OpenStax-CNX module: m22000
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Graphing Linear Equations and
Inequalities: Graphing Equations in
∗
Slope-Intercept Form
Wade Ellis
Denny Burzynski
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0†
Abstract
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. In this chapter
the student is shown how graphs provide information that is not always evident from the equation alone.
The chapter begins by establishing the relationship between the variables in an equation, the number
of coordinate axes necessary to construct its graph, and the spatial dimension of both the coordinate
system and the graph. Interpretation of graphs is also emphasized throughout the chapter, beginning
with the plotting of points. The slope formula is fully developed, progressing from verbal phrases to
mathematical expressions. The expressions are then formed into an equation by explicitly stating that a
ratio is a comparison of two quantities of the same type (e.g., distance, weight, or money). This approach
benets students who take future courses that use graphs to display information. The student is shown
how to graph lines using the intercept method, the table method, and the slope-intercept method, as
well as how to distinguish, by inspection, oblique and horizontal/vertical lines. This module contains an
overview of the chapter "Graphing Linear Equations and Inequalities in One and Two Variables".
1 Overview
• Using the Slope and Intercept to Graph a Line
2 Using the Slope and Intercept to Graph a Line
When a linear equation is given in the general form, ax + by = c, we observed that an ecient graphical
approach was the intercept method. We let x = 0 and computed the corresponding value of y , then let y = 0
and computed the corresponding value of x.
When an equation is written in the slope-intercept form, y = mx + b, there are also ecient ways of
constructing the graph. One way, but less ecient, is to choose two or three x-values and compute to nd
the corresponding y -values. However, computations are tedious, time consuming, and can lead to errors.
Another way, the method listed below, makes use of the slope and the y -intercept for graphing the line. It
is quick, simple, and involves no computations.
∗ Version
1.4: Jun 1, 2009 11:04 am +0000
† http://creativecommons.org/licenses/by/3.0/
http://legacy.cnx.org/content/m22000/1.4/
OpenStax-CNX module: m22000
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Graphing Method
1. Plot the y -intercept(0, b).
2. Determine another point by using the slope m.
3. Draw a line through the two points.
1
Recall that we dened the slope m as the ratio xy22 −y
−x1 . The numerator y2 − y1 represents the number of units
that y changes and the denominator x2 − x1 represents the number of units that x changes. Suppose m = pq .
Then p is the number of units that y changes and q is the number of units that x changes. Since these
changes occur simultaneously, start with your pencil at the y -intercept, move p units in the appropriate
vertical direction, and then move q units in the appropriate horizontal direction. Mark a point at this
location.
3 Sample Set A
Graph the following lines.
Example 1
y = 43 x + 2
Step 1: The y -intercept is the point (0, 2). Thus the line crosses the y -axis 2 units above the origin.
Mark a point at (0, 2).
Step 2: The slope, m, is 43 . This means that if we start at any point on the line and move our pencil
3 units up and then 4 units to the right, we'll be back on the line. Start at a known point,
the y -intercept (0, 2). Move up 3 units, then move 4 units to the right. Mark a point at this
location. (Note also that 34 = −3
−4 . This means that if we start at any point on the line and
move our pencil 3 units down and 4 units to the left, we'll be back on the line. Note also
3
that 43 = 14 . This means that if we start at any point on the line and move to the right 1
unit, we'll have to move up 3/4 unit to get back on the line.)
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Step 3: Draw a line through both points.
Example 2
y = − 12 x +
7
2
Step 1: The y -intercept is thepoint 0, 72 . Thus the line crosses the y -axis 72 units above the origin.
Mark a point at 0, 27 , or 0, 3 12 .
Step 2: The slope, m, is − 12 . We can write − 12 as −1
2 . Thus, we start at a known point, the
y -intercept 0, 3 12 , move down one unit (because of the −1), then move right 2 units. Mark
a point at this location.
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OpenStax-CNX module: m22000
Step 3: Draw a line through both points.
Example 3
y = 25 x
Step 1: We can put this equation into explicit slope-intercept by writing it as y = 52 x + 0.
The y -intercept is the point (0, 0), the origin. This line goes right through the origin.
Step 2: The slope, m, is 25 . Starting at the origin, we move up 2 units, then move to the right 5 units.
Mark a point at this location.
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Step 3: Draw a line through the two points.
Example 4
y = 2x − 4
Step 1: The y -intercept is the point (0, − 4). Thus the line crosses the y -axis 4 units below the origin.
Mark a point at (0, − 4).
Step 2: The slope, m, is 2. If we write the slope as a fraction, 2 = 21 , we can read how to make the
changes. Start at the known point (0, − 4), move up 2 units, then move right 1 unit. Mark
a point at this location.
Step 3: Draw a line through the two points.
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4 Practice Set A
Use the y -intercept and the slope to graph each line.
Exercise 1
y=
−2
3 x
(Solution on p. 14.)
+4
Exercise 2
(Solution on p. 14.)
y = 43 x
5 Excercises
For the following problems, graph the equations.
Exercise 3
y = 32 x + 1
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(Solution on p. 14.)
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Exercise 4
y = 14 x − 2
Exercise 5
y = 5x − 4
Exercise 6
y = − 56 x − 3
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(Solution on p. 14.)
OpenStax-CNX module: m22000
Exercise 7
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(Solution on p. 15.)
y = 23 x − 5
Exercise 8
y = 15 x + 2
Exercise 9
y = − 38 x + 4
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(Solution on p. 15.)
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Exercise 10
y = − 10
3 x+6
Exercise 11
y = 1x − 4
Exercise 12
y = −2x + 1
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(Solution on p. 15.)
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Exercise 13
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(Solution on p. 16.)
y =x+2
Exercise 14
y = 35 x
Exercise 15
y = − 34 x
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(Solution on p. 16.)
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Exercise 16
y=x
Exercise 17
y = −x
Exercise 18
3y − 2x = −3
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(Solution on p. 16.)
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Exercise 19
6x + 10y = 30
Exercise 20
x+y =0
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(Solution on p. 17.)
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6 Excersise for Review
Exercise 21
( here1 ) Solve
the inequality 2 − 4x ≥ x − 3.
Exercise 22
( here2 ) Graph
the inequality y + 3 > 1.
Exercise 23
( here3 ) Graph
the equation y = −2.
Exercise 24
( here4 ) Determine
Exercise 25
( here5 ) Find
(Solution on p. 17.)
(Solution on p. 17.)
the slope and y -intercept of the line −4y − 3x = 16.
(Solution on p. 17.)
the slope of the line passing through the points (−1, 5) and (2, 3).
1 "Solving Linear Equations and Inequalities: Linear Inequalities in One Variable"
<http://legacy.cnx.org/content/m21979/latest/>
2 "Graphing Linear Equations and Inequalities:
<http://legacy.cnx.org/content/m18877/latest/>
3 "Graphing Linear Equations and Inequalities:
<http://legacy.cnx.org/content/m21995/latest/>
4 "Graphing Linear Equations and Inequalities:
<http://legacy.cnx.org/content/m22014/latest/>
5 "Graphing Linear Equations and Inequalities:
<http://legacy.cnx.org/content/m22014/latest/>
http://legacy.cnx.org/content/m22000/1.4/
Graphing Linear Equations and Inequalities in One Variable"
Graphing Linear Equations in Two Variables"
The Slope-Intercept Form of a Line"
The Slope-Intercept Form of a Line"
OpenStax-CNX module: m22000
Solutions to Exercises in this Module
Solution to Exercise (p. 6)
Solution to Exercise (p. 6)
Solution to Exercise (p. 6)
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Solution to Exercise (p. 7)
Solution to Exercise (p. 8)
Solution to Exercise (p. 8)
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Solution to Exercise (p. 9)
Solution to Exercise (p. 10)
Solution to Exercise (p. 10)
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Solution to Exercise (p. 11)
Solution to Exercise (p. 12)
Solution to Exercise (p. 13)
x≤1
Solution to Exercise (p. 13)
Solution to Exercise (p. 13)
m=
−2
3
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