cnc 2025 Pearson Education, Inc
Copyright ◯ c n cn cn cn
,Chapter 1 c n
Graphs, Functions, and Models c n c n c n
To graph (−1, 4) we move from the origin 1
c n c n cn c n c n c n c n c n c n c n
the
c n
Check Your Understanding Section 1.1
cn cn cn cn
left of the y- c n c n c n
axis. Then we move 4 units up from the
c n c n c n c n c n c n c n c n c n
1. The point (— 5, 0) is on an axis, so it is not in any quadra
cn cn c n cn cn cn cn cn cn cn cn cn cn cn cn
x-axis.
nt. The statement is false.
cn cn cn cn To graph (0, 2) we do not move to the right or the l
cn cn cn cn cn cn cn cn cn cn cn cn cn
nthe y- cn
2. The ordered pair (1,— 6) is located 1 unit right of the orig
axis since the first coordinate is 0. From the origin
cn cn cn c n cn cn cn cn cn cn cn cn
in and 6 units below it. The ordered pair—( 6, 1) is locate
cn cn cn cn cn cn c n cn cn
cn cn cn cn cn cn cn cn cn c n cn cn cn
move 2 units up. cn cn cn
d 6 units left of the origin and 1 unit above it. Thus, (1,
cn cn cn— cn cn cn cn cn cn cn cn c n cn c
6) —
n and ( 6, 1) do not name the same point. Th
c n c n c n cn c n c n c n c n c n c n c n
To graph (2, —2) we move from the origin 2 units to
cn cn c n cn cn cn cn cn cn cn cn
e statement is false.
c n cn cn
ight of the y- cn cn cn
y
axis. Then we move 2 units down from the x-ax
c n cn cn cn cn cn cn cn cn
3. True; the first coordinate of a point is also called the absc
cn cn cn cn cn cn cn cn cn cn cn
( 1, 4)
issa.
cn
c n 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
c n c n c n c n c n c n c n c n c n c n c n c n c n
c n
2c n (0,cn2)
(4, 0
7 units above it.
cn
cn cn cn c n )
2 (2, c n c n
2)
5. True; the second coordinate of a point is also called the o
cn cn cn cn cn cn cn cn cn cn cn
( 3, c n c n 5)
4
rdinate.
6. False; the point (0, −3) is on the y-axis.
c n c n cn cn cn c n c n cn
5. To graph ( 5, 1) we move from the origin 5
c n c n c n c n cn c n c n c n c n c n c n c n
to the left—of the y-
n c n cn cn cn cn
axis. Then we move 1 unit up from the x-axis.
c n cn cn cn cn cn cn cn cn
Exercise Set 1.1 cn cn
To graph (5, 1) we move from the origin 5 units to th
cn cn cn cn cn cn cn cn cn cn cn cn
ht of the y-
cn c n c n
1. Point A is located 5 units to the left of the y-
c n c n c n c n c n c n c n c n c n c n c n axis. Then we move 1 unit up from the x-ax
c n c n c n c n c n c n c n c n c n
axis and 4 units up from the x-
c n c n c n c n c n c n c n
To graph (2, 3) we move from the origin 2 units to th
cn cn cn cn cn cn cn cn cn cn cn cn
axis, so its coordinates are (−5, 4).
c n c n c n c n c n cn
ht of the y-
cn c n c n
Point B is located 2 units to the right of the y-
c n c n c n c n c n c n c n c n c n c n c n axis. Then we move 3 units up from the x-ax
c n c n c n c n c n c n c n c n c n
axis and 2 units down from the x-
c n cn cn cn cn cn cn
To graph (2, —1) we move from the origin 2 units to
cn cn c n cn cn cn cn cn cn cn cn
axis, so its coordinates are (2, −2).
cn cn cn cn cn cn
ight of the y- c n c n c n
Point C is located 0 units to the right or left of the y-
cn cn cn cn cn cn cn cn cn cn cn cn cn axis. Then we move 1 unit down from the x-ax
c n c n c n c n c n c n c n c n cn
axis and 5 units down from the x-
cn cn cn cn cn cn cn
To graph (0, 1) we do not move to the right or the
cn cn cn cn cn cn cn cn cn cn cn cn cn
axis, so its coordinates are (0, −5).
cn cn cn cn cn cn
nthe y- cn
Point D is located 3 units to the right of the y-
c n c n c n c n c n c n c n c n c n c n c n axis since the first coordinate is 0. From the origin
cn cn cn cn cn cn c n cn cn
axis and 5 units up from the x-
c n cn cn cn cn cn c n move 1 unit up. cn cn cn
axis, so its coordinates are (3, 5).
cn cn cn cn cn cn
y
Point E is located 5 units to the left of the y-
c n c n c n c n c n c n c n c n c n c n c n
axis and c n
4
4 units down from the x-
c n c n c n c n c n
2
(2, 3) cn
axis, so its coordinates are (−5, −4).
c n c n c n c n c n cn ( 5, cn (0, 1) cn (5, 1) cn
1)
Point F is located 3 units to the right of the y-
c n c n c n c n c n c n c n c n c n c n c n
4 2 2c n (2,c n c n
axis and 0 units up or down from the x-
c n cn cn cn cn cn cn cn cn 1)
axis, so its coordinates are (3, 0).
cn cn cn cn cn cn 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
c n cn cn cn cn cn cn cn cn cn cn cn cn cn
cn cn cn cn cn cn cn cn cn
ht of the y-
sponding second coordinate represents the number
cn cn cn
cn cn cn cn cn cn c
axis. Since the second coordinate is 0, we do not move u
es served by Southwest Airlines. The ordered pairs
c n cn cn cn cn cn cn cn cn cn cn
p or down from the x-axis.
cn cn cn cn c n cn cn
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2
cn cn cn cn cn
cn c n c n c n c n c n c n c n
To graph ( − 3, 5) we move from the origin 3 units to t
cn cn c nc n c n
c n cn cn cn cn cn cn cn cn cn
72),
he left of −
the y-
cn c n c n c n and (2021, 121). cn cn
axis. Then we move 5 units down from the x-axis.
c n c n c n c n c n c n c n c n cn
c 2025 Pearson Education, Inc
Copyright ◯ cn c n cn cn cn
,
, 14 Chapter 1: cn c n c n Graphs, Functions, and cn cn c
els
9. To determine whether (−1, −9) is a solution, subs
c n c n c n c n cn c n c n c n c n
2a + 5b = 3 cn cn cn cn cn
titute 3
−1 for x and −9 for y 2·0+5· ? 3
5
cn cn cn cn cn cn cn cn cn cn cn c n c n cn
c n
. y = 7x − 2
cn cn cn cn cn
−9 ?¯ 7(−1) − c n cn cn 0+3 ¯ cn cn cn
¯ 2 3 ¯ 3 TRUE c n
cn
c n
−7 − 2 cn cn
³ 3´ cn
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
cn
c n
cn
TRUE cn cn cn cn cn cn cn cn cn cn cn
The equation −9 = −9 is true, so (−1, −9) is a sol
c n c n cn c n c n c n c n c n cn c n c n c n
5
ution. To determine whether (0, 2) is a solution, substit
cn cn cn cn cn cn cn cn cn
15. To determine whether (−0.75, 2.75) is a solution
c n c n c n cn c n c n c n
sti- tute −0.75 for x and 2.75 for y.
cn cn cn cn cn cn cn
ute 0 for
cn
cn cn
x and 2 for y.
cn cn cn cn
x2 − y2 = 3 cn cn cn cn cn
y = 7x − 2
c n cn cn cn
2 ? 7·0− c n c n cn cn cn c
(−0.75)2 − (2.75)2 ?¯ 3 cn
cn
c n
cn
cn
2
n
¯ 0— 2 0.5625 − 7.5625 cn cn
¯
¯
2 −2 FALSE cn −7 ¯ 3 c n
cn
c n FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution
cn cn cn cn cn cn cn cn cn cn cn cn cn
The equation − c n −
. 0.75, 2.75 7 = 3 is false, so (
c n c n c n c n c n c n c n cn
³2 3 ´ 2 cn cn
not a solution.
c n cn
11. To determine whether , is a solution, substitute
cn cn
cn cn cn cn cn cn cn cn c n
To determine whether (2, −1) is a solution, su
e 2
c n c n c n cn c n c n c n c n
c n
3 4 3
3
c n
for x and −1 for y.
for x and for y.
c n cn cn cn cn cn
cn cn c n cn
4 x − y2 = 3
2
cn cn cn cn cn
6x − 4y = 1
22 − (−1)2 ?¯ 3
cn cn cn cn cn
cn c n
cn cn
cn
2 3 4— 1
61· −4· ? ¯
3 4
cn c n cn cn c n c
n cn
3 ¯ 3 TRUE
cn c n
4
cn
c n c n
¯ 3 c n
— The equation 3 = 3 is true, so (2, −1) is a solut
cn c n cn cn c n c n c n cn cn c n cn c n
1 ¯ 1 TRUE cn
c n
³2 3 ´ cn 17. Graph 5x − 3y = −15.
cn cn cn cn cn cn
c n cn
The equation 1 = 1 is³ true, , is a solutio
xTo find the x-
c cn n
3 4
cn cn cn cn cn cn cn cn cn
3´
cn c n c n
so n. substitute 1 for
cn cn
To determine whether 1, is a solution, intercept we replace y with 0 and solve for
cn
cn cn cn cn cn cn cn cn cn c n c n c n c n c n c n c n c n
2 . 5x − 3 · 0 = −15 cn cn cn cn cn cn
3
x and for y. 5x = −15 cn cn
2
cn cn cn
cn
x = −3
6x − 4y = 1
cn cn
cn cn cn cn cn
The x-intercept is (−3, 0).
c n c n c n
3
cn
6·1−4· ? 1
c n
cn cn cn cn cn c n
2
c n cn
To find the y-
c n c n c n
intercept we replace x with 0 and solve for
c n c n c n cn c n c n c n c n
6−6 ¯ cn cn
y.
0 ¯ 1 FALSE c n
cn
c n
5 · 0 − 3y = −15
cn cn cn cn cn cn
³ c n
3´
−3y = −15
cn
cn cn
The equation 0 = 1 is false, so 1, y = 5
2
cn cn cn cn cn cn cn c n c n cn cn cn cn
is not a solution.
cn cn cn The y-intercept is (0, 5).
cn cn cn cn
³ 1 4´ We plot the intercepts and draw the line th
cn cn
13. To determine whether − , − is a solution, substitute
cn cn
c n c n c n c n c n c n c n c n
cn cn cn cn cn cn cn cn cn
tains
cn
2 5
1 4 them. We could find a third point as a ch
− for a and − for b.
c n c n c n c n c n c n c n c n c n c n
2 5 hat the were found correctly.
n
c cn cn cn cn
intercepts
cn
c n c n c n cn cn cn
2a + 5b = 3 cn cn cn cn cn
³ 1´ ³ 4´ cn cn cn cn
2 − +5 − ? 3
cn cn
cn cn cn c n cn
2 5
−1 − 4 cn cn
c 2025 Pearson Education, Inc
Copyright ◯ cn c n cn cn cn
Copyright ◯ c n cn cn cn
,Chapter 1 c n
Graphs, Functions, and Models c n c n c n
To graph (−1, 4) we move from the origin 1
c n c n cn c n c n c n c n c n c n c n
the
c n
Check Your Understanding Section 1.1
cn cn cn cn
left of the y- c n c n c n
axis. Then we move 4 units up from the
c n c n c n c n c n c n c n c n c n
1. The point (— 5, 0) is on an axis, so it is not in any quadra
cn cn c n cn cn cn cn cn cn cn cn cn cn cn cn
x-axis.
nt. The statement is false.
cn cn cn cn To graph (0, 2) we do not move to the right or the l
cn cn cn cn cn cn cn cn cn cn cn cn cn
nthe y- cn
2. The ordered pair (1,— 6) is located 1 unit right of the orig
axis since the first coordinate is 0. From the origin
cn cn cn c n cn cn cn cn cn cn cn cn
in and 6 units below it. The ordered pair—( 6, 1) is locate
cn cn cn cn cn cn c n cn cn
cn cn cn cn cn cn cn cn cn c n cn cn cn
move 2 units up. cn cn cn
d 6 units left of the origin and 1 unit above it. Thus, (1,
cn cn cn— cn cn cn cn cn cn cn cn c n cn c
6) —
n and ( 6, 1) do not name the same point. Th
c n c n c n cn c n c n c n c n c n c n c n
To graph (2, —2) we move from the origin 2 units to
cn cn c n cn cn cn cn cn cn cn cn
e statement is false.
c n cn cn
ight of the y- cn cn cn
y
axis. Then we move 2 units down from the x-ax
c n cn cn cn cn cn cn cn cn
3. True; the first coordinate of a point is also called the absc
cn cn cn cn cn cn cn cn cn cn cn
( 1, 4)
issa.
cn
c n 4
4. True; the point ( 2 7) is 2 units left of the origin and
− ,
c n c n c n c n c n c n c n c n c n c n c n c n c n
c n
2c n (0,cn2)
(4, 0
7 units above it.
cn
cn cn cn c n )
2 (2, c n c n
2)
5. True; the second coordinate of a point is also called the o
cn cn cn cn cn cn cn cn cn cn cn
( 3, c n c n 5)
4
rdinate.
6. False; the point (0, −3) is on the y-axis.
c n c n cn cn cn c n c n cn
5. To graph ( 5, 1) we move from the origin 5
c n c n c n c n cn c n c n c n c n c n c n c n
to the left—of the y-
n c n cn cn cn cn
axis. Then we move 1 unit up from the x-axis.
c n cn cn cn cn cn cn cn cn
Exercise Set 1.1 cn cn
To graph (5, 1) we move from the origin 5 units to th
cn cn cn cn cn cn cn cn cn cn cn cn
ht of the y-
cn c n c n
1. Point A is located 5 units to the left of the y-
c n c n c n c n c n c n c n c n c n c n c n axis. Then we move 1 unit up from the x-ax
c n c n c n c n c n c n c n c n c n
axis and 4 units up from the x-
c n c n c n c n c n c n c n
To graph (2, 3) we move from the origin 2 units to th
cn cn cn cn cn cn cn cn cn cn cn cn
axis, so its coordinates are (−5, 4).
c n c n c n c n c n cn
ht of the y-
cn c n c n
Point B is located 2 units to the right of the y-
c n c n c n c n c n c n c n c n c n c n c n axis. Then we move 3 units up from the x-ax
c n c n c n c n c n c n c n c n c n
axis and 2 units down from the x-
c n cn cn cn cn cn cn
To graph (2, —1) we move from the origin 2 units to
cn cn c n cn cn cn cn cn cn cn cn
axis, so its coordinates are (2, −2).
cn cn cn cn cn cn
ight of the y- c n c n c n
Point C is located 0 units to the right or left of the y-
cn cn cn cn cn cn cn cn cn cn cn cn cn axis. Then we move 1 unit down from the x-ax
c n c n c n c n c n c n c n c n cn
axis and 5 units down from the x-
cn cn cn cn cn cn cn
To graph (0, 1) we do not move to the right or the
cn cn cn cn cn cn cn cn cn cn cn cn cn
axis, so its coordinates are (0, −5).
cn cn cn cn cn cn
nthe y- cn
Point D is located 3 units to the right of the y-
c n c n c n c n c n c n c n c n c n c n c n axis since the first coordinate is 0. From the origin
cn cn cn cn cn cn c n cn cn
axis and 5 units up from the x-
c n cn cn cn cn cn c n move 1 unit up. cn cn cn
axis, so its coordinates are (3, 5).
cn cn cn cn cn cn
y
Point E is located 5 units to the left of the y-
c n c n c n c n c n c n c n c n c n c n c n
axis and c n
4
4 units down from the x-
c n c n c n c n c n
2
(2, 3) cn
axis, so its coordinates are (−5, −4).
c n c n c n c n c n cn ( 5, cn (0, 1) cn (5, 1) cn
1)
Point F is located 3 units to the right of the y-
c n c n c n c n c n c n c n c n c n c n c n
4 2 2c n (2,c n c n
axis and 0 units up or down from the x-
c n cn cn cn cn cn cn cn cn 1)
axis, so its coordinates are (3, 0).
cn cn cn cn cn cn 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
c n cn cn cn cn cn cn cn cn cn cn cn cn cn
cn cn cn cn cn cn cn cn cn
ht of the y-
sponding second coordinate represents the number
cn cn cn
cn cn cn cn cn cn c
axis. Since the second coordinate is 0, we do not move u
es served by Southwest Airlines. The ordered pairs
c n cn cn cn cn cn cn cn cn cn cn
p or down from the x-axis.
cn cn cn cn c n cn cn
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2
cn cn cn cn cn
cn c n c n c n c n c n c n c n
To graph ( − 3, 5) we move from the origin 3 units to t
cn cn c nc n c n
c n cn cn cn cn cn cn cn cn cn
72),
he left of −
the y-
cn c n c n c n and (2021, 121). cn cn
axis. Then we move 5 units down from the x-axis.
c n c n c n c n c n c n c n c n cn
c 2025 Pearson Education, Inc
Copyright ◯ cn c n cn cn cn
,
, 14 Chapter 1: cn c n c n Graphs, Functions, and cn cn c
els
9. To determine whether (−1, −9) is a solution, subs
c n c n c n c n cn c n c n c n c n
2a + 5b = 3 cn cn cn cn cn
titute 3
−1 for x and −9 for y 2·0+5· ? 3
5
cn cn cn cn cn cn cn cn cn cn cn c n c n cn
c n
. y = 7x − 2
cn cn cn cn cn
−9 ?¯ 7(−1) − c n cn cn 0+3 ¯ cn cn cn
¯ 2 3 ¯ 3 TRUE c n
cn
c n
−7 − 2 cn cn
³ 3´ cn
−9 ¯ −9 The equation 3 = 3 is true, so 0, is a solution.
cn
c n
cn
TRUE cn cn cn cn cn cn cn cn cn cn cn
The equation −9 = −9 is true, so (−1, −9) is a sol
c n c n cn c n c n c n c n c n cn c n c n c n
5
ution. To determine whether (0, 2) is a solution, substit
cn cn cn cn cn cn cn cn cn
15. To determine whether (−0.75, 2.75) is a solution
c n c n c n cn c n c n c n
sti- tute −0.75 for x and 2.75 for y.
cn cn cn cn cn cn cn
ute 0 for
cn
cn cn
x and 2 for y.
cn cn cn cn
x2 − y2 = 3 cn cn cn cn cn
y = 7x − 2
c n cn cn cn
2 ? 7·0− c n c n cn cn cn c
(−0.75)2 − (2.75)2 ?¯ 3 cn
cn
c n
cn
cn
2
n
¯ 0— 2 0.5625 − 7.5625 cn cn
¯
¯
2 −2 FALSE cn −7 ¯ 3 c n
cn
c n FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution
cn cn cn cn cn cn cn cn cn cn cn cn cn
The equation − c n −
. 0.75, 2.75 7 = 3 is false, so (
c n c n c n c n c n c n c n cn
³2 3 ´ 2 cn cn
not a solution.
c n cn
11. To determine whether , is a solution, substitute
cn cn
cn cn cn cn cn cn cn cn c n
To determine whether (2, −1) is a solution, su
e 2
c n c n c n cn c n c n c n c n
c n
3 4 3
3
c n
for x and −1 for y.
for x and for y.
c n cn cn cn cn cn
cn cn c n cn
4 x − y2 = 3
2
cn cn cn cn cn
6x − 4y = 1
22 − (−1)2 ?¯ 3
cn cn cn cn cn
cn c n
cn cn
cn
2 3 4— 1
61· −4· ? ¯
3 4
cn c n cn cn c n c
n cn
3 ¯ 3 TRUE
cn c n
4
cn
c n c n
¯ 3 c n
— The equation 3 = 3 is true, so (2, −1) is a solut
cn c n cn cn c n c n c n cn cn c n cn c n
1 ¯ 1 TRUE cn
c n
³2 3 ´ cn 17. Graph 5x − 3y = −15.
cn cn cn cn cn cn
c n cn
The equation 1 = 1 is³ true, , is a solutio
xTo find the x-
c cn n
3 4
cn cn cn cn cn cn cn cn cn
3´
cn c n c n
so n. substitute 1 for
cn cn
To determine whether 1, is a solution, intercept we replace y with 0 and solve for
cn
cn cn cn cn cn cn cn cn cn c n c n c n c n c n c n c n c n
2 . 5x − 3 · 0 = −15 cn cn cn cn cn cn
3
x and for y. 5x = −15 cn cn
2
cn cn cn
cn
x = −3
6x − 4y = 1
cn cn
cn cn cn cn cn
The x-intercept is (−3, 0).
c n c n c n
3
cn
6·1−4· ? 1
c n
cn cn cn cn cn c n
2
c n cn
To find the y-
c n c n c n
intercept we replace x with 0 and solve for
c n c n c n cn c n c n c n c n
6−6 ¯ cn cn
y.
0 ¯ 1 FALSE c n
cn
c n
5 · 0 − 3y = −15
cn cn cn cn cn cn
³ c n
3´
−3y = −15
cn
cn cn
The equation 0 = 1 is false, so 1, y = 5
2
cn cn cn cn cn cn cn c n c n cn cn cn cn
is not a solution.
cn cn cn The y-intercept is (0, 5).
cn cn cn cn
³ 1 4´ We plot the intercepts and draw the line th
cn cn
13. To determine whether − , − is a solution, substitute
cn cn
c n c n c n c n c n c n c n c n
cn cn cn cn cn cn cn cn cn
tains
cn
2 5
1 4 them. We could find a third point as a ch
− for a and − for b.
c n c n c n c n c n c n c n c n c n c n
2 5 hat the were found correctly.
n
c cn cn cn cn
intercepts
cn
c n c n c n cn cn cn
2a + 5b = 3 cn cn cn cn cn
³ 1´ ³ 4´ cn cn cn cn
2 − +5 − ? 3
cn cn
cn cn cn c n cn
2 5
−1 − 4 cn cn
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