home screen of your calculator.
The area of original triangle is 1024 units2.
,
stage 0
,
stage 1
, . . .
stage 2
a. What is the area of one of the smallest triangles in stage 1?
b. What is the area of one of the smallest triangles in stage 2?
c. What is the area of one of the smallest triangles in stage 9?
d. If the area of one of the smallest triangles is
1
, what is the
4
stage number?
e. Write the recursive routine that you used.
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3. The length of the original segment is 27 units long. The
1
length of each new segment is the length of the previous
3
segment.
Stage 0
Stage 1
Stage 2
a. What is the total length of the figure in stage 1?
b. What is the total length of the figure in stage 2?
c. What is the total length of the figure in stage 9?
d. If the length of the figure is approximately 2020 units, what
is the stage number?
e. Write the recursive routine that you used.
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3.2 Exponential Growth and Decay: Leaders’ Notes
3.2 Exponential Growth and Decay
Overview:
Participants find models for exponential growth and decay situations.
Objective:
Algebra I TEKS
algebraic methods.
Terms:
growth, decay, multiplier, rate, quotient, ratio, percent increase, percent
decrease
Materials:
Procedures:
Recall the exponential relationships explored in the Student Activity:
Recursion Again. If you did not do these problems with participants, divide
into three groups and assign each of the groups one of the problems to
investigate.
Use Transparency 1 to discuss the relationship between the constant multiplier
and the percent increase or decrease.
You can also use Transparency 2 to develop the relationship live, using
Transparency 1 as a guide.
• What is the relationship between the constant multiplier and the
percent increase or decrease? [The constant multiplier is one plus the
percent increase for exponential growth. It is one minus the percent
decrease for exponential decay.]
• What does this form, writing the constant multiplier as 1+ percent
increase or 1 – percent decrease, emphasize? [The percent increase or
percent decrease.]
• What is the multiplier for the interest earning problem? [1.08]
• How can we write the constant multiplier for the interest earning
problem? [1.08 or 1+0.08]
• So what is the percent increase? [8%]
• What is the multiplier for the triangle problem? [0.25]
• How can we write the constant multiplier for the triangle problem?
[1 − 0.75 ]
• So what is the percent decrease? [75%]
4
• What is the multiplier for the snowflake problem? [ ]
3
• How can we write the constant multiplier for the snowflake problem?
1
[1 + ]
3
• So what is the percent increase? [33.3%]
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In the previous 3.1 Exponential Relationships, participants were given a
situation and, using a table, they found exponential functions to model the
situation. Now we will give participants percent increase or percent decrease,
along with an initial amount and they can find exponential functions to model
situations without relying on a table.
Use Transparency 3 to bridge this gap from the table to the function:
x
y = starting amount ⋅ (rate) , where rate = 1 + percent increase for
exponential growth or rate = 1 − percent decrease for exponential decay.
Activity 1: Exponential Growth
Have participants work together on the Activity and then discuss, using the
following.
1. Write a function for each offer.
x
A. y = 1000(1 + 0.04)
x
B. y = 1000(1 + 0.03)
x
C. y = 1000(1 + 0.065)
2. Compare the three offers in a table.
Offer A – 4%
1 year
$1040.00
2 years
$1081.60
5 years
$1216.65
10 years
$1480.24
20 years
$2191.12
30 years
$3243.40
Offer B – 3%
$1030.00
$1060.90
$1159.27
$1343.91
$1806.11
$2427.26
Offer C – 6.5%
$1065.00
$1134.23
$1370.09
$1877.14
$3523.65
$6614.37
3. Compare the three offers graphically.
Ask participants to generate a list of possible questions to ask students about
the three offers. These might include:
• How long does it take to double your money with each offer?
• How long does it take you to save $XX with each offer?
• Compare the three offers over time.
Have participants solve one of their questions in at least three ways and share
their strategies.
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Ask participants:
• What information does the graph give you that the equations do not?
[The relative growth of the money over time.]
• How could you find the multiplier for each offer by looking only at the
table of values? [You would divide the amount at the end of year 2 by
the amount at the end of year 1. This ratio is the common multiplier
you use to get the amount for the next year.]
• Could you divide any amount by the previous amount in this particular
table? [No, because the values in the input column do not change by 1
consistently. In order to divide an outcome by a previous outcome to
get the multiplier, the input values must increase by 1.]
• For example, to find the multiplier for Offer A, can you divide the
amount after year 5, $1216.65, by the amount after year 2, $1081.60?
1216.65
[No,
≈ 1.12 which is the percent increase over 3 years, not the
1081.60
annual percent increase.]
Activity 2: Exponential Decay
Explain the scenario to participants. As you eat some substances, the amount
in the bloodstream eventually reaches a maximum amount. Thereafter, the
substance is flushed from the bloodstream, in these cases by a certain percent
every hour. We are interested in how much of the substance is remaining in
the bloodstream t hours after the substance has reached the maximum level.
1. Write a function for the amount of substance remaining in the bloodstream
t hours after the maximum level is reached.
Encourage participants to write the functions below in both forms shown
to help them later recognize both forms.
t
t
A. y = 30 ⋅ (1 − 0.25) = 30 ⋅ (0.75)
t
t
B. y = 30 ⋅ (1 − 0.15) = 30 ⋅ (0.85)
t
t
C. y = 30 ⋅ (1 − 0.2) = 30 ⋅ (0.8)
2. Compare the three situations in a table.
Time after
Caffeine – Child Caffeine – Adult
Maximum
Level
1 hour
22.5
25.5
2 hours
16.88
21.68
3 hours
12.66
18.42
4 hours
9.49
15.66
5 hours
7.12
13.31
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Vitamin
24
19.2
15.36
12.29
9.83
422
3. Compare the three situations graphically.
Ask participants to generate a list of possible questions to ask students about
the three situations. These might include:
• How long does it take to halve the amount in the bloodstream with
each situation?
• How long does it take you to have XX amount in the bloodstream with
each situation?
• When will you have less than 1 mg of each substance in the
bloodstream?
Have participants solve one of their questions in at least three ways and share
their strategies.
Ask participants:
• How could you find the multiplier for each offer by looking only at the
table of values? [You would divide the amount present at the end of
the second hour by the amount at the end of the first hour. This ratio is
the common multiplier you use to get the amount for the next hour.]
• Could you divide any amount by the previous amount in this particular
table? [Yes, because the values in the input column change by 1
consistently. Any quotient of successive output values in this table
will give you the multiplier for this situation.]
1. y4, b
2. y3, a
3. y1, c
4. y2, d
Summary:
5.
6.
7.
8.
ii, b
iv, c
i, a
iii, d
The big idea in these activities is for participants to write exponential models
for situations given an initial amount and a percent increase or percent
decrease.
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3.2 Exponential Growth and Decay: Transparency 1
Transparency 1: The Common Multiplier
Sometimes it is more useful to write the multiplier as:
exponential growth: (1 + percent increase)
exponential decay: (1 – percent decrease)
Earning 8% interest on $1000
1 + 0.08
Area of the smallest triangle
,
stage 0
,
stage 1
, . . .
stage 2
0.25 = 1 − 0.75
Total length of the segment
Stage 2
4
1
= 1+
3
3
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Transparency 2: The Common Multiplier
Write the multiplier in terms of percent increase or decrease:
exponential growth:
exponential decay:
Earning 8% interest on $1000
Area of the smallest triangle
,
stage 0
,
stage 1
, . . .
stage 2
Total length of the segment
Stage 2
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Transparency 3: Interest
Suppose you have $500 in a savings account earning 6% annual
interest.
We can think of 6% annual interest as a 6% annual growth rate.
Balance
0
1
2
3
4
t
Process
500
500(1 + .06)
500(1 + .06)(1 + .06) = 500(1 + .06)2
500(1 + .06)(1 + .06)(1 + .06) = 500(1 + .06)3
Total
Amount
$500.00
$530.00
$561.80
$595.51
500(1 + .06)t
y = starting amount ⋅ (rate) x
where rate = 1 + percent increase for exponential growth and
rate = 1 − percent decrease for exponential decay.
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3.2 Exponential Growth and Decay: Activity 1
Activity 1: Exponential Growth
Suppose you have $1000 to invest. To simplify the
comparisons, consider only interest compounded annually.
A. Your credit union’s savings account is offering 4% interest.
B. The corner bank’s savings account is offering 3% interest.
C. A Certificate of Deposit (CD) is offering 6.5% interest.
1. Write a function for each offer.
A.
B.
C.
2. Compare the three offers in a table.
1 year
2 years
5 years
10 years
20 years
30 years
3. Compare the three offers graphically.
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3.2 Exponential Growth and Decay: Activity 2
Activity 2: Exponential Decay
A person drinks a caffeinated soda and takes a vitamin. The
caffeine and the vitamin in the bloodstream each reach a
maximum level of 30 milligrams.
A. Caffeine is flushed out of a child’s blood stream at a rate of
about 25% an hour.
B. Caffeine is flushed out of an adult’s blood stream at a rate of
about 15% an hour.
C. The vitamin is flushed out of a person’s blood stream at a
rate of about 20% an hour.
1. Write a function for the amount of substance remaining in the
bloodstream t hours after the maximum level is reached.
A.
B.
C.
2. Compare the three situations in a table.
Time after
Maximum
Level
1 hour
2 hours
3 hours
4 hours
5 hours
3. Compare the three situations graphically.
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3.2 Exponential Growth and Decay: Reflect and Apply
Reflect and Apply
Match each function with a table and a graph:
x
y1
y2
y3
b
y4
1. y = 20 ⋅ (0.75)
-2 16.5 8.89 24.7 35.6
-1 18.2 13.3 22.2 26.7
0
20
20
20
20
1
22
30
18
15
2
24.2
45
x
a
2. y = 20 ⋅ (0.9) x
3. y = 20 ⋅ (1.1) x
c
d
16.2 11.3
4. y = 20 ⋅ (1.5) x
Match each function with a recursive routine and a graph:
i. 1000, ENTER
ANS*(1+0.03),
ENTER . . .
5. y = 1000 ⋅ (1.09) x
ii. 1000, ENTER
ANS*(1+0.09),
a
b
ENTER . . .
x
c
6. y = 500 ⋅ (0.7)
iii. 500, ENTER
ANS*(0.9),
7. y = 1000 ⋅ (1.03) x
ENTER . . .
iv. 500, ENTER
ANS*(0.7),
ENTER . . .
d
8. y = 500 ⋅ (0.9) x
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3.2 Exponential Growth and Decay: Student Activity
Student Activity: On the Wall
Overview:
Students use different sized paper to model exponential growth and decay
graphs.
Objective:
Algebra I TEKS
algebraic methods.
Terms:
exponential function, growth, decay
Materials:
sticky notes, poster boards, large blank paper (for coordinate axes), markers,
tape
Procedures:
Do this as a whole class activity.
One half of the class will split into groups 1 – 4 and construct the respective
graphs using sticky notes. The other half of the class will split into groups 1 –
4 and construct the same graphs using poster board or other similar sized
paper.
Ask the class to predict the difference in the graphs made with sticky notes
and the graphs made with poster boards. Have them discuss their construction
strategies.
Note: These constructions only simulate exponential growth and decay. They
are not accurate graphs of exponential functions. The domain for these
physical models is integers. The domain for exponential functions is all real
numbers. The idea here is an attempt to give students a geometric feel for
exponential growth and decay by having students take slips of paper and
physically double them, triple them, halve them, and cut them into thirds.
Then they place the slips on a large coordinate axes system to get a concrete
feel for exponential graphs.
After the graphs are completed and hanging on the wall, have students do a
gallery tour and note similarities and differences among the graphs. Then
have a class discussion about the activity.
Some points to bring out in the discussion include the following.
• Why did the graphs of y = 2 x and y = 3x using sticky notes contain
more in the first quadrant than did the graphs using poster board?
[The poster board is so big that it did not allow students to stack many
in the first quadrant. The sticky notes are small enough to allow many
more to be stacked in the first quadrant.]
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•
•
Where else did the large size of the poster board compared to the
x
x
1
1
sticky notes limit the graph? [In the graphs of y =   and y =   ,
 2
 3
the graphs in the second quadrant were limited.]
Why did the graphs of y = 2 x and y = 3x using poster board contain
more in the second quadrant than did the graphs using sticky notes?
[The poster board is big enough that it allowed students to halve and
third many more times in the second quadrant. The sticky notes are so
small that students could not split the slips into as many pieces in the
second quadrant.]
Sample graphs for y = 2 x :
Summary:
Students gain intuition about exponential growth and decay as they physically
produce simulations of exponential functions using concrete materials.
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Student Activity: On the Wall
Use the height of your slip of paper as a unit. Simulate the
graph of the following exponential functions by using pieces of
the paper and partial pieces of the paper affixed to a large grid
on the wall.
Sketch your graph here:
1. y = 2 x
2. y = 3 x
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1
3. y =  
 2
x
1
4. y =  
 3
x
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3.3 Exponential Models: Leaders’ Notes
3.3 Exponential Models
Overview:
Participants find exponential models for given data sets.
Objective:
Algebra I TEKS
algebraic methods.
Terms:
growth, decay, ratio, multiplier, quotient
Materials:
Procedures:
Participants will apply what they have learned about exponential equations to
find exponential models for data.
Activity 1: Population Growth
Work through Exercises 1 – 4 with participants.
1. Determine the growth rate by finding successive quotients as shown
below:
Year
Population (billions)
Quotients
1980
0
4.46
4.53/4.46=1.0157
1981
1
4.53
4.61/4.53=1.0177
1982
2
4.61
4.69/4.61=1.0174
1983
3
4.69
If you find quotients on the graphing calculator, you can average the
quotients.
So the growth rate is about 1.0172, which is to say that the population is
growing at about 1.72% per year.
2. To find an equation to model the growth using the rule
t
y = starting amount ⋅ (rate) , use years from 0 to 9 for 1980 to 1989. Then
the starting amount (y-intercept) is 4.46. The rate is 1.0172.
x
Therefore, the equation to model the growth is y = 4.46 ⋅ (1.0172) .
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•
How can you use the table on your calculator to check the accuracy of
the model? [See below. Look to see that the model returns close to
the same y-values as in the original data set.]
3. To predict the population in 1999, remember that we substituted years
0 – 9 for 1980 – 1989. So the year 1999 is year 19 for our model. Find
the prediction in a few ways.
4. The year 2010 is year 30 for our model. Find the prediction in a few
ways.
Have participants work on Exercises 5 – 9. Make sure participants understand
that they are to create a new model, using only the data in the new table.
Discuss, using the following. Ask a group to present their work.
5. Determine the growth rate by finding successive quotients as shown
below:
Year
Population (billions)
Quotients
1980
0
5.28
5.37/5.28=1.0170
1981
1
5.37
5.45/5.37=1.0149
1982
2
5.45
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If you find quotients on the graphing calculator, you can average the
quotients.
So the growth rate is about 1.0143, which is to say that the population is
growing at about 1.43% per year.
6. To find an equation to model the growth using the rule
t
y = starting amount ⋅ (rate) , use years from 0 to 9 for 1990 to 1999. Then
the starting amount (y-intercept) is 5.28. The rate is 1.0143.
x
So the equation to model the growth is y = 5.28 ⋅ (1.0143) .
•
How can you use the table on your calculator to check the accuracy of
the model? [See below. Look to see that the model returns close to
the same y-values as in the original data set.]
7. To predict the population in 2010, remember that we substituted years 0 –
9 for 1990 – 1999. So the year 2010 is year 20 for our model. Find the
prediction in a few ways.
8. Using the year 1998 as year 0 for our model, the year 2010 is year 12 in
our model. The population reported in 1998 was 5.92 billion so this is the
x
starting point. The model is y = 5.92 ⋅ (1.0133) . Find the predictions in a
few ways.
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•
•
•
Why do the three predictions for the population in 2010 differ? [Each
prediction was found using a model with a different growth rate.]
Compare the growth rate in the 80’s with the rate in the 90’s. [The
growth rate in the 80’s was about 1.72%. The growth rate in the 90’s
was about 1.43%. The growth rate decreased.]
How does the growth rate reported by the UN in 1998 compare with
the two rates you found? [The growth rate reported in 1998 was 1.33
percent, which is less than the two that we found. In fact the world’s
population growth rate has been declining for a few years. See below
and the Transparency.]
Year
1962
1970
1980
1990
1998
•
•
Population
Growth Rate
3,136,197,751
2.19
3,707,610,112
2.07
4,456,705,217
1.70
5,283,755,345
1.56
5,924,574,901
1.31
How do the growth rates shown in the Transparency compare to the
rates you found?
What might be some factors influencing the UN to predict lower
growth rates in the future?
Activity 2: Cooling Down
Have participants do the Activity. Then discuss, using the following. Ask a
group to present their work.
1. Find the cooling rate by taking successive quotients.
The temperature is decreasing by about 4% each minute. The cooling rate
is about 0.96.
2. An equation to model the temperature decrease is y = 46 ⋅ (0.96) .
x
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3. Use the model to predict when the temperature will be 5 ˚C.
1. A linear model for 1980-1989 is y = 4.46 + 0.0822 x .
2. Compare the model and the data. The model returns slightly higher values
(in billions.)
3. A linear model for 1990-1999 is y = 5.28 + 0.08 x .
4. Compare the model and the data. The model returns slightly lower values
(in billions.)
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5. A linear model for the cooling data is y = 46 − 1.56 x .
6. The model returns slightly higher temperatures.
Summary:
The big idea in these activities is to find the multiplier from data from
exponential situations by taking successive quotients. If successive quotients
are constant, then an exponential model is reasonable. With the multiplier and
the y-intercept, participants can find exponential models for data.
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3.3 Exponential Models: Transparency
Transparency: World Population Trends
World Population Size:
past estimates and medium-, high- and low fertility variants,
1950-2050 (billions)
*Source: United Nations Population Division, World Population Prospects: The 1998 Revision, forthcoming
.
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3.3 Exponential Models: Activity 1
Activity 1: Population Growth
The Population of the World (in billions)
Year
1980
1981
1982
1983
1984
1985
1986
1987
1988
1989
0
1
2
3
4
5
6
7
8
9
Population
(billions)
4.46
4.53
4.61
4.69
4.77
4.85
4.94
5.02
5.11
5.20
*1998 Revision of the official United Nations world population estimates and projections
1. Determine the growth rate.
2. Find an equation to model the population growth.
3. Use your model to predict the population in 1999.
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The Population of the World (in billions)
Year
1990
1991
1992
1993
1994
1995
1996
1997
1998
1999
0
1
2
3
4
5
6
7
8
9
Population
(billions)
5.28
5.37
5.45
5.53
5.61
5.69
5.77
5.85
5.92
6.00
*1998 Revision of the official United Nations world population estimates and projections
Create a new model, using the data in this table.
5. Determine the growth rate.
6. Find an equation to model the population growth.
8. In 1998, the United Nations reported the current population
growth rate as 1.33 per cent. Create a model using this
information and the population for 1998 in the table. Use
this model to predict the population in 2010.
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Activity 2: Cooling Down
A student placed a hot cup of water in the freezer and recorded
the following temperatures at the indicated times.
Time (min)
0 1 2 3 4
Temperature 46 44 42 41 39
(C˚)
5
6
37.5 36
7
34.5
8
33
9
32
1. Find the cooling rate.
2. Find an equation to model the temperature decrease.
3. Use your model to predict when the water will be about 5˚C.
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3.3 Exponential Models: Reflect and Apply
Reflect and Apply
Although population and cooling data is often modeled with
exponential models, a linear model can be quite accurate for
short sections of data.
1. Find a linear model for the world population data for 19801989.
2. Use the table on your graphing calculator to check the model
for accuracy for the years for 1980-1989.
3. Find a linear model for the world population data for 19901999.
4. Use the table on your graphing calculator to check the model
for accuracy for the years for 1990-1999.
5. Find a linear model for the cooling data.
6. Use the table on your graphing calculator to check the model.
7. What do you think about the appropriateness of either model?
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4.1 Bounce It: Leaders’ Notes
4.1 Bounce It!
Overview:
Collecting three sets of data from a bouncing ball experiment, participants
find appropriate models and justify their choices.
Objective:
Algebra I TEKS
relationships.
algebraic methods.
Terms:
linear function, quadratic function, exponential function, parameter,
acceleration
Materials:
balls, data collection devices, motion detectors, graphing calculators
Procedures:
Participants should be seated at tables in groups of 3 – 4. Do all of the
Activities with participants, using the data that you collect. After you have
worked through an Activity with them using your data, have them complete
the Activity with the data that they collect.
Activity 1: Collect the Data
Describe the procedure for collecting the data and demonstrate by dropping a
ball under a motion detector.
1. Encourage participants to predict the graph of the ball’s distance from the
floor versus time.
Math Note: The data collected by the motion detector is actually the distance
from the motion detector versus the time. The program we used then
transformed the data to the distance from the floor versus the time.
2. It is important that the graph shows at least 5 good bounces.
3. This sample data is from a racket ball.
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Activity 2: A Bounce
Do the Activity with participants, using your collected data.
1. A quadratic function is an appropriate model. To answer the question
why, you can look at second differences in your data, which should be
fairly constant. There will probably be some glitches around where the
ball hits the floor, because you are finding differences between two sets of
quadratic data (2 different bounces).
2. Find a model with participants.
• How can we fit a quadratic to the first complete bounce? [A sample
follows.]
Graph y = x 2 over the original graph. Reflect over the x-axis by graphing
y = −x2 .
Trace to the vertex of the first complete bounce. Shift right 0.86 (the xcoordinate of the vertex). Shift up 2.241 (the y-coordinate of the vertex).
Now guess and check the stretch factor.
Does -16 have any particular significance? It is because of the
acceleration due to the force of gravity in the physics position equation
1
d = at 2 + vt + d , where a is the acceleration due to gravity, which is
2
equal to -32 ft /sec2 or -9.8 m/sec2.
4. We reflected the parent function y = x 2 over the x-axis, horizontally
translated it, vertically translated it, and vertically stretched it.
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5. Now have participants find quadratics to model some of the other bounces.
You may want to graph equations for all the bounces if time permits.
Have participants complete the Activity using the data they collected in their
groups.
Activity 3: Bounce Height versus Bounce Number
1. Do not fill in the maximum height of bounce number 0. We will figure
that height later.
Sample data:
Bounce
Maximum Height of
Number
Bounce
0
1
2.241
2
1.709
3
1.336
4
1.01
5
0.778
6
0.607
2. Do not erase the original data, as you may need it later. Put the data into 2
different lists. If your calculator allows you to name lists, this may be a
time to do so.
3. An exponential function is an appropriate model.
Have participants look at first and second differences to determine that the
data is neither linear nor quadratic. Note that both the first and second
differences decrease, they are not constant.
Take successive quotients. Note that the quotients do not continually
increase or decrease, but sort of cluster around 0.77.
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4. To find an appropriate exponential model, we need to figure the height of
bounce number 0. By finding the quotients above, we found the common
multiplier.
• How can you use the common multiplier to work backwards to find
the height of bounce number 0? [Find
maximum height of bounce # 2
2.241
. For our sample data,
= 2.91 .]
multiplier
0.77
So for our sample data, the starting point, is 2.91 and the common
x
multiplier is 0.77, so our model is y = 2.91 ⋅ (0.77)
Have participants complete the Activity using the data they collected in their
groups.
Activity 4: Bounce Height versus Drop Height
1. Use your data from the table in Activity 3. Use the height you found for
bounce number 0 for the first drop height.
Drop Height
Bounce Height
2.91
2.241
2.241
1.709
1.709
1.336
1.336
1.01
1.01
0.778
0.778
0.607
2. Have participants predict what the scatter plot will look like before you
graph it.
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3. To find an appropriate model, first note that the domain values do not have
∆y
should be
a common difference. Remember, for data to be linear,
∆x
∆y
∆y
constant. Find
for consecutive points and you should find
to be
∆x
∆x
fairly constant.
∆y
Our sample data found
in L6.
∆x
4. So, since
∆y
is fairly constant, we use a point (2.91, 2.241) and
∆x
∆y
= 0.76 (from mean(L6)) in y = m( x − x1 ) + y1 to get
∆x
y = 0.76( x − 2.91) + 2.241. (Stretch the line y = x by 0.76, horizontally
translate it right 2.91 and vertically translate it up 2.241).
Have participants complete the Activity using the data they collected in
their groups.
Discuss Activities 2 – 4.
How do you determine if a linear model is appropriate for data? [Look
∆y
, by taking first differences.]
for a constant
∆x
How do you determine if a quadratic model is appropriate for data?
[Look for constant second differences.]
How do you determine if an exponential model is appropriate for data?
[Look for constant quotients.]
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Note: Our purpose here is for participants to learn more about linear,
quadratic, and exponential functions. Participants must know the important
differences between linear, quadratic, and exponential data, use these
differences to choose a model, and then they adjust parameters in the models
to approximate good fits for data. This helps participants make connections
between and differentiate among the three functions.
Our purpose is not to teach statistical analysis. Do not get bogged down in
statistical discussions.
After you have completed all 4 Activities, split the group into smaller groups
and work on the Student Activities as time allows. Have groups present their
work.
Note: The Student Activities 1 – 2: Pattern Blocks and Throw Up! are
finding quadratic models for data. The Student Activities 3 –4: Radioactive
Decay and Pendulum Decay are finding exponential models for data.
1. y = 0.0526 x 2 . The model is quadratic because, as seen below, the second
differences are relatively constant.
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2. y = 23000(1 − 0.15) or y = 23000(0.85) . The model is exponential
because the first and second differences are not constant, but the quotients
are, as seen below.
x
x
3. y = 16.7 − 0.007 x . The model is linear because first differences are
constant.
Summary:
Collecting three sets of data from a bouncing ball experiment, participants
find appropriate models and justify their choices.
Spring 2001
451
4.1 Bounce It: Activity 1
Activity 1: Collect the Data
1. Read the directions below.
Predict the graph of the distance
of the ball from the ground versus
time.
0.5 m
2. Using a motion detector, a data collection device, and an
appropriate program, do the following.
Hold the motion detector at least 0.5 meters above the
ball.
Drop the ball and let it bounce under the motion detector.
Collect distance data for about 5 seconds.
Collect data for a least 5 good bounces.
Repeat if necessary.
3. Sketch the resulting graph:
Spring 2001
452
Activity 2: A Bounce
1. Choose the first complete bounce on the graph. What kind of
function would be an appropriate model for this data? Why?
2. Find a model. Write the function.
3. Sketch the data and the model.
4. List the transformations you used to find the model.
5. Choose another complete bounce and find a model for that
data.
Spring 2001
453
Activity 3: Bounce Height versus Bounce Number
1. Use the trace feature to find the maximum height for each
full bounce. Do not fill in the height for bounce number 0.
Bounce
Number
0
1
2
3
4
5
6
Maximum Height
of Bounce
2. Make a scatter plot of (bounce number, maximum height) in
an appropriate viewing window. Sketch it. (Do not lose the
original data!)
3. What kind of function would be an appropriate model?
Why?
4. Find an appropriate model. Write the function. Sketch it
above.
Spring 2001
454
Activity 4: Bounce Height versus Drop Height
1. Using the data from Activity 3, fill in the table.
Drop Height
Bounce Height
2. Make a scatter plot of (drop height, bounce height) in an
appropriate viewing window. Sketch it. (Do not lose the
original data!)
3. What kind of function would be an appropriate model?
Why?
4. Find an appropriate model. Write the function. Sketch it
above.
Spring 2001
455
4.1 Bounce It: Reflect and Apply
Reflect and Apply
Find a model for each data set and justify your choice.
1. When a car stops, the
braking distance depends
on the speed of the car.
Speed Braking
(mph) Distance
(ft)
10
5
20
21
30
47
40
84
50
132
60
189
70
258
2. A car loses value each
year. This is called
depreciation.
Year
Car’s
Value
0
1
2
3
4
5
$23,000
$19,550
$16,618
$14125
$12,006
$10,205
*after the driver has observed an
obstacle and has begun braking
3. The air temperature outside of a plane depends on the altitude
of the plane.
Altitude Temperature
(m)
(C˚)
0
16.7
1000
9.7
2000
2.7
3000
−4.3
4000
−11.3
Spring 2001
456
4.1 Bounce It: Student Activity 1
Student Activity 1: Pattern Blocks
Overview:
Students identify functional relationships using pattern blocks.
Objective:
Algebra I TEKS
Terms:
trapezoid, rhombus, triangle, hexagon
Materials:
pattern blocks, graphing calculator
Procedures:
Explain to students that the Transparency of Hexagons is to be used with each
of the patterns.
Build the first Hexagon with trapezoids. Then have students complete the rest
of the Activity.
Note: The hexagon number is also the unit length of a side.
1. It takes 2 trapezoids to build the first hexagon.
2. It takes 8 trapezoids to build the second hexagon.
3.
Hexagon Number
Number of Trapezoids
1
2
2
8
3
18
n
2n2
4. Scatter plot:
5. y = 2 x 2 . To determine the parent function for the model, students can find
second differences to determine that a quadratic function is an appropriate
model, because second differences are constant.
Have students use the homescreen, tables, and graphs to find the answers.
6. 128 trapezoids
7. 14th hexagon
Spring 2001
457
8.
Hexagon Number
1
2
3
4
5
n
9. Scatter plot:
Number of Rhombi
3
12
27
48
75
3n 2
10. y = 3 x 2 . Again, look at second differences to determine that a quadratic
function is an appropriate model because second differences are constant.
11. 243 trapezoids
12. 13th hexagon
13.
Hexagon Number
Number of Triangles
1
6
2
24
3
54
4
96
5
150
n
6n 2
14. Scatter plot:
15. y = 6 x 2 . Again, look at second differences to determine that a quadratic
function is an appropriate model because second differences are constant.
16. 294 triangles
17. 15th hexagon
Summary:
Using pattern blocks to build hexagons, students find patterns and write
function rules to model the situation.
Spring 2001
458
Student Activity: Transparency of Hexagons
Spring 2001
459
Student Activity 1: Pattern Blocks
1. How many trapezoids does it take to build the first hexagon?
2. How many trapezoids does it take to build the second
hexagon?
3. Fill in the table.
Hexagon Number
1
2
3
Number of Trapezoids
n
4. Create a scatter plot of (hexagon number, number of
trapezoids) and sketch.
5. Find a model for the data.
Use your model to find the following in at least 3 ways:
6. How many trapezoids do you need to build the 8th hexagon?
7. If you use 392 trapezoids, what hexagon number did you
build?
Spring 2001
460
How many rhombi does it take to build the hexagons?
Hexagon Number
1
2
3
Number of Rhombi
n
9. Create a scatter plot of (hexagon number, number of rhombi)
and sketch.
11. How many rhombi do you need to build the 9th hexagon?
12. If you use 507 rhombi, what hexagon number did you build?
Spring 2001
461
How many triangles does it take to build the hexagons?
Hexagon Number
1
2
3
Number of Triangles
n
14. Create a scatter plot of (hexagon number, number of
triangles) and sketch.
16. How many triangles do you need to build the 7th hexagon?
17. If you use 1350 triangles, what hexagon number did you
build?
Spring 2001
462
Student Activity 2: Throw Up!
Overview:
Students explore their own projectile motion problem and find an equation to
model the height of the ball versus time.
Objective:
Algebra I TEKS
Terms:
acceleration, velocity, speed,
Materials:
balls, stop watches, graphing calculators
Procedures:
Students should work in groups of 3 – 4.
Discuss each team member’s responsibilities as you demonstrate the Activity.
Throw the ball in the air and note that the timer should start timing when the
ball leaves the thrower’s hand and should end timing when the ball hits the
ground.
Have students estimate, before they throw:
How long do you think the ball will be in the air? [Students’ estimates
will vary, but they will probably be much too high.]
How high do you think the ball will go?
With what initial velocity do you think you threw the ball?
Send the groups out to collect the data. When they return, help groups
understand how to guess and check their initial velocity given the time they
collected. An example follows.
1. When we threw a racket ball, we timed that it was in the air for 2.55
seconds.
2. Substituting 5 for v0 , h = −16t 2 + v0 t + 5
3. We estimated that our thrower threw at 50 ft/sec, so h = −16t 2 + 50t + 5 .
This initial guess is just to give students a starting place.
4. Students will need to find an appropriate viewing window and may need
to adjust it as well as the initial velocity.
Spring 2001
463
5. The maximum height of our throw was 28.58 feet. Students can use the
calculator to find the maximum of the function, use the graph to zoom in
on the maximum, use the table to zoom in on the maximum, or they could
even use the home screen to guess and check until they found the
maximum.
•
What is another method of finding the initial velocity, given the time?
[The equation to solve is 0 = −16t 2 + v0 t + 5 , for t = 2.55 . Solve
ft
2
]
0 = −16(2.55) + v0 (2.55) + 5 for v0 . v0 ≈ 38.839
sec
Extension Questions:
Assuming the initial height is 5 feet, answer the following for other throws.
• How high did the ball go if its airtime was 4.5 seconds?
• What was the initial velocity if the maximum height was 90 ft?
• What will be the maximum height of a ball that reaches an altitude of
30 feet after half a second?
Spring 2001
464
•
•
•
•
•
•
•
How long would the ball be in the air if Mark McGuire hit a ball
vertically and it went up to a height of 1200 feet?
How high would the ball go if Sammy Sosa hit the ball with an initial
velocity of 200 feet per second?
What would be the initial velocity of a ball hit vertically by Juan
Gonzalez if it had an airtime of 10 seconds?
Which ball went higher, Mark’s, Sammy’s, or Juan’s?
What initial velocity would you need if you want the ball to go as high
as the Sears tower (approximately 1454 feet)?
How long would it take for this ball to complete its flight?
Does it take longer for the ball to go up to its maximum height or to
come down from its maximum height?
Collect the times and maximum heights from the various groups. Note that
the longer the ball was in the air, the higher its maximum height.
Congratulate the highest thrower.
Summary:
Using technology, students approximate the velocity with which they threw a
ball. This helps students gain intuition for velocity and also for the quadratic
model.
Spring 2001
465
Student Activity 2: Throw Up!
Go outside or to a place with a high ceiling.
The thrower throws a ball straight into the air as high as
he/she can.
The timer times how long the ball is in the air, from the
time it leaves the thrower’s hand until it hits the ground.
The recorder records the time.
The position equation for the height of the ball is
1
ft
h = − at 2 + v0 t + h0 , where a = 32 2 , v0 is the initial velocity,
2
sec
and h0 is the initial height.
1. How long was the ball in the air?
2. Assuming that the thrower released the ball at a height of 5
ft., what is the position equation for the throw?
3. How fast do you think the thrower threw the ball? In other
words, what was the ball’s initial velocity? Estimate and
substitute this value into the position equation.
4. Graph your equation in Exercise 3. Does it show the ball in
the air for the correct amount of time? Adjust your estimate
for the ball’s initial velocity until it shows the ball in the air
for approximately the correct amount of time. Write your
equation and sketch the graph.
5. Find the maximum height of the ball. Describe your method.
Spring 2001
466
Student Activity 3: Radioactive Decay
Overview:
Students use graphing calculators to simulate the radioactive decay of radon
gas. They find a model for the data and use the model to predict.
Objective:
Algebra I TEKS
algebraic methods.
Terms:
decay, radioactive
Materials:
Procedures:
Discuss briefly radioactive decay – that radioactive substances decay over a
period of time. This means that it turns into a different material. Different
radioactive materials decay at different rates over time. Not all of the material
decays at the same time, but a certain percentage decays in a certain period of
time. Carbon-14, a radioactive isotope found in living material, decays so
slowly that scientists use it to date fossils. The half-life of Carbon-14 is 5730
years.
Explain to students that they will be simulating the radioactive decay of the
gas radon, which decays at a rate of about 16.7% per day. Students will find
17 random integers out of 100 to approximate the 16.7% decay rate.
Explain the data collection procedure to students. Have students work
through the activity in groups.
Activity:
1. Encourage students to predict.
2. Sample data.
First find 17 random integers between 0 – 99 for Day 1.
Spring 2001
467
0
10
20
30
40
50
60
70
80
90
1
11
21
31
41
51
61
71
81
91
2
12
22
32
42
52
62
72
82
92
3
13
23
33
43
53
63
73
83
93
4
14
24
34
44
54
64
74
84
94
5
15
25
35
45
55
65
75
85
95
6
16
26
36
46
56
66
76
86
96
7
17
27
37
47
57
67
77
87
97
8
18
28
38
48
58
68
78
88
98
9
19
29
39
49
59
69
79
89
99
Day
0
1
2
3
4
5
6
7
8
9
10
Units Remaining
100
85
Next find 17 random integers between 0 – 99 for Day 2. Another way to
do this is to store the generated integers into a list and then sort the list.
This makes it easier to cross off the numbers because they are in order and
also it makes it easier to spot duplicates, which are ignored.
0
10
20
30
40
50
60
70
80
90
1
11
21
31
41
51
61
71
81
91
2
12
22
32
42
52
62
72
82
92
3
13
23
33
43
53
63
73
83
93
4
14
24
34
44
54
64
74
84
94
5
15
25
35
45
55
65
75
85
95
6
16
26
36
46
56
66
76
86
96
7
17
27
37
47
57
67
77
87
97
8
18
28
38
48
58
68
78
88
98
9
19
29
39
49
59
69
79
89
99
Spring 2001
Day
0
1
2
3
4
5
6
7
8
9
10
Units Remaining
100
85
72
468
After repeating, our sample data looked like:
Day
Units
Remaining
0
100
1
85
2
72
3
64
4
56
5
51
6
43
7
34
8
31
9
28
10
23
3. For our sample data:
4. For our sample data:
5. For our sample data, the multiplier is 0.8642 and the number of units of
radon at day 0 is 100. So our model is y = 100 ⋅ 0.8642 x . Compare the
table values for the model to the data.
•
How does this compare to the 17% that we used to simulate the 16.7%
decay rate? [It should be fairly close. Try the model y = 100 ⋅ 0.17 x .
Graph and compare table values.]
Spring 2001
469
•
What do you think would happen if we combined all of the data in the
class? [The data should get closer to the theoretical model with a
decay rate of closer to 17% because of the law of large numbers. Try
it and see.]
6. Have students answer using at least 3 methods.
7. Have students answer using at least 3 methods.
8. Originally we had 100 units of radon. So we are looking for when there
are half of that or 50 units of radon. For our sample data, there are 50
units left between the 4th and the 5th days. So the half life is between the
4th and the 5th days. The real half life for radon gas is approximately 3.8
days.
Summary:
Students use the power of technology to simulate radioactive decay. They
find a model for the exponential decay data and use the model to answer
questions.
Spring 2001
470
Student Activity 3: Radioactive Decay
Radon is a radioactive gas that decays at a rate of about 16.67%
a day. This means that after a day, the amount of radon gas
present is 83.33% of the original amount.
Simulate the decay of 100 units of radon gas as follows.
Use your graphing calculator to randomly choose 17
integers between 0 and 99. Mark off each listed integer in
the grid. These represent the decayed units of gas. Ignore
any repeated integers.
Count the remaining units (not crossed off) and record this
in the table.
Continue to do this until you have completed the table.
1. Predict the graph of (day, units remaining).
Spring 2001
471
Day
0
1
2
3
4
5
6
7
8
9
10 11
12 13
14 15 16
17 18 19
20 21
22 23
24 25 26
27 28 29
30 31
32 33
34 35 36
37 38 39
40 41
42 43
44 45 46
47 48 49
50 51
52 53
54 55 56
57 58 59
60 61
62 63
64 65 66
67 68 69
70 71
72 73
74 75 76
77 78 79
80 81
82 83
84 85 86
87 88 89
90 91
92 93
94 95 96
97 98 99
0
1
2
3
4
5
6
7
8
9
10
Units
Remaining
100
3. Create a scatter plot of (day, units remaining).
4. Find the multiplier for the decay of radon by taking
successive quotients.
5. Find a model for the data using the multiplier and the number
of units of radon at day 0.
Spring 2001
472
Use your model to find the following in at least 3 ways.
6. How much radon is left after 15 days?
7. If there are 5 units of radon left, how many days have
passed?
8. What is the half-life for radon gas? In other words, after how
many days will there be half of the original amount of radon
left?
Spring 2001
473
Student Activity 4: Pendulum Decay
Overview:
Students collect data of the maximum swing of a pendulum versus time and
find a model for the data.
Objective:
Algebra I TEKS
algebraic methods.
Terms:
decay, pendulum, exponential function, quotients
Materials:
graphing calculators, data collection devices, motion detectors, soda cans,
string, meter sticks
Procedures:
Demonstrate the procedure for the experiment.
Points for the experiment:
The soda can should hang straight, not crooked.
The motion detector should be at least 1.5 feet from the maximum
swing of the pendulum.
The motion detector should be at the same height from the floor as the
hanging can.
Notes on the program:
The program collects data about 30 times for about 4 seconds each. It finds
the maximum distance that the pendulum swings in each of those periods. It
calls each of the periods “a swing” therefore there are about 30 “swings”
graphed.
Have students complete the data collection in their groups. Have them repeat
if necessary.
1. Encourage students to predict.
2. Sample data:
3. The successive quotients cluster around 0.9. For our sample data, the
mean of the successive quotients is 0.90853 and the maximum distance for
Spring 2001
474
swing number 0 is 0.14857. So the model for our sample data is
x
y = 0.1486 ⋅ (0.9085) .
4. In other words, how many swings would you expect before the distance is
half of the original distance?
For our sample data, the original distance was 0.1486. So,
0.1486 ÷ 2 = 0.0743 . Thus we need to find for what swing number was
the distance was 0.0743. Solve 0.0743 = 0.1486 ⋅ 0.9085 x graphically, on
the home screen, in a table, etc.
The half life occurs between the 7th and the 8th swing.
Summary:
As a pendulum swings, its swing decays. After collecting this distance data,
students apply their knowledge of exponential functions to find a model for
the data.
Spring 2001
475
Student Activity 4: Pendulum Decay
Construct a pendulum by tying a piece of string about 1 meter
long through the pull-tab of a half-empty soda can and fixing the
other end of the string to a solid point 1.5 – 2 meters off the
floor.
Place a motion detector about 0.5 meters from the maximum
swing of the pendulum at the same height as the pendulum at
rest.
Run the program PENDULUM, which will collect data and plot
(swing number, maximum swing of the pendulum). Gently pull
back the pendulum about 50 cm and release it.
1. Predict a graph of the results.
2. Sketch the resulting graph.
4. What is the half-life of the pendulum swing?
Spring 2001
476
Calculator Programs
LINETBL
FnOff
PlotsOff
ClrDraw
GridOn:AxesOn
randInt(–5,5)Y
randInt(–5,5)X
randInt(–3,3)B
randInt(–10,10)TblStart
randInt(1,5)∆Tbl
(Y/X)A
"B+AX"Y⁄
DispTable
LINEGRPH
FnOff
PlotsOff
ClrDraw
GridOn:AxesOn
randInt(–5,5)Y
randInt(–5,5)X
randInt(–3,3)B
(Y/X)A
"B+AX"Y⁄
ZDecimal
ACTSCRS
FnOff
PlotsOff
ClrDraw
GridOn:AxesOn
randInt(–5,5)Y
randInt(–5,5)X
randInt(–3,3)B
(Y/X)A
"B+AX"Y⁄
ZDecimal
PENDULUM
31N:ClrHome
Send({0})
Send({1,11,2,0,0,0})
Disp "SWING, ENTER"
ClrList L€
Pause
For(J,1,N)
Send({3,.04,99,0})
Get(L⁄)
min(L⁄)L€(J)
End
1-Var Stats L€
Q⁄-(Med-Q⁄)¯/(Q‹-2Med+Q⁄)K
K-L€L€
seq(X,X,0,N-1)L⁄
Plot1(Scatter,L⁄,L€,␣)
ZoomStat
477
Calculator Programs
JUMPIT
Full
ClrHome
Disp "CHECKING THE"
Disp "CALCULATOR-CBL"
Disp "LINK CONNECTION."
{1,0}L⁄
Send(L⁄)
{0}L€
Lbl M
{7}L⁄
Send(L⁄)
Get(L€)
If dim(L€)=1 and L€(1)=0
Then
ClrHome
Disp "***LINK ERROR***"
Disp "RE-INSERT THE"
Disp "LINK CORD
Disp "CONNECTORS"
Disp "FIRMLY"
Disp ""
Disp "THEN PUSH ENTER"
Pause
Goto M
End
Disp ""
ClrHome
Output(6,1," STATUS: O.K."
Output(8,10,"[ENTER]")
Pause
Full
ClrHome
GridOff:FnOff :PlotsOff
Disp "TURN ON THE CBL"
Disp ""
Disp "STAND BETWEEN","PROBE AND LIGHT","ON FLOOR."
Disp ""
Disp "PRESS [ENTER]"
Pause
ClrHome
Disp "JUMP WHEN READY"
{1,0}L⁄
Send(L⁄)
{1,1,1}L⁄
Send(L⁄)
{3,0.01,88,2,1,.2,10,0,1}L⁄
Send(L⁄)
Get(L€)
Get(L⁄)
round(L⁄,2)L⁄
ZoomStat
478
Calculator Programs
CMOVE
Full
ClrHome
Output(4,1,"
CBR MOTION")
Output(8,10,"[ENTER]")
Pause
Normal
Connected
Full
Func
Float
RectGC
GridOff
LabelOff
PlotsOff
FnOff
ClrDraw
CoordOn
AxesOn
ClrHome
Menu("COLLECT
DATA?","YES",H,"QUIT",3)
Lbl 3
ClrHome
Stop
Lbl H
1D
Disp "MOVE IN FRONT OF"
Disp "THE CBR TO MAKE"
Disp "A DISTANCE-TIME"
Disp "PLOT."
Output(8,1,"
[ENTER]")
Pause
{1,0}L⁄
Send(L⁄)
{0}L€
Lbl M
{7}L⁄
Send(L⁄)
Get(L€)
If dim(L€)=1 and L€(1)=0
Then
ClrHome
Disp "***LINK ERROR***"
Disp "PUSH IN THE LINK"
Disp "CORD CONNECTORS"
Disp "FIRMLY THEN HIT"
Disp "[ENTER]."
Pause
Goto M
Else
Full
ClrHome
Full
PlotsOff
FnOff
Func
AxesOn
0Xmin
40Xmax
0Ymin
10Ymax
10Xscl
1Yscl
ClrHome
Lbl 0
{1,0}L⁄
Send(L⁄)
{1,11,3}L⁄
Send(L⁄)
ClrDraw
Text(4,1,"D(FT)")
Text(51,81,"T(S)")
Text(1,30,"HIT [ENTER]")
Text(7,34,"TO START")
Pause
Text(1,30,"
")
Text(7,34,"
")
{3,.1,–1,0}L⁄
Send(L⁄)
40dim(L›)
For(I,1,40,1)
Get(L›(I))
Pt-On(I,L›(I))
End
Send({6,0})
L›L€
ClrList L›
seq(X,X,0,3.9,.1)L⁄
Lbl S
0Xmin
0Ymin
10Ymax
1Yscl
4Xmax
1Xscl
PlotsOff
FnOff
ClrDraw
DispGraph
Text(4,1,"D(FT)")
Text(51,81,"T(S)")
StoreGDB GDB6
0U
0V
Stop
479

Algebra 1: 2000 and Beyond - Ector County Independent School

Transcription

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