Section 5.4
Graphing an Equation
Pre-Activity
Preparation
Sportstime Pizza Parlor offers pizzas with various toppings. Prices begin with a plain
cheese pizza for $7.50. Toppings cost $1.25 each. People can choose their favorite
toppings from a wide variety: green peppers, onions, mushrooms, roasted garlic,
spinach, artichokes, green olives, black olives, anchovies, banana peppers, pineapple,
ham, bacon, pepperoni, sausage, Italian sausage, hamburger, and barbecued chicken.
Customers want to know how much their pizzas cost with various numbers of
toppings, so management has decided to list the cost on the menu.
Cheese pizza
The Big Ten
The Dirty Dozen
Number of
Toppings
Cost
0
$ 7.50
1
$ 8.75
2
$ 10.00
3
$ 11.25
4
$ 12.50
5
$ 13.75
6
$ 15.00
7
$ 16.25
8
$ 17.50
9
$ 18.75
10
$ 20.00
11
$ 21.25
12
$ 22.50
Cost
One enterprising young server decided that a picture would work better for customers
who are more visually aware. She graphed the information on a poster and hung it over
the cash register.
$ 23.00
$ 22.00
$ 21.00
$ 20.00
$ 19.00
$ 18.00
$ 17.00
$ 16.00
$ 15.00
$ 14.00
$ 13.00
$ 12.00
$ 11.00
$ 10.00
$ 9.00
$ 8.00
$ 7.00
$ 6.00
$ 5.00
$ 4.00
$ 3.00
$ 2.00
$ 1.00
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Number of Toppings
Learning Objectives
• Understand the relationship between a graph and its equation
• Use the equation of a line to generate a T-chart
• Use the information contained in a T-chart to create the equation for the associated line
• Graph the equation of a line by plotting points
423
Chapter 5 — Graphing
424
Terminology
New Terms
Previously Used
to
Learn
constant
horizontal line equation
intercept
T-chart
ordered pair solution
vertical line equation
slope
variable
Building Mathematical Language
Linear Equations
The following table summarizes types of linear equations.
Equation
Ax + By = C
Standard form
Information
Examples
In the standard form equation for a line, A, B, and C are
constants and can be any value with one exception: A and B
cannot both be zero at the same time.
2x + 3y = 10
Variables are x and y in general, but any appropriate letter
representing given data may be used.
–x – 4y = 8
5x – y = 6
Recall that linear equations are in two variables, generally
x and y, and that neither variable can be raised to a power
higher than 1.
y = mx + b
Slope-intercept
form
x=c
Vertical line
equation
y=c
Horizontal line
equation
In the slope-intercept equation for a line, constant values
m and b have special meaning for the line. The letter m
represents the slope of the line and the letter b is the ycoordinate of the y-intercept (the point where the line
crosses the y-axis).
y = 3x – 2
There is no y term and the letter c indicates a constant.
x = c is a vertical line (a line parallel to the y-axis) which
intersects the x-axis at the point (c, 0).
x=3
The slope of a vertical line is undefined. You can prove this
by calculating the slope, using the slope formula.
There is no x term and the letter c indicates a constant.
y= c is a horizontal line (a line parallel to the x-axis) which
intersects the y-axis at the point (0, c).
The slope of a horizontal line is zero. You can prove this by
calculating the slope, using the slope formula.
y = –11x + 4
x = –2
x=8
y=4
y = –9
y = 100
Section 5.4 — Graphing an Equation
425
Ordered Pair Solutions in a T-chart
There are infinitely many solutions to many types of equations in two variables, whether the equation
is linear or non-linear. A T-chart, so named because in its basic form it is shaped like a T, is helpful
in organizing solutions to a linear equation. In its most basic form a T-chart has two columns, one for
x-values and one for y-values. More columns can be added as needed so that your work is neat and well
documented.
x
y
A basic T-chart (at right) for two solutions to the equation x + y = 3 might record the
0
3
x and y-intercepts. This T-chart can be read as follows, “When the value of x is 0,
3
0
the value of y is 3. When the value of x is 3, the value of y is 0.”
A T-chart with more information (more x- and y-values) may also show the equation and the ordered pairs
(points):
x
x+y=3
y
(x, y)
x
y = –x + 3
y
(x, y)
0
0+y=3
3
(0, 3)
0
y=0+3
3
(0, 3)
3
3+y=3
0
(3, 0)
3
y = –3 + 3
0
(3, 0)
1
1+y=3
2
(1, 2)
1
y = –1 + 3
2
(1, 2)
or
Notice that either form of the equation works well. Use the form that is most convenient. Once the ordered
pairs are listed, plot them to the graph the equation.
y
(0, 3)
(1, 2)
(3, 0)
Next, add a few more points to the T-chart. Choose additional
values for x and solve for y.
When the T-chart is organized so the x-values are in order from
smallest to largest, a pattern emerges. For this particular equation,
as the x-values get larger, the y-values get smaller. The T-chart is
useful in organizing information, but it can also be used to help
you predict additional solutions to the equation. If x = 4, then
y = –1; if x = 5, then y = –2 and so on.
x
x
x+y=3
y
(x, y)
–2
–2 + y = 3
5
(–2, 5)
–1
–1 + y = 3
4
(–1, 4)
0
0+y=3
3
(0, 3)
1
1+y=3
2
(1, 2)
2
2+y=3
1
(2, 1)
3
3+y=3
0
(3, 0)
Chapter 5 — Graphing
426
Graphing the Equation for a Line
Any equation may be graphed by plotting enough points to determine its shape. We already know that
the shape of a linear equation is a straight line. Recall from Chapter 3 that a line may be defined by any
two points. We can therefore graph a linear equation by finding any two points (x and associated y-values
which make the equation true) and plotting those points. If we draw a line connecting those two points, we
have a graphical representation of the linear equation. We can validate our graph by finding a third solution
to the equation (an additional ordered pair) and graphing that point. If the third point is on the line, then
we have graphed the linear equation correctly. The points most often used to graph the equation of a line
are the x-and y-intercepts, because they are usually easy to calculate. Any two distinct points, however,
may be used to graph the line.
Methodologies
Graphing a Linear Equation
►
►
Example 1: Graph the linear equation: –3x – 4y = 8
Example 2:
Steps in the Methodology
Step 1
Rearrange
the
equation, as
needed, into
a workable
form.
Step 2
Create a Tchart for the
equation.
Step 3
Find the yintercept
for the
equation.
Try It!
Graph the linear equation: 2x + 3y = 10
Simplify the
equation to either
the standard or
slope-intercept
form.
Example 1
Example 2
The equation is already in the
standard form:
–3x – 4y = 8
Set up the Tchart, including
all necessary
information.
x
–3x – 4y = 8
Let x = 0 and solve
for y.
x
0
–3x – 4y = 8 y (x, y)
–3(0) – 4y = 8 –2 (0, –2)
y
(x, y)
Section 5.4 — Graphing an Equation
427
Steps in the Methodology
Step 4
Find the xintercept
for the
equation.
Let y = 0 and solve
for x.
Example 1
x
0
-
Example 2
–3x – 4y = 8
–3(0) – 4y = 8
y (x, y)
–2 (0, –2)
8
–3x – 4(0) = 8
3
 8 
0  − ,0
 3 
In this case the x-value is not a
whole number, and not easily
plotted.
Step 5
Find an
alternate
point.
A third point will be
used to validate the
graph of the line. In
the case where the
x- or y-intercept is
not a whole number
and/or cannot be
easily plotted,
calculate another
point to plot.
Choose another value for x or
y to get another point to graph.
Try to select an x or y value that
will result in a whole number
coordinate pair.* Note that this
may not always be possible.
Let x = 4.
x
0
-
–3x – 4y = 8
–3(0) – 4y = 8
y (x, y)
–2 (0, –2)
8
–3x – 4(0) = 8
3
 8 
0  − ,0
 3 
4
Step 6
Graph the
line, based
upon the
information
in the Tchart.
Plot two of the
points from the
T-chart. Connect
the points using a
straight edge.
–3(4) – 4y = 8
–5 (4, –5)
y
x
(0, -2)
(4, -5)
* Upon closer examination, whatever value we choose for x is multiplied by –3, added to –8, then divided by 4:
-3 x - 8
4 .
Multiplying –3 by any odd number will obviously produce a fraction. So we would logically want to start testing values
such as 2 or 4 for x. Though there is nothing to stop us from randomly selecting values for x, if we stop and think through
what it is we are trying to do (in this case produce a number that is a multiple of the coefficient of y), we can very often
narrow our choices, saving ourselves both time and frustration.
Chapter 5 — Graphing
428
Steps in the Methodology
Step 7
Validate.
Example 1
Validate by
choosing another
value for either x or
y. If the resulting
point is on the line,
the line is correctly
graphed.
If you used a third
point instead of
either the x- or yintercept to draw
your line, you will
need to find another
point to use for
the purposes of
validation.
Example 2
Let x = – 4. Solve for y and plot
the resulting point.
x
0
-
–3x – 4y = 8
–3(0) – 4y = 8
y (x, y)
–2 (0, –2)
8
–3x – 4(0) = 8
3
 8 
0  − ,0
 3 
4 –3(4) – 4y = 8 –5 (4, –5)
–4 –3(–4) – 4y = 8 1 (–4, 1)
y
(-4, 1)
x
(0, -2)
(4, -5)
The point (–4, 1) is on the graph
of the line. 
Models
Model 1
Graph the following equation: y = 2x –1
Step 1
Equation form
Step 2
T-chart
The equation is already in the slope-intercept form.
x
y = 2x –1
y
(x, y)
Section 5.4 — Graphing an Equation
Step 3
y-intercept
429
Because the equation is in the slope-intercept form, we can easily
determine the y-intercept:
Shortcut:
x
y
(x, y)
y = 2x –1
If the equation is in the slope0
y = 2(0) –1 –1 (0, –1)
intercept form (y = mx + b), read
the b value from the equation,
remembering that the the yintercept is the point (0, b).
Step 4
x-intercept
x
0
y = 2x –1
y = 2(0) –1
y
–1
(x, y)
(0, –1)
1
2
0 = 2x –1
0
1 
 ,0
2 
The x-intercept is not a whole number and cannot be graphed with
precision. We will use an alternate point for our second point.
Step 5
Step 6
Alternate point
Graph
x
0
y = 2x –1
y = 2(0) –1
y
–1
(x, y)
(0, –1)
1
2
0 = 2x –1
0
1 
 ,0
2 
1
y = 2(1) –1
1
(1, 1)
y
(1, 1)
x
(0, -1)
Chapter 5 — Graphing
430
Step 7
Validate
choose
x
0
y = 2x –1
y = 2(0) –1
y
–1
(x, y)
(0, –1)
1
2
0 = 2x –1
0
1 
 ,0
2 
1
2
y = 2(1) –1
y = 2(2) –1
1
3
(1, 1)
(2, 3)
to validate
y
(2, 3)
(1, 1)
x
(0, -1)
The point (2, 3) is on the graph of the line. 
Model 2
Graph the following equation: 3x + y = 0
Step 1
Equation form
Step 2
T-chart
The equation is already in standard form.
x
3x + y = 0
y
(x, y)
Section 5.4 — Graphing an Equation
431
Step 3
y-intercept
x
0
3x + y = 0
3(0) + y = 0
y
0
(x, y)
(0, 0)
Step 4
x-intercept
x
0
0
3x + y = 0
3(0) + y = 0
3x + 0 = 0
y
0
0
(x, y)
(0, 0)
(0, 0)
Notice that when a line passes through the point (0, 0),
the x- and y-intercepts are that same point. We therefore
need an additional point in order to graph the line.
Step 5
Alternate point
Step 6
Graph
x
0
0
1
y
0
0
–3
3x + y = 0
3(0) + y = 0
3x + 0 = 0
3(1) + y = 0
(x, y)
(0, 0)
(0, 0)
(1, –3)
y
(0, 0)
(1, -3)
x
Chapter 5 — Graphing
432
Step 7
Validate
x
y
3x + y = 0
0 3(0) + y = 0 0
0
0
3x + 0 = 0
1 3(1) + y = 0 –3
choose –1 3(–1) + y = 0 3
(x, y)
(0, 0)
(0, 0)
(1, –3)
(–1, 3) to validate
y
(-1, 3)
(0, 0)
x
(1, -3)
The point (–1, 3) is on the graph of the line. 
Model 3
Using the T-chart provided at right, graph the equation, determine the slopeintercept form of the equation, and complete the T-chart. Validate your work.
The T-chart gives us enough information to graph the line by plotting two points:
y
(-1, 2)
(0, 0)
x
x
–3
–2
–1
0
1
2
3
y
6
(x, y)
(–3, 6)
2
0
(–1, 2)
(0, 0)
–4
–6
(2, –4)
(3, –6)
Section 5.4 — Graphing an Equation
433
We have at least two points, so we can use the Slope Formula to calculate the slope of the line.
Point 1: (0, 0) Point 2: (–1, 2)
y2 - y1 2 - 0
=
= -2
x2 - x1 -1 - 0
Because we know the y-intercept (0, 0) and the slope (–2), we can write an equation for the line:
y = –2x + 0 or y = –2x
Using this equation, we can now fill in the missing information
in the T-chart:
We can validate that we have correctly graphed the line,
calculated the correct slope, and determined the correct equation
by plotting the new points (–2, 4) and (1, –2). If both points are
on the line we have already graphed, our work is validated.
y
x
–3
–2
–1
0
1
2
3
y = –2x
y = –2(–2)
y = –2(1)
y
6
4
2
0
–2
–4
–6
(x, y)
(–3, 6)
(–2, 4)
(–1, 2)
(0, 0)
(1, –2)
(2, –4)
(3, –6)
(-2, 4)
(-1, 2)
(0, 0)
x
The new points (–2, 4) and
(1, –2) are both on the line. 
(1, -2)
Model 4
Plot the following two equations on the same coordinate plane: 2x = 6 and 3y = 12. If the lines intersect,
determine the point of intersection.
Step 1
Both equations can be simplifed:
Equation form
x=3
y=4
This is obviously a vertical line
(x = c) and for any value of y, the
value of x will always be 3. There
is no reason to create a T-chart.
We can go directly to Step 6 and
graph the equation.
This is obviously a horizontal
line (y = c) and for any value of
x, the value of y will always be 4.
There is no reason to create a Tchart. We can go directly to Step
6 and graph the equation.
Chapter 5 — Graphing
434
Step 6
Graph
y
We can see that the point
of intersection is (3, 4).
y=4
x
x=3
We do not need to rely on what we see in the graph; we can reason our way to the answer:
We know that for the horizontal line, y is always 4, and for the vertical line, x is always 3. Recall from
Chapter 3, that if any two lines in a plane are not parallel and are not the same line (coincident), they
must eventually intersect. A vertical line will always intersect a horizontal line, at some point in the
coordinate plane. Because they intersect and we know the values of x and y, the point of intersection
must be (3, 4).
y
Model 5
 3 3
Determine whether or not the point  − ,  lies on the
 2 2
graph of the equation y = x + 3.
While we could make a guess based on the graph, it is always
preferable to verify algebraically that the x- and y-values
that make up an ordered pair (point) represent a solution to
the equation.
(0, 3)
(-3, 0)
x
Substitute both values into the equation and evaluate.
3? 3
= - +3
2
2
3? 3 6
=- +
2
2 2
3 3
=
2 2
The result is a true statement.
3
3
This means that x = - , y = is a
2
2
solution to the equation, so the point
 3 3
 − ,  does lie on the graph of
 2 2
that equation y = x + 3.
Note that you can also validate that the proposed point lies on the graph of the line (is a solution to the equation) by
substituting either the x- or y-value of the ordered pair into the equation and solving for the other value. Either method
sufficiently validates that the x- and y-values given for the point are a solution to the equation.
Section 5.4 — Graphing an Equation
435
Addressing Common Errors
Incorrect
Process
Issue
Validating
a solution,
intersection,
or point
graphically,
rather than
algebraically
y
Does the point (3, 7) lie on the graph of
the following equation?
y = 3x – 2
(1, 1)
x
No. The line only extends
to a y-value of about 4.
(-1, -5)
Resolution
Remember that in order for a point to lie on a graph of a line, it must be a solution
to the equation. Do not rely on what you think you see in a graph; validate points,
as solutions, to the underlying equation. Also recall that lines, by definition, are
infinite in length. Any graph will therefore show only a very small part of the line.
Correct Process
Substitute both values into the equation and evaluate:
y = 3x - 2
?
7 = 3(3) - 2
7 =7
Issue
Incorrect
Process
Failing to
recognize a
horizontal or
vertical line
Find the x- and yintercepts of the
following equation:
3x = –9
x-intercept:
0 = –3x –9
(3, 0)
y-intercept:
y = –3(0) – 9
(0, –9)
The point (3, 7) is a solution to the
equation, so it does lie on the graph
of the line.
Resolution
Remember that
horizontal lines have
only a y-value
(y = c) and that
vertical lines have only
an x-value (x = c).
Do not insert
variables simply to
make the equation
fit the standard or
slope-intercept form.
The equations for
horizontal and vertical
lines contain only one
variable.
Correct
Process
Simplify the equation:
x = –3
Because the equation has only an xvalue, it is a vertical line.
The line will not have an yintercept because the line will never
cross the y-axis (where the value of
x is zero); the value of x is always
–3.
The line does cross the x-axis, with
a y-value of 0, at the point (–3, 0).
The x-intercept is the point (–3, 0).
Chapter 5 — Graphing
436
Incorrect
Process
Issue
Using a
fractional
coordinate
when plotting
a point
y
Graph the following equation: 2x – y = 5
x
0
2x – y = 5
2(0) – y = 5
y
–5
(x, y)
(0, –5)
5
2
2x – (0) = 5
0
5 
 ,0
2 
 5 ,0
2 


x
(0, -5)
Resolution
When graphing on a grid that is divided into whole number units or squares, it is
difficult to graph fractional coordinates with much precision. You should continue
filling out the T-chart in an attempt to find whole number values for the coordinate
points.
Correct Process
Validation
While the graph may look correct, you
should always try to find two whole
number coordinate points, when graphing
on a whole number grid.
Validate by finding an additional point:
x
0
2x – y = 5
2(0) – y = 5
y
–5
(x, y)
(0, –5)
x
0
2x – y = 5
2(0) – y = 5
y
–5
(x, y)
(0, –5)
5
2
2x – (0) = 5
0
5 
 ,0
2 
5
2
2x – (0) = 5
0
5 
 ,0
2 
1
2(1) – y = 5
–3
(1, –3)
3
2x – (1) = 5
1
(3, 1)
1
2(1) – y = 5
–3
(1, –3)
y
y
(3, 1)
x
x
(1, -3)
(1, -3)
(0, -5)
(0, -5)
Note that the dotted line (the line drawn
using a fractional coordinate) differs from
the validated line.
Section 5.4 — Graphing an Equation
Preparation Inventory
Before proceeding, you should be able to perform each of the following tasks:
Generate a T-chart that shows the x- and y-intercepts and additional points, as needed
Plot ordered pairs from the information in a T-chart
Determine whether or not a given point is a solution to a linear equation
Use the information provided in a T-chart to determine the slope-intercept form of the related
equation
437
Section 5.4
Activity
Graphing an Equation
Performance Criteria
• Use the equation of a line to generate a T-chart
• Graph the equation of a line by plotting points
– T-chart contains at least three points
– correct and appropriate use of a T-chart
– T-chart is filled in correctly
– appropriate validation
• Use the information contained in a T-chart to
create the equation for the associated line
– correct identification of the y-intercept
– correct calculation of slope
Critical Thinking Questions
1. What is the relationship between a point on the graph of a line and a solution to the equation represented
by that graph?
2. How do you use the x- and y-intercepts in graphing an equation?
3. Of what use is a T-chart in graphing an equation?
438
Section 5.4 — Graphing an Equation
439
4. Why is it not necessary to create a T-chart in order to create a graph for a horizontal or vertical-line
equation?
5. Imagine that you are given a T-chart that contains the x-intercept and two other points, neither of which is
the y-intercept. You know you can calculate the slope of the line; can you also calculate the y-intercept? If
so, explain how. If not, explain why not.
6. Imagine you are given the point (1, 1) and told that the slope of the line passing through this point is 2.
Can you write an equation for that line? Explain.
Tips
for
Success
• When building a T-chart, try to use values that make whole number ordered pairs because they are easiest
to graph
Chapter 5 — Graphing
440
Demonstrate Your Understanding
1. Determine, without graphing, whether or not the points below would lie on the graph of the equation
y = –2x + 5.
Point
a)
(3, –1)
b)
5 
 ,0
2 
c)
(–3, 12)
d)
2 
 ,3 
3 
e)
(–1, 7)
Work
Would the point lie
on the graph of the
equation?
Section 5.4 — Graphing an Equation
441
2. Graph the following two equations on
the same coordinate plane. Determine
and label their point of intersection.
y
4y = 20 and 3x = –6
x
Point of intersection: _____________
3. Using the T-chart provided at right, determine the slope-intercept form
of the underlying equation. Do not graph the line.
x
y
(x, y)
–2
12
(–2, 12)
–1
10
(–1, 10)
0
8
(0, 8)
1
6
(1, 6)
2
4
(2, 4)
Slope-intercept form of the equation: _____________________________
Chapter 5 — Graphing
442
4. Graph the equation, 2y = x + 4. Use the T-chart provided
at right. Validate your work. Circle the point used for
validation (on the chart, as well as on the graph.)
x
2y = x + 4
y
(x, y)
0
0
y
x
5. Graph the following equation on the coordinate plane above: 3x = –6
Determine the point of intersection between this line and the line in the previous problem.
Point of intersection: _____________________________
Section 5.4 — Graphing an Equation
443
6. Given the graph below, complete the T-chart provided at right (with at least
four points, including both intercepts) and determine the slope-intercept
form of the equation. Validate your work.
y
x
y
(x, y)
0
0
(6, 1)
(3, 0)
x
(–6, –3)
Slope-intercept form of the equation: _____________________________
Chapter 5 — Graphing
444
7. Using the T-chart provided at right, graph the equation, determine
the slope-intercept form of the equation, and complete the T-chart.
Validate your work.
y
x
y
(x, y)
–3
6
(–3, 6)
4
(–1, 4)
2
(1, 2)
0
(3, 0)
–2
–1
0
1
2
3
x
Slope-intercept form of the equation: _____________________________
Section 5.4 — Graphing an Equation
Identify
and
445
Correct
the
Errors
In the second column, identify the error(s) in the worked solution or validate its answer. If the worked solution
is incorrect, solve the problem correctly.
Identify Errors
or Validate
Worked Solution
1) Graph the following equation:
4x + y = 2
x
0
4x + y = 2
4(0) + y = 2
y
2
(x, y)
(0, 2)
1
2
4x – (0) = 2
0
1 
 ,0
2 
y
(0, 2)
 1 ,0
2 


2)
x
Find the x- and y-intercepts of the
following equation:
4y = 16
x-intercept: 4(0) = x + 16
x = –16
(–16, 0)
y-intercept: 4y = 0 + 16
4y = 16, y = 4
(0, 4)
Correct Process
Chapter 5 — Graphing
446
Identify Errors
or Validate
Worked Solution
3)
Does the point (–3, –6) lie on the
graph of the following equation?
y = –3x + 3
y
x
? –3(–3) + 3
–6 =
?9+3
–6 =
? 12
–6 =
–6 ≠ 12
No. (–3, –6) is not a solution to the
equation, so the point cannot lie on
the graph of the line.
Correct Process