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Chapter 1
Graphs, Functions, and Models
To graph (−1, 4) we move from the origin 1 unit to the
Check Your Understanding Section 1.1 left of the y-axis. Then we move 4 units up from the
x-axis.
1. The point (−5, 0) is on an axis, so it is not in any quadrant. To graph (0, 2) we do not move to the right or the left of
The statement is false. the y-axis since the first coordinate is 0. From the origin
we move 2 units up.
2. The ordered pair (1, −6) is located 1 unit right of the origin
and 6 units below it. The ordered pair (−6, 1) is located 6 To graph (2, −2) we move from the origin 2 units to the
units left of the origin and 1 unit above it. Thus, (1, −6) right of the y-axis. Then we move 2 units down from the
and (−6, 1) do not name the same point. The statement x-axis.
is false. y
3. True; the first coordinate of a point is also called the
( 1, 4) 4
abscissa.
2 (0, 2)
(4, 0)
4. True; the point (−2, 7) is 2 units left of the origin and
4 2 2 4 x
7 units above it. 2 (2, 2)
5. True; the second coordinate of a point is also called the ( 3, 5) 4
ordinate.
6. False; the point (0, −3) is on the y-axis. 5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
To graph (5, 1) we move from the origin 5 units to the right
Exercise Set 1.1 of the y-axis. Then we move 1 unit up from the x-axis.
To graph (2, 3) we move from the origin 2 units to the right
1. Point A is located 5 units to the left of the y-axis and of the y-axis. Then we move 3 units up from the x-axis.
4 units up from the x-axis, so its coordinates are (−5, 4).
To graph (2, −1) we move from the origin 2 units to the
Point B is located 2 units to the right of the y-axis and right of the y-axis. Then we move 1 unit down from the
2 units down from the x-axis, so its coordinates are (2, −2). x-axis.
Point C is located 0 units to the right or left of the y-axis To graph (0, 1) we do not move to the right or the left of
and 5 units down from the x-axis, so its coordinates are the y-axis since the first coordinate is 0. From the origin
(0, −5). we move 1 unit up.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). y
Point E is located 5 units to the left of the y-axis and 4
4 units down from the x-axis, so its coordinates are (2, 3)
2
(−5, −4). ( 5, 1) (0, 1) (5, 1)
4 2 4 x
Point F is located 3 units to the right of the y-axis and 2 (2, 1)
0 units up or down from the x-axis, so its coordinates are
4
(3, 0).
3. To graph (4, 0) we move from the origin 4 units to the right
7. The first coordinate represents the year and the corre-
of the y-axis. Since the second coordinate is 0, we do not
sponding second coordinate represents the number of cities
move up or down from the x-axis.
served by Southwest Airlines. The ordered pairs are
To graph (−3, −5) we move from the origin 3 units to the (1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),
left of the y-axis. Then we move 5 units down from the and (2021, 121).
x-axis.
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14 Chapter 1: Graphs, Functions, and Models
9. To determine whether (−1, −9) is a solution, substitute 2a + 5b = 3
−1 for x and −9 for y. 3
2·0+5·
? 3
y = 7x − 2 5
−9 ? 7(−1) − 2 0+3
−7 − 2 3 3 TRUE
3
−9 −9
TRUE
The equation 3 = 3 is true, so 0, is a solution.
The equation −9 = −9 is true, so (−1, −9) is a solution. 5
To determine whether (0, 2) is a solution, substitute 0 for 15. To determine whether (−0.75, 2.75) is a solution, substi-
x and 2 for y. tute −0.75 for x and 2.75 for y.
y = 7x − 2 x2 − y 2 = 3
2 ? 7 · 0 − 2 (−0.75)2 − (2.75)2 ? 3
0−2 0.5625 − 7.5625
2 −2
FALSE −7 3 FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution. The equation −7 = 3 is false, so (−0.75, 2.75) is not a
2 3 solution.
2
11. To determine whether , is a solution, substitute To determine whether (2, −1) is a solution, substitute 2
3 4 3
3 for x and −1 for y.
for x and for y.
4 x2 − y 2 = 3
6x − 4y = 1
22 − (−1)2 ? 3
2 3 4−1
6· −4· ? 1
3 4
3 3 TRUE
4−3 The equation 3 = 3 is true, so (2, −1) is a solution.
1 1 TRUE
2 3 17. Graph 5x − 3y = −15.
The equation 1 = 1 is true, so , is a solution. To find the x-intercept we replace y with 0 and solve for
3 4
3 x.
To determine whether 1, is a solution, substitute 1 for 5x − 3 · 0 = −15
2
3
x and for y. 5x = −15
2
x = −3
6x − 4y = 1
The x-intercept is (−3, 0).
3
6·1−4· ? 1 To find the y-intercept we replace x with 0 and solve for
2
y.
6−6
5 · 0 − 3y = −15
0 1 FALSE
3 −3y = −15
The equation 0 = 1 is false, so 1, is not a solution. y=5
2
1 4 The y-intercept is (0, 5).
13. To determine whether − , − is a solution, substitute We plot the intercepts and draw the line that contains
2 5
1 4 them. We could find a third point as a check that the
− for a and − for b.
2 5 intercepts were found correctly.
2a + 5b = 3
y
1 4 (0, 5)
2 − +5 − ? 3 4
2 5
2 5x 3y 15
−1 − 4 ( 3, 0)
4 2 2 4 x
−5 3 FALSE 2
1 4
4
The equation −5 = 3 is false, so − , − is not a solu-
2 5
tion.
3
To determine whether 0, is a solution, substitute 0 for
5
3
a and for b.
5
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Exercise Set 1.1 15
19. Graph 2x + y = 4. When x = 0, y = 3x + 5 = 3 · 0 + 5 = 0 + 5 = 5
To find the x-intercept we replace y with 0 and solve for We list these points in a table, plot them, and draw the
x. graph.
2x + 0 = 4 y
x y (x, y)
2x = 4 6
−3 −4 (−3, −4) y 3x 5
x=2
2
The x-intercept is (2, 0). −1 2 (−1, 2)
4 2 4 x
To find the y-intercept we replace x with 0 and solve for 0 5 (0, 5) 2
y.
2·0+y = 4
25. Graph x − y = 3.
y=4
Make a table of values, plot the points in the table, and
The y-intercept is (0, 4).
draw the graph.
We plot the intercepts and draw the line that contains y
them. We could find a third point as a check that the x y (x, y) x y 3
4
intercepts were found correctly.
−2 −5 (−2, −5) 2
y 4 2 2 4 x
0 −3 (0, −3)
2x y 4 4 (0, 4) 2
3 0 (3, 0) 4
2
(2, 0)
4 2 2 4 x 3
2 27. Graph y = − x + 3.
4
4
By choosing multiples of 4 for x, we can avoid fraction
values for y. Make a table of values, plot the points in the
21. Graph 4y − 3x = 12. table, and draw the graph.
To find the x-intercept we replace y with 0 and solve for y
x. x y (x, y)
4
4 · 0 − 3x = 12 −4 6 (−4, 6) 2
−3x = 12
0 3 (0, 3) 4 2 2 4 x
x = −4 2
3
4 0 (4, 0) y 4x 3
The x-intercept is (−4, 0). 4
To find the y-intercept we replace x with 0 and solve for y.
4y − 3 · 0 = 12 29. Graph 5x − 2y = 8.
4y = 12 We could solve for y first.
y=3 5x − 2y = 8
The y-intercept is (0, 3). −2y = −5x + 8 Subtracting 5x on both sides
We plot the intercepts and draw the line that contains 5 1
y = x−4 Multiplying by − on both
them. We could find a third point as a check that the 2 2
sides
intercepts were found correctly.
By choosing multiples of 2 for x we can avoid fraction
y values for y. Make a table of values, plot the points in the
table, and draw the graph.
4y 3x 12 4
2
(0, 3) y
( 4, 0) x y (x, y)
4 2 2 4 x 4
2 0 −4 (0, −4) 2
4
2 1 (2, 1) 4 2 2 4 x
2
23. Graph y = 3x + 5. 4 6 (4, 6) 4
5x 2y 8
We choose some values for x and find the corresponding
y-values.
When x = −3, y = 3x + 5 = 3(−3) + 5 = −9 + 5 = −4.
When x = −1, y = 3x + 5 = 3(−1) + 5 = −3 + 5 = 2.
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16 Chapter 1: Graphs, Functions, and Models
31. Graph x − 4y = 5. 39. Graph y = x2 + 2x + 3.
Make a table of values, plot the points in the table, and Make a table of values, plot the points in the table, and
draw the graph. draw the graph.
y y
x y (x, y) x y (x, y) 10
4 x 4y 5
8
−3 −2 (−3, −2) 2 −3 6 (−3, 6)
6
1 −1 (1, −1) 2 4 6 x −2 3 (−2, 3)
2 4
y 5 x 2 1 2x 1 3
5 0 (5, 0) 4 −1 2 (−1, 2) 2
24 22 2 4 x
0 3 (0, 3)
33. Graph 2x + 5y = −10.
1 6 (1, 6)
In this case, it is convenient to find the intercepts along
with a third point on the graph. Make a table of values,
plot the points in the table, and draw the graph. 41. Graph (b) is the graph of y = 3 − x.
y
x y (x, y) 43. Graph (a) is the graph of y = x2 + 2x + 1.
4 2x 5y 10
−5 0 (−5, 0) 2
45. Enter the equation, select the standard window, and graph
the equation.
0 −2 (0, −2) 6 2 2 x
y 2x 1
5 −4 (5, −4) 4 10
35. Graph y = −x2 . 10 10
Make a table of values, plot the points in the table, and
draw the graph.
10
y
x y (x, y) 47. First solve the equation for y: y = −4x + 7. Enter the
4 2 2 4 x equation in this form, select the standard window, and
−2 −4 (−2, −4) 2
graph the equation.
4 y x2
−1 −1 (−1, −1)
6 4x y 7
0 0 (0, 0) 8 10
1 −1 (1, −1)
10 10
2 −4 (2, −4)
37. Graph y = x2 − 3. 10
Make a table of values, plot the points in the table, and 49. Enter the equation, select the standard window, and graph
draw the graph. the equation.
y
x y (x, y) 1
6
y 3x 2
−3 6 (−3, 6) 4 10
2 y x2 3
−1 −2 (−1, −2)
4 2 2 4 x 10 10
0 −3 (0, −3) 2
1 −2 (1, −2) 10
3 6 (3, 6)
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Chapter 1
Graphs, Functions, and Models
To graph (−1, 4) we move from the origin 1 unit to the
Check Your Understanding Section 1.1 left of the y-axis. Then we move 4 units up from the
x-axis.
1. The point (−5, 0) is on an axis, so it is not in any quadrant. To graph (0, 2) we do not move to the right or the left of
The statement is false. the y-axis since the first coordinate is 0. From the origin
we move 2 units up.
2. The ordered pair (1, −6) is located 1 unit right of the origin
and 6 units below it. The ordered pair (−6, 1) is located 6 To graph (2, −2) we move from the origin 2 units to the
units left of the origin and 1 unit above it. Thus, (1, −6) right of the y-axis. Then we move 2 units down from the
and (−6, 1) do not name the same point. The statement x-axis.
is false. y
3. True; the first coordinate of a point is also called the
( 1, 4) 4
abscissa.
2 (0, 2)
(4, 0)
4. True; the point (−2, 7) is 2 units left of the origin and
4 2 2 4 x
7 units above it. 2 (2, 2)
5. True; the second coordinate of a point is also called the ( 3, 5) 4
ordinate.
6. False; the point (0, −3) is on the y-axis. 5. To graph (−5, 1) we move from the origin 5 units to the
left of the y-axis. Then we move 1 unit up from the x-axis.
To graph (5, 1) we move from the origin 5 units to the right
Exercise Set 1.1 of the y-axis. Then we move 1 unit up from the x-axis.
To graph (2, 3) we move from the origin 2 units to the right
1. Point A is located 5 units to the left of the y-axis and of the y-axis. Then we move 3 units up from the x-axis.
4 units up from the x-axis, so its coordinates are (−5, 4).
To graph (2, −1) we move from the origin 2 units to the
Point B is located 2 units to the right of the y-axis and right of the y-axis. Then we move 1 unit down from the
2 units down from the x-axis, so its coordinates are (2, −2). x-axis.
Point C is located 0 units to the right or left of the y-axis To graph (0, 1) we do not move to the right or the left of
and 5 units down from the x-axis, so its coordinates are the y-axis since the first coordinate is 0. From the origin
(0, −5). we move 1 unit up.
Point D is located 3 units to the right of the y-axis and
5 units up from the x-axis, so its coordinates are (3, 5). y
Point E is located 5 units to the left of the y-axis and 4
4 units down from the x-axis, so its coordinates are (2, 3)
2
(−5, −4). ( 5, 1) (0, 1) (5, 1)
4 2 4 x
Point F is located 3 units to the right of the y-axis and 2 (2, 1)
0 units up or down from the x-axis, so its coordinates are
4
(3, 0).
3. To graph (4, 0) we move from the origin 4 units to the right
7. The first coordinate represents the year and the corre-
of the y-axis. Since the second coordinate is 0, we do not
sponding second coordinate represents the number of cities
move up or down from the x-axis.
served by Southwest Airlines. The ordered pairs are
To graph (−3, −5) we move from the origin 3 units to the (1971, 3), (1981, 15), (1991, 32), (2001, 59), (2011, 72),
left of the y-axis. Then we move 5 units down from the and (2021, 121).
x-axis.
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14 Chapter 1: Graphs, Functions, and Models
9. To determine whether (−1, −9) is a solution, substitute 2a + 5b = 3
−1 for x and −9 for y. 3
2·0+5·
? 3
y = 7x − 2 5
−9 ? 7(−1) − 2 0+3
−7 − 2 3 3 TRUE
3
−9 −9
TRUE
The equation 3 = 3 is true, so 0, is a solution.
The equation −9 = −9 is true, so (−1, −9) is a solution. 5
To determine whether (0, 2) is a solution, substitute 0 for 15. To determine whether (−0.75, 2.75) is a solution, substi-
x and 2 for y. tute −0.75 for x and 2.75 for y.
y = 7x − 2 x2 − y 2 = 3
2 ? 7 · 0 − 2 (−0.75)2 − (2.75)2 ? 3
0−2 0.5625 − 7.5625
2 −2
FALSE −7 3 FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution. The equation −7 = 3 is false, so (−0.75, 2.75) is not a
2 3 solution.
2
11. To determine whether , is a solution, substitute To determine whether (2, −1) is a solution, substitute 2
3 4 3
3 for x and −1 for y.
for x and for y.
4 x2 − y 2 = 3
6x − 4y = 1
22 − (−1)2 ? 3
2 3 4−1
6· −4· ? 1
3 4
3 3 TRUE
4−3 The equation 3 = 3 is true, so (2, −1) is a solution.
1 1 TRUE
2 3 17. Graph 5x − 3y = −15.
The equation 1 = 1 is true, so , is a solution. To find the x-intercept we replace y with 0 and solve for
3 4
3 x.
To determine whether 1, is a solution, substitute 1 for 5x − 3 · 0 = −15
2
3
x and for y. 5x = −15
2
x = −3
6x − 4y = 1
The x-intercept is (−3, 0).
3
6·1−4· ? 1 To find the y-intercept we replace x with 0 and solve for
2
y.
6−6
5 · 0 − 3y = −15
0 1 FALSE
3 −3y = −15
The equation 0 = 1 is false, so 1, is not a solution. y=5
2
1 4 The y-intercept is (0, 5).
13. To determine whether − , − is a solution, substitute We plot the intercepts and draw the line that contains
2 5
1 4 them. We could find a third point as a check that the
− for a and − for b.
2 5 intercepts were found correctly.
2a + 5b = 3
y
1 4 (0, 5)
2 − +5 − ? 3 4
2 5
2 5x 3y 15
−1 − 4 ( 3, 0)
4 2 2 4 x
−5 3 FALSE 2
1 4
4
The equation −5 = 3 is false, so − , − is not a solu-
2 5
tion.
3
To determine whether 0, is a solution, substitute 0 for
5
3
a and for b.
5
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Exercise Set 1.1 15
19. Graph 2x + y = 4. When x = 0, y = 3x + 5 = 3 · 0 + 5 = 0 + 5 = 5
To find the x-intercept we replace y with 0 and solve for We list these points in a table, plot them, and draw the
x. graph.
2x + 0 = 4 y
x y (x, y)
2x = 4 6
−3 −4 (−3, −4) y 3x 5
x=2
2
The x-intercept is (2, 0). −1 2 (−1, 2)
4 2 4 x
To find the y-intercept we replace x with 0 and solve for 0 5 (0, 5) 2
y.
2·0+y = 4
25. Graph x − y = 3.
y=4
Make a table of values, plot the points in the table, and
The y-intercept is (0, 4).
draw the graph.
We plot the intercepts and draw the line that contains y
them. We could find a third point as a check that the x y (x, y) x y 3
4
intercepts were found correctly.
−2 −5 (−2, −5) 2
y 4 2 2 4 x
0 −3 (0, −3)
2x y 4 4 (0, 4) 2
3 0 (3, 0) 4
2
(2, 0)
4 2 2 4 x 3
2 27. Graph y = − x + 3.
4
4
By choosing multiples of 4 for x, we can avoid fraction
values for y. Make a table of values, plot the points in the
21. Graph 4y − 3x = 12. table, and draw the graph.
To find the x-intercept we replace y with 0 and solve for y
x. x y (x, y)
4
4 · 0 − 3x = 12 −4 6 (−4, 6) 2
−3x = 12
0 3 (0, 3) 4 2 2 4 x
x = −4 2
3
4 0 (4, 0) y 4x 3
The x-intercept is (−4, 0). 4
To find the y-intercept we replace x with 0 and solve for y.
4y − 3 · 0 = 12 29. Graph 5x − 2y = 8.
4y = 12 We could solve for y first.
y=3 5x − 2y = 8
The y-intercept is (0, 3). −2y = −5x + 8 Subtracting 5x on both sides
We plot the intercepts and draw the line that contains 5 1
y = x−4 Multiplying by − on both
them. We could find a third point as a check that the 2 2
sides
intercepts were found correctly.
By choosing multiples of 2 for x we can avoid fraction
y values for y. Make a table of values, plot the points in the
table, and draw the graph.
4y 3x 12 4
2
(0, 3) y
( 4, 0) x y (x, y)
4 2 2 4 x 4
2 0 −4 (0, −4) 2
4
2 1 (2, 1) 4 2 2 4 x
2
23. Graph y = 3x + 5. 4 6 (4, 6) 4
5x 2y 8
We choose some values for x and find the corresponding
y-values.
When x = −3, y = 3x + 5 = 3(−3) + 5 = −9 + 5 = −4.
When x = −1, y = 3x + 5 = 3(−1) + 5 = −3 + 5 = 2.
Copyright c 2025 Pearson Education, Inc.
hs
th aand
beecher
models
Solution
judith
7thmanual
edition
a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!####################%%%%%%%%%%%%%%%%%%%%
for
by college
marvin lalgebra
bittingergraphs
judith aand
beecher
models
judith
7th edition
a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!####################
by marvin l bittinger judith a beecher judith a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!
,hs
th aand
beecher
models
Solution
judith
7thmanual
edition
a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!####################%%%%%%%%%%%%%%%%%%%%
for
by college
marvin lalgebra
bittingergraphs
judith aand
beecher
models
judith
7th edition
a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!####################
by marvin l bittinger judith a beecher judith a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!
16 Chapter 1: Graphs, Functions, and Models
31. Graph x − 4y = 5. 39. Graph y = x2 + 2x + 3.
Make a table of values, plot the points in the table, and Make a table of values, plot the points in the table, and
draw the graph. draw the graph.
y y
x y (x, y) x y (x, y) 10
4 x 4y 5
8
−3 −2 (−3, −2) 2 −3 6 (−3, 6)
6
1 −1 (1, −1) 2 4 6 x −2 3 (−2, 3)
2 4
y 5 x 2 1 2x 1 3
5 0 (5, 0) 4 −1 2 (−1, 2) 2
24 22 2 4 x
0 3 (0, 3)
33. Graph 2x + 5y = −10.
1 6 (1, 6)
In this case, it is convenient to find the intercepts along
with a third point on the graph. Make a table of values,
plot the points in the table, and draw the graph. 41. Graph (b) is the graph of y = 3 − x.
y
x y (x, y) 43. Graph (a) is the graph of y = x2 + 2x + 1.
4 2x 5y 10
−5 0 (−5, 0) 2
45. Enter the equation, select the standard window, and graph
the equation.
0 −2 (0, −2) 6 2 2 x
y 2x 1
5 −4 (5, −4) 4 10
35. Graph y = −x2 . 10 10
Make a table of values, plot the points in the table, and
draw the graph.
10
y
x y (x, y) 47. First solve the equation for y: y = −4x + 7. Enter the
4 2 2 4 x equation in this form, select the standard window, and
−2 −4 (−2, −4) 2
graph the equation.
4 y x2
−1 −1 (−1, −1)
6 4x y 7
0 0 (0, 0) 8 10
1 −1 (1, −1)
10 10
2 −4 (2, −4)
37. Graph y = x2 − 3. 10
Make a table of values, plot the points in the table, and 49. Enter the equation, select the standard window, and graph
draw the graph. the equation.
y
x y (x, y) 1
6
y 3x 2
−3 6 (−3, 6) 4 10
2 y x2 3
−1 −2 (−1, −2)
4 2 2 4 x 10 10
0 −3 (0, −3) 2
1 −2 (1, −2) 10
3 6 (3, 6)
Copyright c 2025 Pearson Education, Inc.
hs
th aand
beecher
models
Solution
judith
7thmanual
edition
a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!####################%%%%%%%%%%%%%%%%%%%%
for
by college
marvin lalgebra
bittingergraphs
judith aand
beecher
models
judith
7th edition
a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!####################
by marvin l bittinger judith a beecher judith a penna.pdf@@@@@@@@@@@@@@@@@@@@@@!!!!!!!!!!!!!!!!!!
