brc 2025 Pearson Education, Inc.
Copyright ◯ b r br br br
,Chapter 1 b r
Graphs, Functions, and Models b r b r b r
To graph (−1, 4) we move from the origin 1
b r b r br b r b r b r b r b r b r b r
the
b r
Check Your Understanding Section 1.1
br br br br
left of the y- b r b r b r
axis. Then we move 4 units up from the
b r b r b r b r b r b r b r b r b r
1. The point (— 5, 0) is on an axis, so it is not in any quadra
br br b r br br br br br br br br br br br br
x-axis.
nt. The statement is false.
br br br br To graph (0, 2) we do not move to the right or the l
br br br br br br br br br br br br br
rthe y- br
2. The ordered pair (1,— 6) is located 1 unit right of the orig
axis since the first coordinate is 0. From the origin
br br br b r br br br br br br br br
in and 6 units below it. The ordered pair—( 6, 1) is locate
br br br br br br b r br br
br br br br br br br br br b r br br br
move 2 units up. br br br
d 6 units left of the origin and 1 unit above it. Thus, (1,
br br —
br br br br br br br br br b r br b
6) —
r and ( 6, 1) do not name the same point. The
b r b r b r br b r b r b r b r b r b r b r
To graph (2, —2) we move from the origin 2 units to
br br b r br br br br br br br br
statement is false.
b r br br
ght of the y- br br br
y
axis. Then we move 2 units down from the x-ax
b r br br br br br br br br
3. True; the first coordinate of a point is also called the absc
br br br br br br br br br br br
( 1, 4)
issa.
br b
r 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
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br
2b r (0,br2)
(4, 0
7 units above it.
br
br br br b r )
2 (2, b r b r
2)
5. True; the second coordinate of a point is also called the o
br br br br br br br br br br br
( 3, b r b r 5)
4
rdinate.
6. False; the point (0, −3) is on the y-axis.
b r b r b r br b r b r b r b r
5. To graph ( 5, 1) we move from the origin 5
b r b r b r b r br b r b r b r b r b r b r b r
to the left —
of the y-
b r br br br br
axis. Then we move 1 unit up from the x-axis.
b r br br br br br br br br
Exercise Set 1.1 br br
To graph (5, 1) we move from the origin 5 units to th
br br br br br br br br br br br br
t of the y-
br b r b r
1. Point A is located 5 units to the left of the y-
b r b r b r b r b r b r b r b r b r b r b r axis. Then we move 1 unit up from the x-axi
b r b r b r b r b r b r b r b r b r
axis and 4 units up from the x-
b r b r b r b r b r b r b r
To graph (2, 3) we move from the origin 2 units to th
br br br br br br br br br br br br
axis, so its coordinates are (−5, 4).
b r b r b r b r b r br
t of the y-
br b r b r
Point B is located 2 units to the right of the y-
b r b r b r b r b r b r b r b r b r b r b r axis. Then we move 3 units up from the x-ax
b r b r b r b r b r b r b r b r b r
axis and 2 units down from the x-
b r br br br br br br
To graph (2, —1) we move from the origin 2 units to
br br b r br br br br br br br br
axis, so its coordinates are (2, −2).
br br br br br br
ght of the y- b r b r b r
Point C is located 0 units to the right or left of the y-
br br br br br br br br br br br br br axis. Then we move 1 unit down from the x-ax
b r b r b r b r b r b r b r b r br
axis and 5 units down from the x-
br br br br br br br
To graph (0, 1) we do not move to the right or the l
br br br br br br br br br br br br br
axis, so its coordinates are (0, −5).
br br br br br br
rthe y- br
Point D is located 3 units to the right of the y-
b r b r b r b r b r b r b r b r b r b r b r axis since the first coordinate is 0. From the origin
br br br br br br b r br br
axis and 5 units up from the x-
b r br br br br br b r move 1 unit up. br br br
axis, so its coordinates are (3, 5).
br br br br br br
y
Point E is located 5 units to the left of the y-
b r b r b r b r b r b r b r b r b r b r b r
axis and b r
4
4 units down from the x-
b r b r b r b r b r
2
(2, 3) br
axis, so its coordinates are (−5, −4).
b r b r b r b r b r br ( 5, 1 br (0, 1) br (5, 1) br
)
Point F is located 3 units to the right of the y-
b r b r b r b r b r b r b r b r b r b r b r
4 2 2b r (2,b r b r
axis and 0 units up or down from the x-
b r br br br br br br br br 1)
axis, so its coordinates are (3, 0).
br br br br br br 4
3. To graph (4, 0) we move from the origin 4 units to the rig
7. The first coordinate represents the year and the cor
b r br br br br br br br br br br br br br
br br br br br br br br br
ht of the y-
sponding second coordinate represents the number
br br br
br br br br br br b
axis. Since the second coordinate is 0, we do not move u
es served by Southwest Airlines. The ordered pairs
b r br br br br br br br br br br
p or down from the x-axis.
br br br br b r br br
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2
br br br br br
br b r b r b r b r b r b r b r
To graph ( − 3, 5) we move from the origin 3 units to th
br br b r b r b br r br br br br br br br br br
72),
br
− y-
e left of the b r b r b r and (2021, 121). br br
axis. Then we move 5 units down from the x-axis.
b r b r b r b r b r b r b r b r br
c 2025 Pearson Education, Inc.
Copyright ◯ br b r br br br
,
, 14 Chapter 1: br b r b r Graphs, Functions, and br br b
els
9. To determine whether (−1, −9) is a solution, subs
b r b r b r b r br b r b r b r b r
2a + 5b = 3 br br br br br
titute 3
−1 for x and −9 for y 2·0+5· ? 3
5
br br br br br br br br br br br b r b r br
b r
. y = 7x − 2
br br br br br
−9 ?¯ 7(−1) − 2b r br br 0+3 ¯ br br br
¯ −7 − 2 br br 3 ¯ 3 TRUE ³ 3´ b r
br
b r
br
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
br
TRUE
b r
br
br br br br br br br br br br br
The equation −9 = −9 is true, so (−1, −9) is a sol
b r b r br b r b r b r b r b r br b r b r b r
5
ution. To determine whether (0, 2) is a solution, substit
br br br br br br br br br
15. To determine whether (−0.75, 2.75) is a solution
b r b r b r br b r b r b r
sti- tute −0.75 for x and 2.75 for y.
br br br br br br br
ute 0 for
br
br br
x and 2 for y.
br br br br
x2 − y2 = 3 br br br br br
y = 7x − 2 b r br br br
2 ? 7·0− b r b r br br br b
(−0.75)2 − (2.75)2 ?¯ 3 br
br
b r
br
br
2
r
¯ 0— 2 0.5625 − 7.5625 br br
¯
¯
2 −2 FALSE br −7 ¯ 3 b r
br
b r FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution.
br br br br br br br br br br br br br
The equation −
b r −
³2 3 ´ 2 br br 0.75, 2.75 7 = 3 is false, so (
b r b r b r b r b r b r b r br
11. To determine whether , is a solution, substitute
br br
br br br br br br br br b r
not a solution.
b r br
To determine whether (2, −1) is a solution, su
2
b r b r b r br b r b r b r b r
b r
3 4 3
3
b r
for x and −1 for y.
for x and for y.
b r br br br br br
br br b r br
4 x − y2 = 32
br br br br br
6x − 4y = 1
22 − (−1)2 ?¯ 3
br br b r br br
br b r
br br
br
2 3 4— 1
61· −4· ? ¯
3 4
br b r br br b r b
r br
3 ¯ 3 TRUE
br b r
4
br
b r b r
¯ 3 b r
— The equation 3 = 3 is true, so (2, −1) is a soluti
br b r br br b r b r b r br br b r br b r
1 ¯ 1 TRUE br
b r
³2 3 ´ br 17. Graph 5x − 3y = −15.
br br br br br br
b r br
The equation 1 = 1 is³ true, , is a solution
xTo find the x-
b br r
3 4
br br br br br br br br br
3´
br b r b r
so
To determine whether 1, is a solution,. substitute 1 for
br br
intercept we replace y with 0 and solve for
br
br br br br br br br br br b r b r b r b r b r b r b r b r
2 . 5x − 3 · 0 = −15 br br br br br br
3
x and for y. 5x = −15 br br
2
br br br
x = −3
br
6x − 4y = 1
br br
br br b r br br
The x-intercept is (−3, 0).
b r b r b r
3
br
6·1−4· ? 1
b r
br br br br br b r
2
b r br
To find the y-
b r b r b r
intercept we replace x with 0 and solve for
b r b r b r br b r b r b r b r
6−6 ¯ br br
y.
0 ¯ 1 FALSE b r
br
b r
5 · 0 − 3y = −15
br br br br br br
³ b r
3´
−3y = −15
br
br br
The equation 0 = 1 is false, so 1, y = 5
2
br br br br br br br b r b r br b r br br
is not a solution.
br br br The y-intercept is (0, 5).
br br br br
³ 1 4´ We plot the intercepts and draw the line th
br br
13. To determine whether − , − is a solution, substitute
br br
b r b r b r b r b r b r b r b r
br br br br br br br br br
tains
br
2 5
1 4 them. We could find a third point as a ch
− for a and − for b.
b r b r b r b r b r b r b r b r b r b r
2 5 hat the were found correctly.
b
r br br b
r br
intercepts
br
b r b r b r br br br
2a + 5b = 3 br br br br br
³ 1´ ³ 4´ br br br br
2 − +5 − ? 3
br br
br br br b r br
2 5
−1 − 4 br br
−5 ¯ 3 FALSE br
b r
c 2025 Pearson Education, Inc.
Copyright ◯ br b r br br br
Copyright ◯ b r br br br
,Chapter 1 b r
Graphs, Functions, and Models b r b r b r
To graph (−1, 4) we move from the origin 1
b r b r br b r b r b r b r b r b r b r
the
b r
Check Your Understanding Section 1.1
br br br br
left of the y- b r b r b r
axis. Then we move 4 units up from the
b r b r b r b r b r b r b r b r b r
1. The point (— 5, 0) is on an axis, so it is not in any quadra
br br b r br br br br br br br br br br br br
x-axis.
nt. The statement is false.
br br br br To graph (0, 2) we do not move to the right or the l
br br br br br br br br br br br br br
rthe y- br
2. The ordered pair (1,— 6) is located 1 unit right of the orig
axis since the first coordinate is 0. From the origin
br br br b r br br br br br br br br
in and 6 units below it. The ordered pair—( 6, 1) is locate
br br br br br br b r br br
br br br br br br br br br b r br br br
move 2 units up. br br br
d 6 units left of the origin and 1 unit above it. Thus, (1,
br br —
br br br br br br br br br b r br b
6) —
r and ( 6, 1) do not name the same point. The
b r b r b r br b r b r b r b r b r b r b r
To graph (2, —2) we move from the origin 2 units to
br br b r br br br br br br br br
statement is false.
b r br br
ght of the y- br br br
y
axis. Then we move 2 units down from the x-ax
b r br br br br br br br br
3. True; the first coordinate of a point is also called the absc
br br br br br br br br br br br
( 1, 4)
issa.
br b
r 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
b r b r b r b r b r b r b r b r b r b r b r b r b r
br
2b r (0,br2)
(4, 0
7 units above it.
br
br br br b r )
2 (2, b r b r
2)
5. True; the second coordinate of a point is also called the o
br br br br br br br br br br br
( 3, b r b r 5)
4
rdinate.
6. False; the point (0, −3) is on the y-axis.
b r b r b r br b r b r b r b r
5. To graph ( 5, 1) we move from the origin 5
b r b r b r b r br b r b r b r b r b r b r b r
to the left —
of the y-
b r br br br br
axis. Then we move 1 unit up from the x-axis.
b r br br br br br br br br
Exercise Set 1.1 br br
To graph (5, 1) we move from the origin 5 units to th
br br br br br br br br br br br br
t of the y-
br b r b r
1. Point A is located 5 units to the left of the y-
b r b r b r b r b r b r b r b r b r b r b r axis. Then we move 1 unit up from the x-axi
b r b r b r b r b r b r b r b r b r
axis and 4 units up from the x-
b r b r b r b r b r b r b r
To graph (2, 3) we move from the origin 2 units to th
br br br br br br br br br br br br
axis, so its coordinates are (−5, 4).
b r b r b r b r b r br
t of the y-
br b r b r
Point B is located 2 units to the right of the y-
b r b r b r b r b r b r b r b r b r b r b r axis. Then we move 3 units up from the x-ax
b r b r b r b r b r b r b r b r b r
axis and 2 units down from the x-
b r br br br br br br
To graph (2, —1) we move from the origin 2 units to
br br b r br br br br br br br br
axis, so its coordinates are (2, −2).
br br br br br br
ght of the y- b r b r b r
Point C is located 0 units to the right or left of the y-
br br br br br br br br br br br br br axis. Then we move 1 unit down from the x-ax
b r b r b r b r b r b r b r b r br
axis and 5 units down from the x-
br br br br br br br
To graph (0, 1) we do not move to the right or the l
br br br br br br br br br br br br br
axis, so its coordinates are (0, −5).
br br br br br br
rthe y- br
Point D is located 3 units to the right of the y-
b r b r b r b r b r b r b r b r b r b r b r axis since the first coordinate is 0. From the origin
br br br br br br b r br br
axis and 5 units up from the x-
b r br br br br br b r move 1 unit up. br br br
axis, so its coordinates are (3, 5).
br br br br br br
y
Point E is located 5 units to the left of the y-
b r b r b r b r b r b r b r b r b r b r b r
axis and b r
4
4 units down from the x-
b r b r b r b r b r
2
(2, 3) br
axis, so its coordinates are (−5, −4).
b r b r b r b r b r br ( 5, 1 br (0, 1) br (5, 1) br
)
Point F is located 3 units to the right of the y-
b r b r b r b r b r b r b r b r b r b r b r
4 2 2b r (2,b r b r
axis and 0 units up or down from the x-
b r br br br br br br br br 1)
axis, so its coordinates are (3, 0).
br br br br br br 4
3. To graph (4, 0) we move from the origin 4 units to the rig
7. The first coordinate represents the year and the cor
b r br br br br br br br br br br br br br
br br br br br br br br br
ht of the y-
sponding second coordinate represents the number
br br br
br br br br br br b
axis. Since the second coordinate is 0, we do not move u
es served by Southwest Airlines. The ordered pairs
b r br br br br br br br br br br
p or down from the x-axis.
br br br br b r br br
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2
br br br br br
br b r b r b r b r b r b r b r
To graph ( − 3, 5) we move from the origin 3 units to th
br br b r b r b br r br br br br br br br br br
72),
br
− y-
e left of the b r b r b r and (2021, 121). br br
axis. Then we move 5 units down from the x-axis.
b r b r b r b r b r b r b r b r br
c 2025 Pearson Education, Inc.
Copyright ◯ br b r br br br
,
, 14 Chapter 1: br b r b r Graphs, Functions, and br br b
els
9. To determine whether (−1, −9) is a solution, subs
b r b r b r b r br b r b r b r b r
2a + 5b = 3 br br br br br
titute 3
−1 for x and −9 for y 2·0+5· ? 3
5
br br br br br br br br br br br b r b r br
b r
. y = 7x − 2
br br br br br
−9 ?¯ 7(−1) − 2b r br br 0+3 ¯ br br br
¯ −7 − 2 br br 3 ¯ 3 TRUE ³ 3´ b r
br
b r
br
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
br
TRUE
b r
br
br br br br br br br br br br br
The equation −9 = −9 is true, so (−1, −9) is a sol
b r b r br b r b r b r b r b r br b r b r b r
5
ution. To determine whether (0, 2) is a solution, substit
br br br br br br br br br
15. To determine whether (−0.75, 2.75) is a solution
b r b r b r br b r b r b r
sti- tute −0.75 for x and 2.75 for y.
br br br br br br br
ute 0 for
br
br br
x and 2 for y.
br br br br
x2 − y2 = 3 br br br br br
y = 7x − 2 b r br br br
2 ? 7·0− b r b r br br br b
(−0.75)2 − (2.75)2 ?¯ 3 br
br
b r
br
br
2
r
¯ 0— 2 0.5625 − 7.5625 br br
¯
¯
2 −2 FALSE br −7 ¯ 3 b r
br
b r FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution.
br br br br br br br br br br br br br
The equation −
b r −
³2 3 ´ 2 br br 0.75, 2.75 7 = 3 is false, so (
b r b r b r b r b r b r b r br
11. To determine whether , is a solution, substitute
br br
br br br br br br br br b r
not a solution.
b r br
To determine whether (2, −1) is a solution, su
2
b r b r b r br b r b r b r b r
b r
3 4 3
3
b r
for x and −1 for y.
for x and for y.
b r br br br br br
br br b r br
4 x − y2 = 32
br br br br br
6x − 4y = 1
22 − (−1)2 ?¯ 3
br br b r br br
br b r
br br
br
2 3 4— 1
61· −4· ? ¯
3 4
br b r br br b r b
r br
3 ¯ 3 TRUE
br b r
4
br
b r b r
¯ 3 b r
— The equation 3 = 3 is true, so (2, −1) is a soluti
br b r br br b r b r b r br br b r br b r
1 ¯ 1 TRUE br
b r
³2 3 ´ br 17. Graph 5x − 3y = −15.
br br br br br br
b r br
The equation 1 = 1 is³ true, , is a solution
xTo find the x-
b br r
3 4
br br br br br br br br br
3´
br b r b r
so
To determine whether 1, is a solution,. substitute 1 for
br br
intercept we replace y with 0 and solve for
br
br br br br br br br br br b r b r b r b r b r b r b r b r
2 . 5x − 3 · 0 = −15 br br br br br br
3
x and for y. 5x = −15 br br
2
br br br
x = −3
br
6x − 4y = 1
br br
br br b r br br
The x-intercept is (−3, 0).
b r b r b r
3
br
6·1−4· ? 1
b r
br br br br br b r
2
b r br
To find the y-
b r b r b r
intercept we replace x with 0 and solve for
b r b r b r br b r b r b r b r
6−6 ¯ br br
y.
0 ¯ 1 FALSE b r
br
b r
5 · 0 − 3y = −15
br br br br br br
³ b r
3´
−3y = −15
br
br br
The equation 0 = 1 is false, so 1, y = 5
2
br br br br br br br b r b r br b r br br
is not a solution.
br br br The y-intercept is (0, 5).
br br br br
³ 1 4´ We plot the intercepts and draw the line th
br br
13. To determine whether − , − is a solution, substitute
br br
b r b r b r b r b r b r b r b r
br br br br br br br br br
tains
br
2 5
1 4 them. We could find a third point as a ch
− for a and − for b.
b r b r b r b r b r b r b r b r b r b r
2 5 hat the were found correctly.
b
r br br b
r br
intercepts
br
b r b r b r br br br
2a + 5b = 3 br br br br br
³ 1´ ³ 4´ br br br br
2 − +5 − ? 3
br br
br br br b r br
2 5
−1 − 4 br br
−5 ¯ 3 FALSE br
b r
c 2025 Pearson Education, Inc.
Copyright ◯ br b r br br br
