zlc 2025 Pearson Education, Inc.
Copyright ◯ z l zl zl zl
,Chapter 1 z l
Graphs, Functions, and Models z l z l z l
To graph (−1, 4) we move from the origin 1 un
z l z l zl z l z l z l z l z l z l z l
he
Check Your Understanding Section 1.1
zl zl zl zl
left of the y- z l z l z l
axis. Then we move 4 units up from the
z l z l z l z l z l z l z l z l z l
1. The point ( — 5, 0) is on an axis, so it is not in any quadrant.
zl zl z l zl zl zl zl zl zl zl zl zl zl zl zl zl
x-axis.
The statement is false.
zl zl zl To graph (0, 2) we do not move to the right or the left
zl zl zl zl zl zl zl zl zl zl zl zl zl
e y- zl
2. The ordered pair (1, — 6) is located 1 unit right of the origin
axis since the first coordinate is 0. From the origin
zl zl zl z l zl zl zl zl zl zl zl zl zl
and 6 units below it. The ordered pair ( 6, — 1) is located 6 u
zl zl zl zl zl zl z l zl zl z l
zl zl zl zl zl zl zl zl z l zl zl zl zl zl
ve 2 units up.
zl zl zl
nits left of the origin and 1 unit above it. Thus, (1, 6) — a
zl zl zl zl zl zl zl zl zl z l zl z l z l
nd ( —6, 1) do not name the same point. The stateme
zl z l zl z l z l z l z l z l z l z l z l
To graph (2, —
2) we move from the origin 2 units to th
zl zl z l zl zl zl zl zl zl zl zl zl
nt is false.
zl zl
ht of the y-
zl zl zl
y
axis. Then we move 2 units down from the x-axis
z l zl zl zl zl zl zl zl zl
3. True; the first coordinate of a point is also called the abscis
zl zl zl zl zl zl zl zl zl zl zl
( 1, 4)
sa.
zl z
l 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
z l z l z l z l z l z l z l z l z l z l z l z l z l
zl
2z l (0,zl2)
(4, 0)
7 units above it.
zl
zl zl zl z l
4 2 2
2 (2, 4 z l z l 2
)
5. True; the second coordinate of a point is also called the ord
zl zl zl zl zl zl zl zl zl zl zl
( 3, z l z l 5)
4
inate.
6. False; the point (0, −3) is on the y-axis.
zl zl zl zl zl zl zl zl
5. To graph ( 5, 1) we move from the origin 5 un
z l z l z l z l zl z l z l z l z l z l z l z l
the left of —
z l the y- zl zl zl zl
axis. Then we move 1 unit up from the x-axis.
z l zl zl zl zl zl zl zl zl
Exercise Set 1.1 zl zl
To graph (5, 1) we move from the origin 5 units to the
zl zl zl zl zl zl zl zl zl zl zl zl
of the y-axis. Then we move 1 unit up from t
z l z l z l z l z l z l z l z l z l z l
1. Point A is located 5 units to the left of the y-
z l z l z l z l z l z l z l z l z l z l z l axis.
axis and 4 units up from the x-
z l z l z l z l z l z l z l
To graph (2, 3) we move from the origin 2 units to the
zl zl zl zl zl zl zl zl zl zl zl zl
axis, so its coordinates are (−5, 4).
z l z l z l z l z l zl
of the y-axis. Then we move 3 units up from t
z l z l z l z l z l z l z l z l z l z l
Point B is located 2 units to the right of the y-
z l z l z l z l z l z l z l z l z l z l z l axis.
axis and 2 units down from the x-
z l zl zl zl zl zl zl
To graph (2, —
1) we move from the origin 2 units to t
zl zl z l zl zl zl zl zl zl zl zl zl
axis, so its coordinates are (2, −2).
zl zl zl zl zl zl
ht of the y-
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Point C is located 0 units to the right or left of the y-
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z l z l z l z l z l z l z l z l zl
axis and 5 units down from the x-
zl zl zl zl zl zl zl
To graph (0, 1) we do not move to the right or the left
zl zl zl zl zl zl zl zl zl zl zl zl zl
axis, so its coordinates are (0, −5).
zl zl zl zl zl zl
e y- zl
Point D is located 3 units to the right of the y-
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zl zl zl zl zl zl z l zl zl z l
axis and 5 units up from the x-
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zl zl zl
axis, so its coordinates are (3, 5).
zl zl zl zl zl zl
y
Point E is located 5 units to the left of the y-
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axis and z l
4
4 units down from the x-
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2
(2, 3) zl
axis, so its coordinates are (−5, −4).
z l z l z l z l z l zl ( 5, 1 zl (0, 1) zl (5, 1) zl
)
Point F is located 3 units to the right of the y-
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axis and 0 units up or down from the x-
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1)
axis, so its coordinates are (3, 0).
zl zl zl zl zl zl 4
3. To graph (4, 0) we move from the origin 4 units to the right
z l zl zl zl zl zl zl zl zl zl zl zl zl zl zl
7. The first coordinate represents the year and the corre
zl zl zl zl zl zl zl zl zl
of the y-
zl zl
sponding second coordinate represents the number o
zl zl zl zl zl zl zl
axis. Since the second coordinate is 0, we do not move up
z l zl zl zl zl zl zl zl zl zl zl zl
served by Southwest Airlines. The ordered pairs are
zl zl zl zl z l zl zl zl
or down from the x-axis.
zl zl zl zl
71, 3), (1981, 15), (1991, 32), (2001, 59), (2011,
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To graph ( − 3, 5) we move from the origin 3 units to the
zl zl z l z l z zl l zl zl zl zl zl zl zl zl zl zl and (2021, 121). zl zl
left of the −y- z l z l z l
axis. Then we move 5 units down from the x-axis.
z l z l z l z l z l z l z l z l zl
c 2025 Pearson Education, Inc.
Copyright ◯ zl z l zl zl zl
,
, 14 Chapter 1: zl z l z l Graphs, Functions, and M zl zl zl
ls
9. To determine whether (−1, −9) is a solution, substi
z l z l z l z l zl z l z l z l z l
2a + 5b = 3 zl zl zl zl zl
tute 3
−1 for x and −9 for y. 2·0+5· ? 3
5
zl zl zl zl zl zl zl zl zl zl zl zl z l z l zl
z l
y = 7x − 2
zl zl zl zl
−9 ?¯ 7(−1) − 2 z l zl zl 0+3 ¯ zl zl zl
¯ −7 − 2 zl zl 3 ¯ 3 TRUE ³ 3´ z l
zl
z l
zl
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
zl
TRUE z l
zl
zl zl zl zl zl zl zl zl zl zl zl
The equation −9 = −9 is true, so (−1, −9) is a solut
z l z l zl z l z l z l z l z l zl z l z l z l
5
ion. To determine whether (0, 2) is a solution, substitute
zl zl zl zl zl zl zl zl zl zl
15. To determine whether (−0.75, 2.75) is a solution,
zl z l z l zl z l z l z l
ti- tute −0.75 for x and 2.75 for y.
zl zl zl zl zl zl zl
0 for
zl
zl
x and 2 for y.
zl zl zl zl
x2 − y2 = 3 zl zl zl zl zl
y = 7x − 2 z l zl zl zl
2 ? 7 ·0−
z l z l zl zl zl zl
(−0.75)2 − (2.75)2 ?¯ 3 zl
zl
z l
zl
zl
2
¯ 0— 2 0.5625 − 7.5625 zl zl
¯
¯
2 −2 zl
FALSE −7 ¯ 3 FALSE z l
zl
z l
The equation 2 = −2 is false, so (0, 2) is not a solution.
zl zl zl zl zl zl zl zl zl zl zl zl zl
The equation − z l −
³ 2 3´ 2 zl zl 0.75, 2.75) 7 = 3 is false, so (
z l z l z l z l z l z l z l zl
11. To determine whether , is a solution, substitute
zl zl
zl zl zl zl zl zl zl zl z l
ot a solution.
z l zl
To determine whether (2, −1) is a solution, sub
2
z l z l z l zl z l z l z l z l
l
3 4 3
3
z l
for x and −1 for y.
for x and for y.
z l zl zl zl zl zl
zl zl z l zl
4 x − y2 = 32
zl zl zl zl zl
6x − 4y = 1
22 − (−1)2 ?¯ 3
zl zl z l zl zl
zl z l
zl zl
zl
2 3 4— 1
6 · −4 · ? 1 ¯
3 4
zl z l zl zl z l z l
zl
3 ¯ 3 TRUE
zl z l
4
zl
z l z l
¯ 3 z l
— The equation 3 = 3 is true, so (2, −1) is a solution
zl zl zl zl zl zl zl zl zl zl zl zl
1 ¯ 1 TRUE zl
z l
³2 3´ zl 17. Graph 5x − 3y = −15.
zl zl zl zl zl zl
z l zl
The equation 1 = 1 is³ true, s , is a solution To
x find the x-
z zl l
3 4
zl zl zl zl zl zl zl zl zl
3´
zl zl zl
o
To determine whether 1, is a solution, .substitute 1 for
zl zl
intercept we replace y with 0 and solve for
zl
zl zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl zl
2 . 5x − 3 · 0 = −15 zl zl zl zl zl zl
3
x and for y. 5x = −15 zl z l
2
zl zl zl
zl
x = −3
6x − 4y = 1
zl zl
zl zl z l zl zl
The x-intercept is (−3, 0).
z l z l z l
3
zl
6 · 1 −4 · ? 1
z l
2
zl zl zl zl zl z l z l zl
To find the y-
zl zl zl
intercept we replace x with 0 and solve for zl zl zl zl zl zl zl zl
6 −6 ¯ zl zl
y.
0 ¯ 1 FALSE z l
zl
z l
5 · 0 − 3y = −15
zl zl zl zl zl zl
³ z l
3´
−3y = −15
zl
zl zl
The equation 0 = 1 is false, so 1, is not a solution. y =5
2
zl zl zl zl zl zl zl z l z l zl zl zl zl z l zl zl
³ 1 4´ zl zl The y-intercept is (0, 5).
zl zl zl zl
13. To determine whether − , − is a solution, substitute
zl zl
We plot the intercepts and draw the line tha
zl zl zl zl zl zl zl zl zl zl
2 5 z l z l z l z l z l z l z l z l
1 4 ins
them. We could find a third point as a chec
− for a and − for b.
z l z l z l z l z l z l z l z l z l z l
2 5 at the
l
z zl zl z
l zl
intercepts were found correctly.
zl
z l z l z l zl zl zl
2a + 5b = 3 zl zl zl zl zl
³ 1´ ³ 4´ zl zl zl zl
2 − +5 − ? 3
zl zl
zl zl zl z l zl
2 5
−1 − 4 zl zl
−5 ¯ 3 FALSE zl
z l
c 2025 Pearson Education, Inc.
Copyright ◯ zl z l zl zl zl
Copyright ◯ z l zl zl zl
,Chapter 1 z l
Graphs, Functions, and Models z l z l z l
To graph (−1, 4) we move from the origin 1 un
z l z l zl z l z l z l z l z l z l z l
he
Check Your Understanding Section 1.1
zl zl zl zl
left of the y- z l z l z l
axis. Then we move 4 units up from the
z l z l z l z l z l z l z l z l z l
1. The point ( — 5, 0) is on an axis, so it is not in any quadrant.
zl zl z l zl zl zl zl zl zl zl zl zl zl zl zl zl
x-axis.
The statement is false.
zl zl zl To graph (0, 2) we do not move to the right or the left
zl zl zl zl zl zl zl zl zl zl zl zl zl
e y- zl
2. The ordered pair (1, — 6) is located 1 unit right of the origin
axis since the first coordinate is 0. From the origin
zl zl zl z l zl zl zl zl zl zl zl zl zl
and 6 units below it. The ordered pair ( 6, — 1) is located 6 u
zl zl zl zl zl zl z l zl zl z l
zl zl zl zl zl zl zl zl z l zl zl zl zl zl
ve 2 units up.
zl zl zl
nits left of the origin and 1 unit above it. Thus, (1, 6) — a
zl zl zl zl zl zl zl zl zl z l zl z l z l
nd ( —6, 1) do not name the same point. The stateme
zl z l zl z l z l z l z l z l z l z l z l
To graph (2, —
2) we move from the origin 2 units to th
zl zl z l zl zl zl zl zl zl zl zl zl
nt is false.
zl zl
ht of the y-
zl zl zl
y
axis. Then we move 2 units down from the x-axis
z l zl zl zl zl zl zl zl zl
3. True; the first coordinate of a point is also called the abscis
zl zl zl zl zl zl zl zl zl zl zl
( 1, 4)
sa.
zl z
l 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
z l z l z l z l z l z l z l z l z l z l z l z l z l
zl
2z l (0,zl2)
(4, 0)
7 units above it.
zl
zl zl zl z l
4 2 2
2 (2, 4 z l z l 2
)
5. True; the second coordinate of a point is also called the ord
zl zl zl zl zl zl zl zl zl zl zl
( 3, z l z l 5)
4
inate.
6. False; the point (0, −3) is on the y-axis.
zl zl zl zl zl zl zl zl
5. To graph ( 5, 1) we move from the origin 5 un
z l z l z l z l zl z l z l z l z l z l z l z l
the left of —
z l the y- zl zl zl zl
axis. Then we move 1 unit up from the x-axis.
z l zl zl zl zl zl zl zl zl
Exercise Set 1.1 zl zl
To graph (5, 1) we move from the origin 5 units to the
zl zl zl zl zl zl zl zl zl zl zl zl
of the y-axis. Then we move 1 unit up from t
z l z l z l z l z l z l z l z l z l z l
1. Point A is located 5 units to the left of the y-
z l z l z l z l z l z l z l z l z l z l z l axis.
axis and 4 units up from the x-
z l z l z l z l z l z l z l
To graph (2, 3) we move from the origin 2 units to the
zl zl zl zl zl zl zl zl zl zl zl zl
axis, so its coordinates are (−5, 4).
z l z l z l z l z l zl
of the y-axis. Then we move 3 units up from t
z l z l z l z l z l z l z l z l z l z l
Point B is located 2 units to the right of the y-
z l z l z l z l z l z l z l z l z l z l z l axis.
axis and 2 units down from the x-
z l zl zl zl zl zl zl
To graph (2, —
1) we move from the origin 2 units to t
zl zl z l zl zl zl zl zl zl zl zl zl
axis, so its coordinates are (2, −2).
zl zl zl zl zl zl
ht of the y-
z l z l z l
Point C is located 0 units to the right or left of the y-
zl zl zl zl zl zl zl zl zl zl zl zl zl axis. Then we move 1 unit down from the x-axis
z l z l z l z l z l z l z l z l zl
axis and 5 units down from the x-
zl zl zl zl zl zl zl
To graph (0, 1) we do not move to the right or the left
zl zl zl zl zl zl zl zl zl zl zl zl zl
axis, so its coordinates are (0, −5).
zl zl zl zl zl zl
e y- zl
Point D is located 3 units to the right of the y-
z l z l z l z l z l z l z l z l z l z l z l axis since the first coordinate is 0. From the origin
zl zl zl zl zl zl z l zl zl z l
axis and 5 units up from the x-
z l zl zl zl zl zl z l ve 1 unit up.
zl zl zl
axis, so its coordinates are (3, 5).
zl zl zl zl zl zl
y
Point E is located 5 units to the left of the y-
z l z l z l z l z l z l z l z l z l z l z l
axis and z l
4
4 units down from the x-
z l z l z l z l z l
2
(2, 3) zl
axis, so its coordinates are (−5, −4).
z l z l z l z l z l zl ( 5, 1 zl (0, 1) zl (5, 1) zl
)
Point F is located 3 units to the right of the y-
z l z l z l z l z l z l z l z l z l z l z l 4 2 2z l (2,z l z l
axis and 0 units up or down from the x-
z l zl zl zl zl zl zl zl zl
1)
axis, so its coordinates are (3, 0).
zl zl zl zl zl zl 4
3. To graph (4, 0) we move from the origin 4 units to the right
z l zl zl zl zl zl zl zl zl zl zl zl zl zl zl
7. The first coordinate represents the year and the corre
zl zl zl zl zl zl zl zl zl
of the y-
zl zl
sponding second coordinate represents the number o
zl zl zl zl zl zl zl
axis. Since the second coordinate is 0, we do not move up
z l zl zl zl zl zl zl zl zl zl zl zl
served by Southwest Airlines. The ordered pairs are
zl zl zl zl z l zl zl zl
or down from the x-axis.
zl zl zl zl
71, 3), (1981, 15), (1991, 32), (2001, 59), (2011,
zl z l zl z l zl z l zl z l
To graph ( − 3, 5) we move from the origin 3 units to the
zl zl z l z l z zl l zl zl zl zl zl zl zl zl zl zl and (2021, 121). zl zl
left of the −y- z l z l z l
axis. Then we move 5 units down from the x-axis.
z l z l z l z l z l z l z l z l zl
c 2025 Pearson Education, Inc.
Copyright ◯ zl z l zl zl zl
,
, 14 Chapter 1: zl z l z l Graphs, Functions, and M zl zl zl
ls
9. To determine whether (−1, −9) is a solution, substi
z l z l z l z l zl z l z l z l z l
2a + 5b = 3 zl zl zl zl zl
tute 3
−1 for x and −9 for y. 2·0+5· ? 3
5
zl zl zl zl zl zl zl zl zl zl zl zl z l z l zl
z l
y = 7x − 2
zl zl zl zl
−9 ?¯ 7(−1) − 2 z l zl zl 0+3 ¯ zl zl zl
¯ −7 − 2 zl zl 3 ¯ 3 TRUE ³ 3´ z l
zl
z l
zl
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
zl
TRUE z l
zl
zl zl zl zl zl zl zl zl zl zl zl
The equation −9 = −9 is true, so (−1, −9) is a solut
z l z l zl z l z l z l z l z l zl z l z l z l
5
ion. To determine whether (0, 2) is a solution, substitute
zl zl zl zl zl zl zl zl zl zl
15. To determine whether (−0.75, 2.75) is a solution,
zl z l z l zl z l z l z l
ti- tute −0.75 for x and 2.75 for y.
zl zl zl zl zl zl zl
0 for
zl
zl
x and 2 for y.
zl zl zl zl
x2 − y2 = 3 zl zl zl zl zl
y = 7x − 2 z l zl zl zl
2 ? 7 ·0−
z l z l zl zl zl zl
(−0.75)2 − (2.75)2 ?¯ 3 zl
zl
z l
zl
zl
2
¯ 0— 2 0.5625 − 7.5625 zl zl
¯
¯
2 −2 zl
FALSE −7 ¯ 3 FALSE z l
zl
z l
The equation 2 = −2 is false, so (0, 2) is not a solution.
zl zl zl zl zl zl zl zl zl zl zl zl zl
The equation − z l −
³ 2 3´ 2 zl zl 0.75, 2.75) 7 = 3 is false, so (
z l z l z l z l z l z l z l zl
11. To determine whether , is a solution, substitute
zl zl
zl zl zl zl zl zl zl zl z l
ot a solution.
z l zl
To determine whether (2, −1) is a solution, sub
2
z l z l z l zl z l z l z l z l
l
3 4 3
3
z l
for x and −1 for y.
for x and for y.
z l zl zl zl zl zl
zl zl z l zl
4 x − y2 = 32
zl zl zl zl zl
6x − 4y = 1
22 − (−1)2 ?¯ 3
zl zl z l zl zl
zl z l
zl zl
zl
2 3 4— 1
6 · −4 · ? 1 ¯
3 4
zl z l zl zl z l z l
zl
3 ¯ 3 TRUE
zl z l
4
zl
z l z l
¯ 3 z l
— The equation 3 = 3 is true, so (2, −1) is a solution
zl zl zl zl zl zl zl zl zl zl zl zl
1 ¯ 1 TRUE zl
z l
³2 3´ zl 17. Graph 5x − 3y = −15.
zl zl zl zl zl zl
z l zl
The equation 1 = 1 is³ true, s , is a solution To
x find the x-
z zl l
3 4
zl zl zl zl zl zl zl zl zl
3´
zl zl zl
o
To determine whether 1, is a solution, .substitute 1 for
zl zl
intercept we replace y with 0 and solve for
zl
zl zl zl zl zl zl zl zl zl zl zl zl z l zl zl zl zl
2 . 5x − 3 · 0 = −15 zl zl zl zl zl zl
3
x and for y. 5x = −15 zl z l
2
zl zl zl
zl
x = −3
6x − 4y = 1
zl zl
zl zl z l zl zl
The x-intercept is (−3, 0).
z l z l z l
3
zl
6 · 1 −4 · ? 1
z l
2
zl zl zl zl zl z l z l zl
To find the y-
zl zl zl
intercept we replace x with 0 and solve for zl zl zl zl zl zl zl zl
6 −6 ¯ zl zl
y.
0 ¯ 1 FALSE z l
zl
z l
5 · 0 − 3y = −15
zl zl zl zl zl zl
³ z l
3´
−3y = −15
zl
zl zl
The equation 0 = 1 is false, so 1, is not a solution. y =5
2
zl zl zl zl zl zl zl z l z l zl zl zl zl z l zl zl
³ 1 4´ zl zl The y-intercept is (0, 5).
zl zl zl zl
13. To determine whether − , − is a solution, substitute
zl zl
We plot the intercepts and draw the line tha
zl zl zl zl zl zl zl zl zl zl
2 5 z l z l z l z l z l z l z l z l
1 4 ins
them. We could find a third point as a chec
− for a and − for b.
z l z l z l z l z l z l z l z l z l z l
2 5 at the
l
z zl zl z
l zl
intercepts were found correctly.
zl
z l z l z l zl zl zl
2a + 5b = 3 zl zl zl zl zl
³ 1´ ³ 4´ zl zl zl zl
2 − +5 − ? 3
zl zl
zl zl zl z l zl
2 5
−1 − 4 zl zl
−5 ¯ 3 FALSE zl
z l
c 2025 Pearson Education, Inc.
Copyright ◯ zl z l zl zl zl
