Graphing Equations T-charts

Graphing Equations
T-charts
One of the most common methods of graphing an equation is by using an x-y “T-chart.” T-charts
use a few points to plot and connect a line. Normally we choose between 3 and 5 x-values to
“plug in” and solve for y (the x-values should be both positive and negative and should stay
close to 0). A typical set that can be used for most problems are x = {-2,-1, 0, 1, 2}.
Example:
Given that y=2x-1. Substituting the x = {-2,-1, 0, 1, 2} to find the
corresponding y values, the T-chart will correspond as follows:
For example, with x=-2, y=2x-1 = 2(-2)-1 = -5. Follow this same procedure for all of the xvalues that you have selected. Plotting the points and connecting them gives the finished graph:
0
-6
-4
-2
0
-1
-2
-3
-4
-5
2
4
6
Graphing Inequalities
To graph an inequality, first look to see if it has a ‘less than or equal to’ sign, ≤, or a ‘greater than
or equal to sign’, ≥. If there is ≥ or ≤ symbol, use a solid line for the graph. If instead it has a
less than or greater than sign, < or >, respectively, use a dashed line.
When plotting an inequality, plot the line like a normal line using the appropriate dashed or solid
line and then shade the appropriate side of the line that describes which side of the graph which
the inequality is greater or less than. To determine which part of the line to shade, choose a test
point [(x,y) coordinate] that is not on the line and put the x and y-values into the equation. If the
statement this creates is true, shade on the side of the plot where you chose the point. If the
statement this creates is incorrect, shade on the opposite side of the plot where you chose the
point. If possible, choosing the origin is a good choice of test point (remember: the test point
cannot fall on the line).
Example: Graph the following inequality:
≤ 2 + 1
First make a T-chart as plugging in x ={-2,-1, 0, 1, 2} to find the corresponding y values
Since, the inequality is less than or equal to, draw a solid line for y=2x+1. Then, to decide which
direction to shade, pick a point (let’s use (0, 0)) and see if that point makes the inequality true or
false. 0 ≤ 2(0) + 1 gives us 0 ≤ 1, which is true. Therefore, shade toward the point (0, 0).
Example 2: Graph the following inequality
>
1
−2
2
First make a T-chart with x = {-4,-2, 0, 2, 4} substitutions and the corresponding y-values.
Since, the inequality is greater than, draw a dashed line for = − 2. Then, to decide which
direction to shade, pick a point (let’s use (0, 0)) and see if that point makes the inequality true or
false. 0 > (0) − 2 gives us 0 > −2, which is true. Therefore, shade toward the point (0, 0).
Distribution and the FOIL Method
The Distributive Property
( + ) = + Multiplying a number across a set of parentheses is the same as multiplying each of the terms
inside the parentheses by that number (and adding or subtracting them).
Example:
3( + 1) = 3 ∙ + 3 ∙ 1 = 3 + 3
Example:
4(2 + 8) = 4 ∙ 2 + 4 ∙ 8 = 8 + 32
The FOIL Method
FOIL (first-outer-inner-last) is a method used to multiply two binomial expressions. Multiply
each pair of terms and add them together.
Example: Expand the following expression (x + 3)(x + 2)
The foil method says to expand it into a simpler form:
"first":
"outer":
"inner":
"last":
(x)(x) = x2
(x)(2) = 2x
(3)(x) = 3x
(3)(2) = 6
Then, add these terms together. The foil method says to expand it into a simpler form:
( + 3)( + 2) = x + 2 + 3 + 6
Combining like terms gives the final answer:
+ 5 + 6
Example: Expand the following expression:
( + 4)( − 1)
( ∙ ) + ∙ (−1) + (4 ∙ ) + (4 ∙ (−1))
− + 4 − 4
Combining like terms gives the final answer:
+ 3 − 4
Factoring Trinomials
Basic Factoring
The simplest way to factor a trinomial is to try and find a common factor in each of the terms
that can be factored out of the expression. Then, factor the resulting expression like normal.
Example:
Factor − − 2
First, factor an x out of each of the terms:
( − − 2)
Next, factor the binomial in the parentheses:
( + 1)( − 2)
So, the factor are x, x+1, and x-2. The zeros are 0, -1, and 2.
Special Trinomials
There are a few types of special trinomials that are easy to factor once you know the pattern.
Perfect square trinomials can be factored by following these rules:
+ 2 + = ( + )
− 2 + = ( − )
Example:
Factor + 2 + 1
First, recognize that this is a perfect square trinomial with a=x, -2ab = 2x, and b = 1
Second, apply the appropriate rule from above to factor.
+ 2 + = ( + ) = ( + 1)
Difference of Squares
Another useful factoring technique involves a difference of squares. It follows this pattern:
− = ( + )( − )
This means that if you have 2 squares subtracted from each other (whether they are variables,
numbers, or a combination of the two doesn’t matter), the expression can be factored.
Example:
Factor − 9
Recognize that this is a difference of squares because ∙ = and 3 = 9.
You then know that a = x and b = 3. So, − 9 = ( + 3)( − 3)
Quadratic Formula
The general form of a quadratic equation is:
+ + ! = 0: #ℎ%&% ≠ 0
The Quadratic Formula to find the root(s) for any quadratic equation is:
− ± √ − 4!
=
2
The various possibilities of roots are determined by what is called the discriminate. The
discriminate is the expression contained under the square root symbol − 4!:
*+ − 4! > 0 → -ℎ%&% &% -#. &%/ &..-0
*+ − 4! = 0 → -ℎ%&% *0 &%/ &%1%-%2 &..*+ − 4! < 0 → -ℎ%&% &% -#. !.41/% &..-0
Example:
Solve 2x2 – 4x – 3 = 0, for x
There are no factors of (2)(–3) = –6 that add up to –4, so this quadratic
cannot be factored; Instead, use the Quadratic Formula. In this case, a = 2,
b = 4 and c = –3. So:
Then the answers are x = –0.58 and x = 2.58, rounded to two decimal places.
Cross Multiplication
Cross multiplication is a way to solve proportions. Below are two equations where cross
multiplication can be used:
4 x
=
16 2
or
6 3
=
y 5
To solve problems with cross multiplication, first draw an ‘X’, of where the multiplication is
going to take place, connecting the diagonal terms as shown below:
4 x
=
16 2
or
6 3
=
y 5
The numbers or variables in the denominator should be multiplied with the diagonal term in the
numerator and set equal to the other set of diagonal terms multiplied together.
(2·4) = (16·x)
or
(6·5) = (y·3)
8 = 16x
or
30 = 3y
The equation can then be solved by dividing to isolate the variable.
x=½
or
y = 10
Cross multiplication can still be used even with binomials in the fraction.
1
( x − 3)
=
3
6
The only difference is that we now have to distribute the 6 to both of the terms in the binomial.
(1·6) = ((3·x) – (3·3))
After the 3 has been distributed, the problem can be solved as above.
6 = 3x - 9
3x = 15
x=5
Systems of Linear Equations:
Systems of linear equations can be used to model and solve lots of interesting, real world
problems. A system of linear equations is made up of two or more equations. In order to
completely solve the system, the number of variables must equal the number of equations. An
example of a system with two equations and two unknowns is shown below:
2x + 4y = 6
(Equation #1)
-3x + 4y = 11
(Equation #2)
There are three ways to solve systems of linear equations: graphing, substitution and elimination.
We will go through an example of each method.
Method 1: Graphing
The first step in the graphing method is to convert the equations to slope-intercept form. This is
done by solving for y in the given equations.
5
5
Doing this gives: = − + and = +
With the equations in slope-intercept form, it is convenient to graph them. If the two lines
intersect at one point (x,y), the system has one solution. If the two lines are parallel, and never
intersect, the system has no solution. If the two lines are exactly the same, the system has an
infinite number of solutions. The plots below show all of the possible types of solutions.
There is one solution at the point (-1,2). The solution to the system is x = -1 and y = 2.
The two lines are parallel. There are no solutions to this system.
The two lines are exactly the same. Since they are right on top of each other there are an infinite
number of solutions to the system.
Method 2: Substitution
We are going to use a similar technique in the substitution method. But in this case, we are only
going to solve one of the equations.
1
3
y =− x+
2
2
− 3 x + 4 y = 11
The next step in the substitution process is to take first equation and plug it into the second
equation. The arrow above shows where to make the substitution (plug in -1/2x + 3/2 for y).
3
 1
− 3 x + 4  − x +  = 11
2
 2
Using the distributive property and combining like terms we can now solve the equation for x.
− 3 x + (− 2 x + 6 ) = 11
− 5 x + 6 = 11
− 5x = 5
x = −1
The last step in the substitution method is plug our newly solved x back into one of the equations
and solve for y. It doesn’t matter which equation; they should both give the same answer.
2x + 4y = 6
-3x + 4y = 11
2(-1) + 4y = 6
-3 (-1) + 4y = 11
-2 +4y = 6
3 + 4y = 11
4y = 8
4y = 8
y=2
y=2
The substitution method has now given us that x = -1 and y = 2. Notice that this is exactly the
same answer we found in the graphing method.
Method 3: Elimination
The final method for solving systems of linear equations is called elimination. In elimination, we
either add or subtract the two equations to or from each other. Doing this eliminates either the x
or y variable and allows us to solve for the other. We are going to start off with a slightly
different set of equations for this example
x + 2y = 3
(Equation #1)
-3x + 4y = 11
(Equation #2)
The first step in elimination is to get either the x or y coefficients the same in both equations. In
this example we have ‘2y’ in the first equation and ‘4y’ in the second equation. In order to get
the same coefficient in both equations we have to multiply the entire first equation by 2.
2*(x + 2y = 3)
2x + 4y = 6
Now we have ’4y’ in both equations. If the equations already have a common coefficient, the
first step can be skipped. Now the second equation is subtracted from the first equation. This
eliminates the y variable and allows us to solve for x.
2x + 4 y = 6
− (−3 x + 4 y = 11)
_______________
5 x + 0 y = −5
5x = -5
x = -1
The last step, as with the substitution method, is to plug the x back into either one of the original
equations and solve.
x + 2y = 3
-3x + 4y = 11
1(-1) + 2y = 3
-3(-1) + 4y = 11
-1 +2y = 3
3 + 4y = 11
2y = 4
4y = 8
y=2
y=2
Just like in the first two methods we get the solution to the system of x = -1 and y = 2.
Percentages
The important thing to remember with percentage problems is to pay attention to the wording.
Example: 18 is what percent of 30?
To go from a sentence to algebra you should look for key words. “Is” is the first key word in this
problem. This is the same as saying “is equal to” and can be replaced by a “=”. So now we have
18 = “what percent” of 30. What percent is the thing you are solving for and can be replaced by
a variable (x). “Of” in this case means to multiply. So now we have 18=x·30 for our equation.
Solving gives x=0.6 which is a decimal answer and needs to be changed into a percentage. To
go from a decimal to a percentage you multiply by 100% and you get x = 60% as the final
answer. This problem can also be set up as a proportion, knowing that a percent is always going
to be out of a total of 100%. In this problem the proportion can be set up as follows:
6
7
=
8
.
77
Then you can cross-multiply 18 ·100= 30 x, solve for x to get 60%.
Example 2: What is 20% of 80?
Again replace “what” with “x”, replace “is” with “=”, and replace “of” with “*” and you get
x=20%*80. Convert the % to a decimal and you get x=.2*80, which gives x=16. Proportions can
8
also be used in this set up: 67 =
x=16.
7
. Then you cross-multiply 20·80=100x and solve for x to get
77