cs t 2025stPearsonst Education,st Inc
Copyrightst◯
,Chapter 1 s t
Graphs, Functions, and Models s t s t s t
To graph (−1, 4) we move from the origin 1
s t s t st s t s t s t s t s t s t s t
o the s t
Check Your Understanding Section 1.1
st st st st
left of the y- s t s t s t
axis. Then we move 4 units up from the
s t s t s t s t s t s t s t s t s t
1. The point (— 5, 0) is on an axis, so it is not in any quadra
st st s t st st st st st st st st st st st st
x-axis.
nt. The statement is false.
st st st st To graph (0, 2) we do not move to the right or the
st st st st st st st st st st st st st
the y-
s t st
2. The ordered pair (1,— 6) is located 1 unit right of the orig
axis since the first coordinate is 0. From the origin
st st st s t st st st st st st st st
st st st st st st s t st st
in and 6 units below it. The ordered pair— ( 6, 1) is locate
st st st st st st st st st s t st st st
move 2 units up. st st st
d 6 units left of the origin and 1 unit above it. Thus, (1,
st st st — st st st st st st st st s t st s
6) —
t and ( 6, 1) do not name the same point. Th
s t st s t st st st st st st st s t
To graph (2, —2) we move from the origin 2 units to
st st s t st st st st st st st st
e statement is false.
st st st
ight of the y- st st st
y
axis. Then we move 2 units down from the x-ax
s t st st st st st st st st
3. True; the first coordinate of a point is also called the abs
st st st st st st st st st st st
( 1,st4)
cissa. s t 4
4. True; the point ( 2 7) is 2 units left of the origin and
s t s t s t s t s t s t s t s t s t s t s t s t s t
− , st 2s t (0,st2)
(4,st0
7 units above it.
st st st s t )
2 (2,s t s t
2)
5. True; the second coordinate of a point is also called the o
st st st st st st st st st st st
( 3,s t s t 5)
4
rdinate.
6. False; the point (0, −3) is on the y-axis.
st st st st st st st st
5. To graph ( 5, 1) we move from the origin 5
s t st st s t st st st st st st st s
to the left—of the y-
t st st st st st
axis. Then we move 1 unit up from the x-axis.
s t st st st st st st st st
Exercise Set 1.1 st st
To graph (5, 1) we move from the origin 5 units to t
st st st st st st st st st st st st
ht of the y-
st s t s t
1. Point A is located 5 units to the left of the y-
s t s t s t s t s t s t s t s t s t s t s t axis. Then we move 1 unit up from the x-ax
s t s t s t s t s t s t s t s t s t
axis and 4 units up from the x-
s t s t st st st st st
To graph (2, 3) we move from the origin 2 units to t
st st st st st st st st st st st st
axis, so its coordinates are (−5, 4).
st st st st st st
ht of the y-
st s t s t
Point B is located 2 units to the right of the y-
s t s t s t s t s t s t s t s t s t s t s t axis. Then we move 3 units up from the x-ax
s t s t s t s t s t s t s t s t s t
axis and 2 units down from the x-
s t st st st st st st
To graph (2, —1) we move from the origin 2 units to
st st s t st st st st st st st st
axis, so its coordinates are (2, −2).
st st st st st st
ight of the y- st st st
Point C is located 0 units to the right or left of the y-
st st st st st st st st st st st st st axis. Then we move 1 unit down from the x-ax
s t st st st st st st st st
axis and 5 units down from the x-
st st st st st st st
To graph (0, 1) we do not move to the right or the
st st st st st st st st st st st st st
axis, so its coordinates are (0, −5).
st st st st st st
the y-
s t st
Point D is located 3 units to the right of the y-
s t s t s t s t s t s t s t s t s t s t s t axis since the first coordinate is 0. From the origin
st st st st st st s t st st
axis and 5 units up from the x-
s t st st st st st s t move 1 unit up. st st st
axis, so its coordinates are (3, 5).
st st st st st st
y
Point E is located 5 units to the left of the y-
s t s t s t s t s t s t s t s t s t s t s t
axis and s t
4
4 units down from the x-
s t s t s t s t s t
2
(2,st3)
axis, so its coordinates are (−5, −4).
s t s t s t s t s t st ( 5,st (0,st1) (5,st1)
1)
Point F is located 3 units to the right of the y-
s t s t s t s t s t s t s t s t s t s t s t
4 2 2s t (2,s t s t
axis and 0 units up or down from the x-
s t st st st st st st st st 1)
axis, so its coordinates are (3, 0).
st st st st st st 4
3. To graph (4, 0) we move from the origin 4 units to the ri
7. The first coordinate represents the year and the cor
s t st st st st st st st st st st st st st
st st st st st st st st st
ght of the y-
sponding second coordinate represents the number
st st st
st st st st st st s
axis. Since the second coordinate is 0, we do not move u
es served by Southwest Airlines. The ordered pairs
s t st st st st st st st st st st
st st st st s t st st
p or down from the x-axis.
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2
st st st st st
st st st st st st st st
To graph ( − 3, 5) we move from the origin 3 units to t
st st s t s t s ts t st st st st st st st st st 72),
he left of −
stthe y- st st st and (2021, 121). st st
axis. Then we move 5 units down from the x-axis.
s t st st st st st st st st
cs t 2025stPearsonst Education,st Inc
Copyrightst◯
,
, 14 Chapter 1: st s ts t Graphs, Functions, and st st s
els
9. To determine whether (−1, −9) is a solution, subs
s t s t s t s t st s t s t s t s t 2a + 5b = 3 st st st st st
titute 3
−1 for x and −9 for y
st st st st st 2·0+5· st st st st st s t ? 3 s t st
5
st
s t
. y = 7x − 2
st st st st st
−9 ?¯ 7(−1) − s t st st 0+3 ¯ st st st
¯ 2 3 ¯ 3 s t
st
s t TRUE
−7 − 2 st st
³ 3´ st
st
−9 ¯ −9 s t
st
TRUE The equation 3 = 3 is true, so 0, is a solution.
st st st st st st st st st st st
The equation −9 = −9 is true, so (−1, −9) is a so
s t s t st s t s t s t s t s t st s t s t s t
5
lution. To determine whether (0, 2) is a solution, substi
st st st st st st st st st
15. To determine whether (−0.75, 2.75) is a solution
st st st st st st st
sti- tute −0.75 for x and 2.75 for y.
st st st st st st st st
tute 0 for st st
x and 2 for y.
st st st st
x2 − y2 = 3st st st st st
y = 7x − 2
st st st st
2 ? 7·0− s t s t st st st
(−0.75)2 − (2.75)2 ?¯ 3 st
st
s t
st
st
2
st
¯ 0— 2 0.5625 − 7.5625 st st
¯
¯
2 −2 FALSE st −7 ¯ 3 s t
st
s t FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution
st st st st st st st st st st st st st
The equation − s t −
. 0.75, 2.75 7 = 3 is false, so (
s t s t s t s t s t s t s t st
³2 3 ´ 2 st
st st
st tnot a solution.s t st
11. To determine whether , is a solution, substitute
st st st st st st st st st
To determine whether (2, −1) is a solution, su
s t s t s t s t s t s t s t
e 2
st
s t
3 s t 4 3
3 st for x and −1 for y.
st st st st
for x and for y.
st
st st st st
4 x − y2 = 3
2
st st st st st
6x − 4y = 1
22 − (−1)2 ?¯ 3
st st st st st st s t
st st
st
2 3 4— 1
61· −4·
st s t ? st st s t s ¯
t
3 st
4 st s t
4 3 ¯ 3 s t
st
s t TRUE
¯ 3 s t
— The equation 3 = 3 is true, so (2, −1) is a solut
st st st st st st st st st st st st
1 ¯ 1 TRUE st
s t
³2 3 ´ st 17. Graph 5x − 3y = −15.
st st st st st st
s t st
The equation 1 = 1 is³ true, , is a solutio
xTo find the x-
s st t
st st st st st st st st st
3 ´ 3 st st
st st st
so st
n. substitute 1 for
To determine whether 1, is a solution, st st
st
st st st st st intercept we replace y with 0 and solve for
st st st st st st st
4
st st
2 . 5x − 3 · 0 = −15 st st st st st st
3
x and for y. st st
5x = −15 st st
2
st
st
x = −3
6x − 4y = 1
st st
st st st st st
The x-intercept is (−3, 0).
s t s t s t st
3 s t
6 ·1 −4 ·st st st st st s t ? 1 s t st
To find the y-
2 st st st
intercept we replace x with 0 and solve for
st st st st st st st st
6−6 ¯ st st
y.
0 ¯ 1 s t
st
s t FALSE 5 · 0 − 3y = −15
st st st st st st
³ s t
3´
−3y = −15
st
st st
The equation 0 = 1 is false, so
st st st st st st st s t s t 1, y = 5
2
st st st st
is not a solution.
st st st The y-intercept is (0, 5).
st st st st
³ st
1 4´ st
13. To determine whether −
st st st st st
st
st st
st
, − is a solution, substitute st st st
We plot the intercepts and draw the line th
s t s t s t s t s t s t s t s t
2 5 tains
1 4 them. We could find a third point as a ch
− for a and − for b.
s t s t s t s t s t s t s t s t s t s t
st st st st st
2 5 that the were found correctly.
st
st s t intercepts s t st st st
2a + 5b = 3 st st st st st
³ 1´ ³ 4´ st
st
st st
st
st
2 − +5 − ? 3 st st st s t st
2 5
−1 − 4 st st
cs t 2025stPearsonst Education,st Inc
Copyrightst◯
Copyrightst◯
,Chapter 1 s t
Graphs, Functions, and Models s t s t s t
To graph (−1, 4) we move from the origin 1
s t s t st s t s t s t s t s t s t s t
o the s t
Check Your Understanding Section 1.1
st st st st
left of the y- s t s t s t
axis. Then we move 4 units up from the
s t s t s t s t s t s t s t s t s t
1. The point (— 5, 0) is on an axis, so it is not in any quadra
st st s t st st st st st st st st st st st st
x-axis.
nt. The statement is false.
st st st st To graph (0, 2) we do not move to the right or the
st st st st st st st st st st st st st
the y-
s t st
2. The ordered pair (1,— 6) is located 1 unit right of the orig
axis since the first coordinate is 0. From the origin
st st st s t st st st st st st st st
st st st st st st s t st st
in and 6 units below it. The ordered pair— ( 6, 1) is locate
st st st st st st st st st s t st st st
move 2 units up. st st st
d 6 units left of the origin and 1 unit above it. Thus, (1,
st st st — st st st st st st st st s t st s
6) —
t and ( 6, 1) do not name the same point. Th
s t st s t st st st st st st st s t
To graph (2, —2) we move from the origin 2 units to
st st s t st st st st st st st st
e statement is false.
st st st
ight of the y- st st st
y
axis. Then we move 2 units down from the x-ax
s t st st st st st st st st
3. True; the first coordinate of a point is also called the abs
st st st st st st st st st st st
( 1,st4)
cissa. s t 4
4. True; the point ( 2 7) is 2 units left of the origin and
s t s t s t s t s t s t s t s t s t s t s t s t s t
− , st 2s t (0,st2)
(4,st0
7 units above it.
st st st s t )
2 (2,s t s t
2)
5. True; the second coordinate of a point is also called the o
st st st st st st st st st st st
( 3,s t s t 5)
4
rdinate.
6. False; the point (0, −3) is on the y-axis.
st st st st st st st st
5. To graph ( 5, 1) we move from the origin 5
s t st st s t st st st st st st st s
to the left—of the y-
t st st st st st
axis. Then we move 1 unit up from the x-axis.
s t st st st st st st st st
Exercise Set 1.1 st st
To graph (5, 1) we move from the origin 5 units to t
st st st st st st st st st st st st
ht of the y-
st s t s t
1. Point A is located 5 units to the left of the y-
s t s t s t s t s t s t s t s t s t s t s t axis. Then we move 1 unit up from the x-ax
s t s t s t s t s t s t s t s t s t
axis and 4 units up from the x-
s t s t st st st st st
To graph (2, 3) we move from the origin 2 units to t
st st st st st st st st st st st st
axis, so its coordinates are (−5, 4).
st st st st st st
ht of the y-
st s t s t
Point B is located 2 units to the right of the y-
s t s t s t s t s t s t s t s t s t s t s t axis. Then we move 3 units up from the x-ax
s t s t s t s t s t s t s t s t s t
axis and 2 units down from the x-
s t st st st st st st
To graph (2, —1) we move from the origin 2 units to
st st s t st st st st st st st st
axis, so its coordinates are (2, −2).
st st st st st st
ight of the y- st st st
Point C is located 0 units to the right or left of the y-
st st st st st st st st st st st st st axis. Then we move 1 unit down from the x-ax
s t st st st st st st st st
axis and 5 units down from the x-
st st st st st st st
To graph (0, 1) we do not move to the right or the
st st st st st st st st st st st st st
axis, so its coordinates are (0, −5).
st st st st st st
the y-
s t st
Point D is located 3 units to the right of the y-
s t s t s t s t s t s t s t s t s t s t s t axis since the first coordinate is 0. From the origin
st st st st st st s t st st
axis and 5 units up from the x-
s t st st st st st s t move 1 unit up. st st st
axis, so its coordinates are (3, 5).
st st st st st st
y
Point E is located 5 units to the left of the y-
s t s t s t s t s t s t s t s t s t s t s t
axis and s t
4
4 units down from the x-
s t s t s t s t s t
2
(2,st3)
axis, so its coordinates are (−5, −4).
s t s t s t s t s t st ( 5,st (0,st1) (5,st1)
1)
Point F is located 3 units to the right of the y-
s t s t s t s t s t s t s t s t s t s t s t
4 2 2s t (2,s t s t
axis and 0 units up or down from the x-
s t st st st st st st st st 1)
axis, so its coordinates are (3, 0).
st st st st st st 4
3. To graph (4, 0) we move from the origin 4 units to the ri
7. The first coordinate represents the year and the cor
s t st st st st st st st st st st st st st
st st st st st st st st st
ght of the y-
sponding second coordinate represents the number
st st st
st st st st st st s
axis. Since the second coordinate is 0, we do not move u
es served by Southwest Airlines. The ordered pairs
s t st st st st st st st st st st
st st st st s t st st
p or down from the x-axis.
(1971, 3), (1981, 15), (1991, 32), (2001, 59), (2
st st st st st
st st st st st st st st
To graph ( − 3, 5) we move from the origin 3 units to t
st st s t s t s ts t st st st st st st st st st 72),
he left of −
stthe y- st st st and (2021, 121). st st
axis. Then we move 5 units down from the x-axis.
s t st st st st st st st st
cs t 2025stPearsonst Education,st Inc
Copyrightst◯
,
, 14 Chapter 1: st s ts t Graphs, Functions, and st st s
els
9. To determine whether (−1, −9) is a solution, subs
s t s t s t s t st s t s t s t s t 2a + 5b = 3 st st st st st
titute 3
−1 for x and −9 for y
st st st st st 2·0+5· st st st st st s t ? 3 s t st
5
st
s t
. y = 7x − 2
st st st st st
−9 ?¯ 7(−1) − s t st st 0+3 ¯ st st st
¯ 2 3 ¯ 3 s t
st
s t TRUE
−7 − 2 st st
³ 3´ st
st
−9 ¯ −9 s t
st
TRUE The equation 3 = 3 is true, so 0, is a solution.
st st st st st st st st st st st
The equation −9 = −9 is true, so (−1, −9) is a so
s t s t st s t s t s t s t s t st s t s t s t
5
lution. To determine whether (0, 2) is a solution, substi
st st st st st st st st st
15. To determine whether (−0.75, 2.75) is a solution
st st st st st st st
sti- tute −0.75 for x and 2.75 for y.
st st st st st st st st
tute 0 for st st
x and 2 for y.
st st st st
x2 − y2 = 3st st st st st
y = 7x − 2
st st st st
2 ? 7·0− s t s t st st st
(−0.75)2 − (2.75)2 ?¯ 3 st
st
s t
st
st
2
st
¯ 0— 2 0.5625 − 7.5625 st st
¯
¯
2 −2 FALSE st −7 ¯ 3 s t
st
s t FALSE
The equation 2 = −2 is false, so (0, 2) is not a solution
st st st st st st st st st st st st st
The equation − s t −
. 0.75, 2.75 7 = 3 is false, so (
s t s t s t s t s t s t s t st
³2 3 ´ 2 st
st st
st tnot a solution.s t st
11. To determine whether , is a solution, substitute
st st st st st st st st st
To determine whether (2, −1) is a solution, su
s t s t s t s t s t s t s t
e 2
st
s t
3 s t 4 3
3 st for x and −1 for y.
st st st st
for x and for y.
st
st st st st
4 x − y2 = 3
2
st st st st st
6x − 4y = 1
22 − (−1)2 ?¯ 3
st st st st st st s t
st st
st
2 3 4— 1
61· −4·
st s t ? st st s t s ¯
t
3 st
4 st s t
4 3 ¯ 3 s t
st
s t TRUE
¯ 3 s t
— The equation 3 = 3 is true, so (2, −1) is a solut
st st st st st st st st st st st st
1 ¯ 1 TRUE st
s t
³2 3 ´ st 17. Graph 5x − 3y = −15.
st st st st st st
s t st
The equation 1 = 1 is³ true, , is a solutio
xTo find the x-
s st t
st st st st st st st st st
3 ´ 3 st st
st st st
so st
n. substitute 1 for
To determine whether 1, is a solution, st st
st
st st st st st intercept we replace y with 0 and solve for
st st st st st st st
4
st st
2 . 5x − 3 · 0 = −15 st st st st st st
3
x and for y. st st
5x = −15 st st
2
st
st
x = −3
6x − 4y = 1
st st
st st st st st
The x-intercept is (−3, 0).
s t s t s t st
3 s t
6 ·1 −4 ·st st st st st s t ? 1 s t st
To find the y-
2 st st st
intercept we replace x with 0 and solve for
st st st st st st st st
6−6 ¯ st st
y.
0 ¯ 1 s t
st
s t FALSE 5 · 0 − 3y = −15
st st st st st st
³ s t
3´
−3y = −15
st
st st
The equation 0 = 1 is false, so
st st st st st st st s t s t 1, y = 5
2
st st st st
is not a solution.
st st st The y-intercept is (0, 5).
st st st st
³ st
1 4´ st
13. To determine whether −
st st st st st
st
st st
st
, − is a solution, substitute st st st
We plot the intercepts and draw the line th
s t s t s t s t s t s t s t s t
2 5 tains
1 4 them. We could find a third point as a ch
− for a and − for b.
s t s t s t s t s t s t s t s t s t s t
st st st st st
2 5 that the were found correctly.
st
st s t intercepts s t st st st
2a + 5b = 3 st st st st st
³ 1´ ³ 4´ st
st
st st
st
st
2 − +5 − ? 3 st st st s t st
2 5
−1 − 4 st st
cs t 2025stPearsonst Education,st Inc
Copyrightst◯
