dtc 2025 Pearson Education, Inc.
Copyright ◯ d t dt dt dt
,Chapter 1 d t
Graphs, Functions, and Models d t d t d t
To graph (−1, 4) we move from the origin 1 u
d t d t dt d t d t d t d t d t d t d t
tthe
Check Your Understanding Section 1.1
dt dt dt dt
left of the y- d t d t d t
axis. Then we move 4 units up from the
d t d t d t d t d t d t d t d t d t
1. The point ( — 5, 0) is on an axis, so it is not in any quadran
dt dt d t dt dt dt dt dt dt dt dt dt dt dt dt
x-axis.
t. The statement is false.
dt dt dt dt To graph (0, 2) we do not move to the right or the le
dt dt dt dt dt dt dt dt dt dt dt dt dt
the y- dt
2. The ordered pair (1, — 6) is located 1 unit right of the origi
axis since the first coordinate is 0. From the origin
dt dt dt d t dt dt dt dt dt dt dt dt
n and 6 units below it. The ordered pair — ( 6, 1) is located
dt dt dt dt dt dt d t dt dt d
dt dt dt dt dt dt dt dt dt d t dt dt dt dt
ove 2 units up. dt dt dt
6 units left of the origin and 1 unit above it. Thus, (1, — 6
dt dt dt dt dt dt dt dt dt dt d t dt d t
) andd— ( 6, 1) do not name the same point. The st
t d t d t dt d t d t d t d t d t d t d t d t
To graph (2, —2) we move from the origin 2 units to
dt dt d t dt dt dt dt dt dt dt dt d
atement is false. dt dt
ght of the y- dt dt dt
y
axis. Then we move 2 units down from the x-axi
d t dt dt dt dt dt dt dt dt
3. True; the first coordinate of a point is also called the absci
dt dt dt dt dt dt dt dt dt dt dt
( 1, 4)
ssa.
dt d
t 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
d t d t d t d t d t d t d t d t d t d t d t d t d t
d t
2d t (0,dt2)
(4, 0)
7 units above it.
dt
dt dt dt d t
4 2 2
2 (2, 4 d t d t
2)
5. True; the second coordinate of a point is also called the or
dt dt dt dt dt dt dt dt dt dt dt
( 3, d t d t 5)
4
dinate.
6. False; the point (0, −3) is on the y-axis.
d t d t d t dt d t d t d t d t
5. To graph ( 5, 1) we move from the origin 5 u
d t d t d t d t dt d t d t d t d t d t d t d t
o the left of— the y-
d t dt dt dt dt
axis. Then we move 1 unit up from the x-axis.
d t dt dt dt dt dt dt dt dt
Exercise Set 1.1 dt dt
To graph (5, 1) we move from the origin 5 units to th
dt dt dt dt dt dt dt dt dt dt dt dt
t of the y-
dt d t d t
1. Point A is located 5 units to the left of the y-
d t d t d t d t d t d t d t d t d t d t d t axis. Then we move 1 unit up from the x-axi
d t d t d t d t d t d t d t d t d t
axis and 4 units up from the x-
d t d t d t d t d t d t d t
To graph (2, 3) we move from the origin 2 units to th
dt dt dt dt dt dt dt dt dt dt dt dt
axis, so its coordinates are (−5, 4).
d t d t d t d t d t dt
t of the y-
dt d t d t
Point B is located 2 units to the right of the y-
d t d t d t d t d t d t d t d t d t d t d t axis. Then we move 3 units up from the x-axi
d t d t d t d t d t d t d t d t d t
axis and 2 units down from the x-
d t dt dt dt dt dt dt
To graph (2, —1) we move from the origin 2 units to
dt dt d t dt dt dt dt dt dt dt dt
axis, so its coordinates are (2, −2).
dt dt dt dt dt dt
ght of the y- d t d t d t
Point C is located 0 units to the right or left of the y-
dt dt dt dt dt dt dt dt dt dt dt dt dt axis. Then we move 1 unit down from the x-axi
d t d t d t d t d t d t d t d t dt
axis and 5 units down from the x-
dt dt dt dt dt dt dt
To graph (0, 1) we do not move to the right or the le
dt dt dt dt dt dt dt dt dt dt dt dt dt
axis, so its coordinates are (0, −5).
dt dt dt dt dt dt
the y- dt
Point D is located 3 units to the right of the y-
d t d t d t d t d t d t d t d t d t d t d t axis since the first coordinate is 0. From the origin
dt dt dt dt dt dt d t dt dt d
axis and 5 units up from the x-
d t dt dt dt dt dt d t ove 1 unit up. dt dt dt
axis, so its coordinates are (3, 5).
dt dt dt dt dt dt
y
Point E is located 5 units to the left of the y-
d t d t d t d t d t d t d t d t d t d t d t
axis and d t
4
4 units down from the x-
d t d t d t d t d t
2
(2, 3) dt
axis, so its coordinates are (−5, −4).
d t d t d t d t d t dt ( 5, 1 dt (0, 1) dt (5, 1)dt
)
Point F is located 3 units to the right of the y-
d t d t d t d t d t d t d t d t d t d t d t
4 2 2d t (2,d t d t
axis and 0 units up or down from the x-
d t dt dt dt dt dt dt dt dt 1)
axis, so its coordinates are (3, 0).
dt dt dt dt dt dt 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 corr
d t dt dt dt dt dt dt dt dt dt dt dt dt dt
dt dt dt dt dt dt dt dt dt
ht of the y-
sponding second coordinate represents the number
dt dt dt
dt dt dt dt dt dt dt
axis. Since the second coordinate is 0, we do not move u
s served by Southwest Airlines. The ordered pairs a
d t dt dt dt dt dt dt dt dt dt dt
p or down from the x-axis.
dt dt dt dt d t dt dt dt
971, 3), (1981, 15), (1991, 32), (2001, 59), (201
dt dt dt dt dt
dt d t d t d t d t d t d t d t
To graph ( − 3, 5) we move from the origin 3 units to th
dt dt d t d t d dt t dt dt dt dt dt dt dt dt dt
),
dt
− y-
e left of the d t d t d t and (2021, 121). dt dt
axis. Then we move 5 units down from the x-axis.
d t d t d t d t d t d t d t d t dt
c 2025 Pearson Education, Inc.
Copyright ◯ dt d t dt dt dt
,
, 14 Chapter 1: dt d t d t Graphs, Functions, and dt dt d
els
9. To determine whether (−1, −9) is a solution, subst
d t d t d t d t dt d t d t d t d t
2a + 5b = 3 dt dt dt dt dt
itute 3
−1 for x and −9 for y. 2·0+5· ? 3
5
dt dt dt dt dt dt dt dt dt dt dt d t d t dt
d t
dt y = 7x − 2dt dt dt dt
−9 ?¯ 7(−1) − 2 d t dt dt 0+3 ¯ dt dt dt
¯ −7 − 2 dt dt 3 ¯ 3 TRUE ³ 3´ d t
dt
d t
dt
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
dt
TRUEd t
dt
dt dt dt dt dt dt dt dt dt dt dt
The equation −9 = −9 is true, so (−1, −9) is a sol
d t d t dt d t d t d t d t d t dt d t d t d t
5
ution. To determine whether (0, 2) is a solution, substit
dt dt dt dt dt dt dt dt dt
15. To determine whether (−0.75, 2.75) is a solution
d t d t d t dt d t d t d t
sti- tute −0.75 for x and 2.75 for y.
dt dt dt dt dt dt dt
ute 0 for
dt
dt dt
x and 2 for y.
dt dt dt dt
x2 − y2 = 3 dt dt dt dt dt
y = 7x − 2 d t dt dt dt
2 ? 7·0− d t d t dt dt dt dt
(−0.75)2 − (2.75)2 ?¯ 3 dt
dt
d t
dt
dt
2
¯ 0— 2 0.5625 − 7.5625 dt dt
¯
¯
2 −2 FALSE dt −7 ¯ 3 d t
dt
d t FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution.
dt dt dt dt dt dt dt dt dt dt dt dt dt
The equation −
d t −
³2 3´ 2 dt dt 0.75, 2.75 7 = 3 is false, so (
d t d t d t d t d t d t d t dt
11. To determine whether , is a solution, substitute
dt dt
dt dt dt dt dt dt dt dt d t
not a solution.
d t dt
To determine whether (2, −1) is a solution, su
2
d t d t d t dt d t d t d t d t
d t
3 4 3
3
d t
for x and −1 for y.
for x and for y.
d t dt dt dt dt dt
dt dt d t dt
4 x − y2 = 32
dt dt dt dt dt
6x − 4y = 1
22 − (−1)2 ?¯ 3
dt dt d t dt dt
dt d t
dt dt
dt
2 3 4— 1
6
1 · 3 −4 · 4 ?
dt d t
dt
dt dt d t d t ¯
3 ¯ 3 TRUE
dt d t
4
dt
d t d t
¯ 3 d t
— The equation 3 = 3 is true, so (2, −1) is a soluti
d t d t dt dt d t d t d t dt dt d t dt d t
1 ¯ 1 TRUE dt
d t
³2 3 ´ dt 17. Graph 5x − 3y = −15.
dt dt dt dt dt dt
d t dt
The equation 1 = 1 is³ true, s , is a solution
xTo find the x-
d dt t
3 4
dt dt dt dt dt dt dt dt dt
3´
dt d t d t
o
To determine whether 1, is a solution,. substitute 1 for
dt dt
intercept we replace y with 0 and solve for
dt
dt dt dt dt dt dt dt dt dt d t d t d t d t d t d t d t d t
2 . 5x − 3 · 0 = −15 dt dt dt dt dt dt
3
x and for y. 5x = −15 dt dt
2
dt dt dt
x = −3
dt
6x − 4y = 1
dt dt
dt dt d t dt dt
The x-intercept is (−3, 0).
d t d t d t
3
dt
6 · 1 −4 · ? 1
d t
dt dt dt dt dt d t
2
d t dt
To find the y-
d t d t d t
intercept we replace x with 0 and solve for
d t d t d t d t d t d t d t d t
6−6 ¯ dt dt
y.
0 ¯ 1 FALSE d t
dt
d t
5 · 0 − 3y = −15
dt dt dt dt dt dt
³ d t
3´
−3y = −15
dt
dt dt
The equation 0 = 1 is false, so 1, y =5
2
dt dt dt dt dt dt dt d t d t dt d t dt dt
is not a solution.
dt dt dt The y-intercept is (0, 5).
dt dt dt dt
³ 1 4´ We plot the intercepts and draw the line th
dt dt
13. To determine whether − , − is a solution, substitute
dt dt
d t d t d t d t d t d t d t d t
dt dt dt dt dt dt dt dt dt
ains
dt
2 5
1 4 them. We could find a third point as a che
− for a and − for b.
d t d t d t d t d t d t d t d t d t d t
2 5 hat the were found correctly.
d
t dt dt d
t dt
intercepts
dt
d t d t d t dt dt dt
2a + 5b = 3 dt dt dt dt dt
³ 1´ ³ 4´ dt dt dt dt
2 − +5 − ? 3
dt dt
dt dt dt d t dt
2 5
−1 − 4 dt dt
−5 ¯ 3 FALSE dt
d t
c 2025 Pearson Education, Inc.
Copyright ◯ dt d t dt dt dt
Copyright ◯ d t dt dt dt
,Chapter 1 d t
Graphs, Functions, and Models d t d t d t
To graph (−1, 4) we move from the origin 1 u
d t d t dt d t d t d t d t d t d t d t
tthe
Check Your Understanding Section 1.1
dt dt dt dt
left of the y- d t d t d t
axis. Then we move 4 units up from the
d t d t d t d t d t d t d t d t d t
1. The point ( — 5, 0) is on an axis, so it is not in any quadran
dt dt d t dt dt dt dt dt dt dt dt dt dt dt dt
x-axis.
t. The statement is false.
dt dt dt dt To graph (0, 2) we do not move to the right or the le
dt dt dt dt dt dt dt dt dt dt dt dt dt
the y- dt
2. The ordered pair (1, — 6) is located 1 unit right of the origi
axis since the first coordinate is 0. From the origin
dt dt dt d t dt dt dt dt dt dt dt dt
n and 6 units below it. The ordered pair — ( 6, 1) is located
dt dt dt dt dt dt d t dt dt d
dt dt dt dt dt dt dt dt dt d t dt dt dt dt
ove 2 units up. dt dt dt
6 units left of the origin and 1 unit above it. Thus, (1, — 6
dt dt dt dt dt dt dt dt dt dt d t dt d t
) andd— ( 6, 1) do not name the same point. The st
t d t d t dt d t d t d t d t d t d t d t d t
To graph (2, —2) we move from the origin 2 units to
dt dt d t dt dt dt dt dt dt dt dt d
atement is false. dt dt
ght of the y- dt dt dt
y
axis. Then we move 2 units down from the x-axi
d t dt dt dt dt dt dt dt dt
3. True; the first coordinate of a point is also called the absci
dt dt dt dt dt dt dt dt dt dt dt
( 1, 4)
ssa.
dt d
t 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
d t d t d t d t d t d t d t d t d t d t d t d t d t
d t
2d t (0,dt2)
(4, 0)
7 units above it.
dt
dt dt dt d t
4 2 2
2 (2, 4 d t d t
2)
5. True; the second coordinate of a point is also called the or
dt dt dt dt dt dt dt dt dt dt dt
( 3, d t d t 5)
4
dinate.
6. False; the point (0, −3) is on the y-axis.
d t d t d t dt d t d t d t d t
5. To graph ( 5, 1) we move from the origin 5 u
d t d t d t d t dt d t d t d t d t d t d t d t
o the left of— the y-
d t dt dt dt dt
axis. Then we move 1 unit up from the x-axis.
d t dt dt dt dt dt dt dt dt
Exercise Set 1.1 dt dt
To graph (5, 1) we move from the origin 5 units to th
dt dt dt dt dt dt dt dt dt dt dt dt
t of the y-
dt d t d t
1. Point A is located 5 units to the left of the y-
d t d t d t d t d t d t d t d t d t d t d t axis. Then we move 1 unit up from the x-axi
d t d t d t d t d t d t d t d t d t
axis and 4 units up from the x-
d t d t d t d t d t d t d t
To graph (2, 3) we move from the origin 2 units to th
dt dt dt dt dt dt dt dt dt dt dt dt
axis, so its coordinates are (−5, 4).
d t d t d t d t d t dt
t of the y-
dt d t d t
Point B is located 2 units to the right of the y-
d t d t d t d t d t d t d t d t d t d t d t axis. Then we move 3 units up from the x-axi
d t d t d t d t d t d t d t d t d t
axis and 2 units down from the x-
d t dt dt dt dt dt dt
To graph (2, —1) we move from the origin 2 units to
dt dt d t dt dt dt dt dt dt dt dt
axis, so its coordinates are (2, −2).
dt dt dt dt dt dt
ght of the y- d t d t d t
Point C is located 0 units to the right or left of the y-
dt dt dt dt dt dt dt dt dt dt dt dt dt axis. Then we move 1 unit down from the x-axi
d t d t d t d t d t d t d t d t dt
axis and 5 units down from the x-
dt dt dt dt dt dt dt
To graph (0, 1) we do not move to the right or the le
dt dt dt dt dt dt dt dt dt dt dt dt dt
axis, so its coordinates are (0, −5).
dt dt dt dt dt dt
the y- dt
Point D is located 3 units to the right of the y-
d t d t d t d t d t d t d t d t d t d t d t axis since the first coordinate is 0. From the origin
dt dt dt dt dt dt d t dt dt d
axis and 5 units up from the x-
d t dt dt dt dt dt d t ove 1 unit up. dt dt dt
axis, so its coordinates are (3, 5).
dt dt dt dt dt dt
y
Point E is located 5 units to the left of the y-
d t d t d t d t d t d t d t d t d t d t d t
axis and d t
4
4 units down from the x-
d t d t d t d t d t
2
(2, 3) dt
axis, so its coordinates are (−5, −4).
d t d t d t d t d t dt ( 5, 1 dt (0, 1) dt (5, 1)dt
)
Point F is located 3 units to the right of the y-
d t d t d t d t d t d t d t d t d t d t d t
4 2 2d t (2,d t d t
axis and 0 units up or down from the x-
d t dt dt dt dt dt dt dt dt 1)
axis, so its coordinates are (3, 0).
dt dt dt dt dt dt 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 corr
d t dt dt dt dt dt dt dt dt dt dt dt dt dt
dt dt dt dt dt dt dt dt dt
ht of the y-
sponding second coordinate represents the number
dt dt dt
dt dt dt dt dt dt dt
axis. Since the second coordinate is 0, we do not move u
s served by Southwest Airlines. The ordered pairs a
d t dt dt dt dt dt dt dt dt dt dt
p or down from the x-axis.
dt dt dt dt d t dt dt dt
971, 3), (1981, 15), (1991, 32), (2001, 59), (201
dt dt dt dt dt
dt d t d t d t d t d t d t d t
To graph ( − 3, 5) we move from the origin 3 units to th
dt dt d t d t d dt t dt dt dt dt dt dt dt dt dt
),
dt
− y-
e left of the d t d t d t and (2021, 121). dt dt
axis. Then we move 5 units down from the x-axis.
d t d t d t d t d t d t d t d t dt
c 2025 Pearson Education, Inc.
Copyright ◯ dt d t dt dt dt
,
, 14 Chapter 1: dt d t d t Graphs, Functions, and dt dt d
els
9. To determine whether (−1, −9) is a solution, subst
d t d t d t d t dt d t d t d t d t
2a + 5b = 3 dt dt dt dt dt
itute 3
−1 for x and −9 for y. 2·0+5· ? 3
5
dt dt dt dt dt dt dt dt dt dt dt d t d t dt
d t
dt y = 7x − 2dt dt dt dt
−9 ?¯ 7(−1) − 2 d t dt dt 0+3 ¯ dt dt dt
¯ −7 − 2 dt dt 3 ¯ 3 TRUE ³ 3´ d t
dt
d t
dt
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
dt
TRUEd t
dt
dt dt dt dt dt dt dt dt dt dt dt
The equation −9 = −9 is true, so (−1, −9) is a sol
d t d t dt d t d t d t d t d t dt d t d t d t
5
ution. To determine whether (0, 2) is a solution, substit
dt dt dt dt dt dt dt dt dt
15. To determine whether (−0.75, 2.75) is a solution
d t d t d t dt d t d t d t
sti- tute −0.75 for x and 2.75 for y.
dt dt dt dt dt dt dt
ute 0 for
dt
dt dt
x and 2 for y.
dt dt dt dt
x2 − y2 = 3 dt dt dt dt dt
y = 7x − 2 d t dt dt dt
2 ? 7·0− d t d t dt dt dt dt
(−0.75)2 − (2.75)2 ?¯ 3 dt
dt
d t
dt
dt
2
¯ 0— 2 0.5625 − 7.5625 dt dt
¯
¯
2 −2 FALSE dt −7 ¯ 3 d t
dt
d t FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution.
dt dt dt dt dt dt dt dt dt dt dt dt dt
The equation −
d t −
³2 3´ 2 dt dt 0.75, 2.75 7 = 3 is false, so (
d t d t d t d t d t d t d t dt
11. To determine whether , is a solution, substitute
dt dt
dt dt dt dt dt dt dt dt d t
not a solution.
d t dt
To determine whether (2, −1) is a solution, su
2
d t d t d t dt d t d t d t d t
d t
3 4 3
3
d t
for x and −1 for y.
for x and for y.
d t dt dt dt dt dt
dt dt d t dt
4 x − y2 = 32
dt dt dt dt dt
6x − 4y = 1
22 − (−1)2 ?¯ 3
dt dt d t dt dt
dt d t
dt dt
dt
2 3 4— 1
6
1 · 3 −4 · 4 ?
dt d t
dt
dt dt d t d t ¯
3 ¯ 3 TRUE
dt d t
4
dt
d t d t
¯ 3 d t
— The equation 3 = 3 is true, so (2, −1) is a soluti
d t d t dt dt d t d t d t dt dt d t dt d t
1 ¯ 1 TRUE dt
d t
³2 3 ´ dt 17. Graph 5x − 3y = −15.
dt dt dt dt dt dt
d t dt
The equation 1 = 1 is³ true, s , is a solution
xTo find the x-
d dt t
3 4
dt dt dt dt dt dt dt dt dt
3´
dt d t d t
o
To determine whether 1, is a solution,. substitute 1 for
dt dt
intercept we replace y with 0 and solve for
dt
dt dt dt dt dt dt dt dt dt d t d t d t d t d t d t d t d t
2 . 5x − 3 · 0 = −15 dt dt dt dt dt dt
3
x and for y. 5x = −15 dt dt
2
dt dt dt
x = −3
dt
6x − 4y = 1
dt dt
dt dt d t dt dt
The x-intercept is (−3, 0).
d t d t d t
3
dt
6 · 1 −4 · ? 1
d t
dt dt dt dt dt d t
2
d t dt
To find the y-
d t d t d t
intercept we replace x with 0 and solve for
d t d t d t d t d t d t d t d t
6−6 ¯ dt dt
y.
0 ¯ 1 FALSE d t
dt
d t
5 · 0 − 3y = −15
dt dt dt dt dt dt
³ d t
3´
−3y = −15
dt
dt dt
The equation 0 = 1 is false, so 1, y =5
2
dt dt dt dt dt dt dt d t d t dt d t dt dt
is not a solution.
dt dt dt The y-intercept is (0, 5).
dt dt dt dt
³ 1 4´ We plot the intercepts and draw the line th
dt dt
13. To determine whether − , − is a solution, substitute
dt dt
d t d t d t d t d t d t d t d t
dt dt dt dt dt dt dt dt dt
ains
dt
2 5
1 4 them. We could find a third point as a che
− for a and − for b.
d t d t d t d t d t d t d t d t d t d t
2 5 hat the were found correctly.
d
t dt dt d
t dt
intercepts
dt
d t d t d t dt dt dt
2a + 5b = 3 dt dt dt dt dt
³ 1´ ³ 4´ dt dt dt dt
2 − +5 − ? 3
dt dt
dt dt dt d t dt
2 5
−1 − 4 dt dt
−5 ¯ 3 FALSE dt
d t
c 2025 Pearson Education, Inc.
Copyright ◯ dt d t dt dt dt
