Answers (Anticipation Guide and Lesson...
30
Answers Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 9 A1 Glencoe Algebra 1 Chapter Resources Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. NAME DATE PERIOD Chapter 9 3 Glencoe Algebra 1 Before you begin Chapter 9 • Read each statement. • Decide whether you Agree (A) or Disagree (D) with the statement. • Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure). STEP 1 A, D, o NS Statement STEP 2 A o D 1. The graph of a quadratic function is a parabola. 2. The graph of 4x 2 – 2x + 7 will be a parabola opening downward since the coefficient of x 2 is positive. 3. A quadratic function’s axis of symmetry is either the x-axis or the y-axis. 4. The graph of a quadratic function opening upward has no maximum value. 5. The x-intercepts of the graph of a quadratic function are the solutions to the related quadratic equation. 6. All quadratic equations have two real solutions. 7. Any quadratic expression can be written as a perfect square by a method called completing the square. 8. The quadratic formula can only be used to solve quadratic equations that cannot be solved by factoring or graphing. 9. A function containing powers is called an exponential function. 10. Receiving compound interest on a bank account is one example of exponential growth. After you complete Chapter 9 • Reread each statement and complete the last column by entering an A or a D. • Did any of your opinions about the statements change from the first column? • For those statements that you mark with a D, use a piece of paper to write an example of why you disagree. 9 Anticipation Guide Quadratic and Exponential Functions Step 1 Step 2 A D D A A D A D D A Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. NAME DATE PERIOD Lesson 9-1 Chapter 9 5 Glencoe Algebra 1 Characteristics of Quadratic Functions Quadratic Function a function described by an equation of the form f(x) = ax 2 + bx + c, where a ≠ 0 Example: y = 2x 2 + 3x + 8 The parent graph of the family of quadratic fuctions is y = x 2 . Graphs of quadratic functions have a general shape called a parabola. A parabola opens upward and has a minimum point when the value of a is positive, and a parabola opens downward and has a maximum point when the value of a is negative. a. Use a table of values to graph y = x 2 - 4x + 1. x y -1 6 0 1 1 -2 2 -3 3 -2 4 1 x y O Graph the ordered pairs in the table and connect them with a smooth curve. b. What is the domain and range of this function? The domain (the x-values) is all real numbers. The range (the y-values) is all real numbers greater than or equal to -3, which is the minimum. a. Use a table of values to graph y = -x 2 - 6x - 7. x y -6 -7 -5 -2 -4 1 -3 2 -2 1 -1 -2 0 -7 x y O Graph the ordered pairs in the table and connect them with a smooth curve. b. What is the domain and range of this function? The domain (the x-values is all real numbers. The range (the y-values) is all real numbers less than or equal to 2, which is the maximum. Exercises Use a table of values to graph each function. Determine the domain and range. 1. y = x 2 + 2 2. y = -x 2 - 4 3. y = x 2 - 3x + 2 x y O x y O x y O D: {x|x is a real D: {x|x is a real D: {x|x is a real number.} R: {y|y ≥ 2} number.} R: {y|y ≤ -4} number.} R: {y|y ≥ - 1 − 4 } 9-1 Study Guide and Intervention Graphing Quadratic Functions Example 1 Example 2 Answers (Anticipation Guide and Lesson 9-1)
Transcript of Answers (Anticipation Guide and Lesson...
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A1 Glencoe Algebra 1
Chapter Resources
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
3
Gle
ncoe A
lgeb
ra 1
Befo
re y
ou
beg
in C
ha
pte
r 9
•
R
ead
each
sta
tem
en
t.
•
D
eci
de w
heth
er
you
Agre
e (
A)
or
Dis
agre
e (
D)
wit
h t
he s
tate
men
t.
•
W
rite
A o
r D
in
th
e f
irst
colu
mn
OR
if
you
are
not
sure
wh
eth
er
you
agre
e o
r d
isagre
e,
wri
te N
S (
Not
Su
re).
ST
EP
1A
, D
, o
NS
Sta
tem
en
tS
TE
P 2
A o
D
1.
Th
e g
rap
h o
f a q
uad
rati
c fu
nct
ion
is
a p
ara
bola
.
2.
Th
e g
rap
h o
f 4x
2 –
2x +
7 w
ill
be a
para
bola
op
en
ing d
ow
nw
ard
si
nce
th
e c
oeff
icie
nt
of
x2 i
s p
osi
tive.
3.
A q
uad
rati
c fu
nct
ion
’s a
xis
of
sym
metr
y i
s eit
her
the x
-axis
or
the y
-axis
.
4.
Th
e g
rap
h o
f a q
uad
rati
c fu
nct
ion
op
en
ing u
pw
ard
has
no
maxim
um
valu
e.
5.
Th
e x
-in
terc
ep
ts o
f th
e g
rap
h o
f a q
uad
rati
c fu
nct
ion
are
th
e
solu
tion
s to
th
e r
ela
ted
qu
ad
rati
c equ
ati
on
.
6.
All
qu
ad
rati
c equ
ati
on
s h
ave t
wo r
eal
solu
tion
s.
7.
An
y q
uad
rati
c exp
ress
ion
can
be w
ritt
en
as
a p
erf
ect
squ
are
by
a m
eth
od
call
ed
com
ple
tin
g t
he
squ
are
.
8.
Th
e q
uad
rati
c fo
rmu
la c
an
on
ly b
e u
sed
to s
olv
e q
uad
rati
c equ
ati
on
s th
at
can
not
be s
olv
ed
by f
act
ori
ng o
r gra
ph
ing.
9.
A f
un
ctio
n c
on
tain
ing p
ow
ers
is
call
ed
an
exp
on
en
tial
fun
ctio
n.
10.
Rece
ivin
g c
om
pou
nd
in
tere
st o
n a
ban
k a
ccou
nt
is o
ne e
xam
ple
of
exp
on
en
tial
gro
wth
.
Aft
er y
ou
com
ple
te C
ha
pte
r 9
•
R
ere
ad
each
sta
tem
en
t an
d c
om
ple
te t
he l
ast
colu
mn
by e
nte
rin
g a
n A
or
a D
.
•
D
id a
ny o
f you
r op
inio
ns
abou
t th
e s
tate
men
ts c
han
ge f
rom
th
e f
irst
colu
mn
?
•
F
or
those
sta
tem
en
ts t
hat
you
mark
wit
h a
D,
use
a p
iece
of
pap
er
to w
rite
an
exam
ple
of
wh
y y
ou
dis
agre
e.
9A
nti
cip
ati
on
Gu
ide
Qu
ad
rati
c a
nd
Exp
on
en
tial
Fu
ncti
on
s
Ste
p 1
Ste
p 2
A D D A A D A D D A
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-1
Ch
ap
ter
9
5
Gle
ncoe A
lgeb
ra 1
Ch
ara
cteri
stic
s o
f Q
uad
rati
c Fu
nct
ion
s
Qu
ad
rati
c
Fu
ncti
on
a f
unction d
escribed b
y a
n e
quation o
f th
e f
orm
f(x
) =
ax
2 +
bx +
c,
where
a ≠
0
Exam
ple
:
y =
2x
2 +
3x +
8
Th
e p
are
nt
gra
ph
of
the f
am
ily o
f qu
ad
rati
c fu
ctio
ns
is y
= x
2.
Gra
ph
s of
qu
ad
rati
c fu
nct
ion
s h
ave a
gen
era
l sh
ap
e c
all
ed
a p
ara
bo
la.
A p
ara
bola
op
en
s u
pw
ard
an
d h
as
a m
inim
um
p
oin
t w
hen
th
e v
alu
e o
f a
is
posi
tive,
an
d a
para
bola
op
en
s d
ow
nw
ard
an
d h
as
a m
ax
imu
m
po
int
wh
en
th
e v
alu
e o
f a
is
negati
ve.
a.
Use a
ta
ble
of
va
lues t
o g
ra
ph
y =
x2 -
4x +
1.
xy
-1
6
01
1-
2
2-
3
3-
2
41
x
y
O
Gra
ph
th
e o
rdere
d p
air
s in
th
e t
able
an
d
con
nect
th
em
wit
h a
sm
ooth
cu
rve.
b.
Wh
at
is t
he d
om
ain
an
d r
an
ge o
f
this
fu
ncti
on
?
Th
e d
om
ain
(th
e x
-valu
es)
is
all
real
nu
mbers
. T
he r
an
ge (
the y
-valu
es)
is
all
re
al
nu
mbers
gre
ate
r t
han
or
equ
al
to
-3,
wh
ich
is
the m
inim
um
.
a.
Use a
ta
ble
of
va
lues t
o g
ra
ph
y =
-x
2 -
6x -
7.
xy
-6
-7
-5
-2
-4
1
-3
2
-2
1
-1
-2
0-
7
x
y
O
Gra
ph
th
e o
rdere
d p
air
s in
th
e t
able
an
d
con
nect
th
em
wit
h a
sm
ooth
cu
rve.
b.
Wh
at
is t
he d
om
ain
an
d r
an
ge o
f th
is
fun
cti
on
?
Th
e d
om
ain
(th
e x
-valu
es
is a
ll r
eal
nu
mbers
. T
he r
an
ge (
the y
-valu
es)
is
all
re
al
nu
mbers
less
th
an
or
equ
al
to 2
, w
hic
h i
s th
e m
axim
um
.
Exerc
ises
Use a
ta
ble
of
va
lues t
o g
ra
ph
ea
ch
fu
ncti
on
. D
ete
rm
ine t
he d
om
ain
an
d r
an
ge.
1. y =
x2 +
2
2. y =
-x
2 -
4
3. y =
x2 -
3x +
2
x
y
O
x
y
O
x
y
O
D
: {x|
x i
s a
real
D: {x|x
is a
real
D
: {x|x
is a
real
n
um
ber.} R
: {y|
y ≥
2}
nu
mb
er.} R
: {y|y
≤ -
4}
nu
mb
er.}
R: {y|y
≥ - 1
−
4 }
9-1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
s
Exam
ple
1Exam
ple
2
Answers (Anticipation Guide and Lesson 9-1)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A2 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
6
Gle
ncoe A
lgeb
ra 1
Sym
metr
y a
nd
Vert
ices
Para
bola
s h
ave a
geom
etr
ic p
rop
ert
y c
all
ed
sy
mm
etr
y.
Th
at
is,
if t
he f
igu
re i
s fo
lded
in
half
, each
half
wil
l m
atc
h t
he o
ther
half
exact
ly.
Th
e v
ert
ical
lin
e
con
tain
ing t
he f
old
lin
e i
s ca
lled
th
e a
xis
of
sy
mm
etr
y.
Th
e a
xis
of
sym
metr
y c
on
tain
s th
e
min
imu
m o
r m
axim
um
poin
t of
the p
ara
bola
, th
e v
erte
x.
a.
Writ
e t
he e
qu
ati
on
of
the a
xis
of
sy
mm
etr
y.
In y
= 2
x2 +
4x +
1,
a =
2 a
nd
b =
4.
Su
bst
itu
te t
hese
valu
es
into
th
e e
qu
ati
on
of
the a
xis
of
sym
metr
y.
x =
-
b
−
2a
x =
-
4
−
2(2
) = -
1
Th
e a
xis
of
sym
metr
y i
s x =
-1.
b.
Fin
d t
he c
oo
rd
ina
tes o
f th
e v
erte
x.
Sin
ce t
he e
qu
ati
on
of
the a
xis
of
sym
metr
y i
s x =
-1 a
nd
th
e v
ert
ex l
ies
on
th
e a
xis
, th
e x
-coord
inate
of
the v
ert
ex
is -
1.
y =
2x
2 +
4x +
1
Origin
al equation
y =
2(-
1)2
+ 4
(-1)
+ 1
S
ubstitu
te.
y =
2(1
) -
4 +
1
Sim
plif
y.
y =
-1
Th
e v
ert
ex i
s at
(-1,
-1).
c.
Iden
tify
th
e v
erte
x a
s a
ma
xim
um
or
a m
inim
um
.
Sin
ce t
he c
oeff
icie
nt
of
the x
2-t
erm
is
posi
tive,
the p
ara
bola
op
en
s u
pw
ard
, an
d
the v
ert
ex i
s a m
inim
um
poin
t.
d.
Gra
ph
th
e f
un
cti
on
. x
y O
( -1,-
1)
x=-
1
Exerc
ises
Co
nsid
er e
ach
eq
ua
tio
n.
Dete
rm
ine w
heth
er t
he f
un
cti
on
ha
s m
axim
um
or
min
imu
m v
alu
e.
Sta
te t
he m
ax
imu
m o
r m
inim
um
va
lue.
Wh
at
are t
he d
om
ain
an
d
ra
ng
e o
f th
e f
un
cti
on
? F
ind
th
e e
qu
ati
on
of
the a
xis
of
sy
mm
etr
y.
Gra
ph
th
e
fun
cti
on
.
9-1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
s
Exam
ple
Axis
of
Sym
metr
y
For
the p
ara
bola
y =
ax
2 +
bx +
c,
where
a ≠
0,
the lin
e x
= -
b
−
2a i
s t
he a
xis
of
sym
metr
y.
Exam
ple
: T
he a
xis
of
sym
metr
y o
f
y =
x2 +
2x +
5 is t
he lin
e x
= -
1.
Co
nsid
er t
he g
ra
ph
of
y =
2x
2 +
4x +
1.
1. y =
x2 +
3
2. y =
-x
2 -
4x -
4
3. y =
x2 +
2x +
3
x
y
O
x
y
O
x
y
O
min
; (0
, 3);
D
: {x|
all r
eals},
R: {y|
y ≥
3};
x = 0
max;
(-2,
0);
D
: {x|
all r
eals},
R: {y|
y ≤
0};
x = -
2
min
; (-
1,
2);
D
: {x
| all r
eals},
R: {y
|y ≥
2}; x
= -
1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-1
Ch
ap
ter
9
7
Gle
ncoe A
lgeb
ra 1
Sk
ills
Pra
ctic
e
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
sU
se a
ta
ble
of
va
lues t
o g
ra
ph
ea
ch
fu
ncti
on
. S
tate
th
e d
om
ain
th
e r
an
ge.
1. y
= x
2 -
4
2. y =
-x
2 +
3
3. y =
x2 -
2x -
6
x
y
O
x
y
O
x
y
O
Fin
d t
he v
erte
x,
the e
qu
ati
on
of
the a
xis
of
sy
mm
etr
y,
an
d t
he y
-in
tercep
t.
4. y =
2x
2 -
8x +
6
5. y =
x2 +
4x +
6
6. y =
-3x
2 -
12x +
3
Co
nsid
er e
ach
eq
ua
tio
n.
a
. D
ete
rm
ine w
heth
er t
he f
un
cti
on
ha
s m
axim
um
or m
inim
um
va
lue.
b. S
tate
th
e m
ax
imu
m o
r m
inim
um
va
lue.
c. W
ha
t a
re t
he d
om
ain
an
d r
an
ge o
f th
e f
un
cti
on
?
7. y =
2x
2
8. y =
x2 -
2x -
5
9. y =
-x
2 +
4x -
1
Gra
ph
ea
ch
fu
ncti
on
.
10. f(
x)
= -
x2 -
2x +
2
11. f(
x)
= 2
x2 +
4x -
2
12. f(
x)
= -
2x
2 -
4x +
6
xO
f(x
)
xO
f(x
)
xO
f(x
)
9-1 D
= a
ll r
eals
R =
{y | y ≥ – 4
} D
= a
ll r
eals
R =
{y | y ≤ 3
} D
= a
ll r
eals
R =
{y | y ≥ – 7
}
(2, -
2) ;
x =
2;
(0, 6
) (-
2,
2) ;
x =
-2;
(0, 6
) (-
2, 15
) ; x
= -
2;
(0, 3
)
min
imu
m; (0
, 0
) ;
D =
all r
eals
,
R = {
y | y ≥ 0
}
min
imu
m; (1
, -
6) ;
D =
all r
eals
,
R = {
y | y
≥ -
6}
maxim
um
; (2
, 3
) ;
D =
all r
eals
,
R = {
y | y
≤ 3
}
Answers (Lesson 9-1)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A3 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
8
Gle
ncoe A
lgeb
ra 1
Practi
ce
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
sU
se a
ta
ble
of
va
lues t
o g
ra
ph
ea
ch
fu
ncti
on
. D
ete
rm
ine t
he d
om
ain
an
d r
an
ge.
1. y =
-x
2 +
2
2. y =
x2 -
6x +
3
3. y =
-2x
2 -
8x -
5
x
y
O
x
y
O
D
: {x|
all r
eals}
D: {x|
all r
eals}
D
: {x|
all r
eals}
R: {y|y
≤ 2}
R: {y|y
≥ -
6}
R
: {y|
y ≤
3}
Fin
d t
he v
erte
x,
the e
qu
ati
on
of
the a
xis
of
sy
mm
etr
y,
an
d t
he y
-in
tercep
t.
4. y =
x2 -
9
5. y =
-2x
2 +
8x -
5
6. 4x
2 -
4x +
1
(
0, -
9);
x =
0;
(0, -
9)
(2,
3);
x =
2;
(0, -
5)
(0.5
, 0);
x =
0.5
; (0
, 1)
Co
nsid
er e
ach
eq
ua
tio
n.
Dete
rm
ine w
heth
er t
he f
un
cti
on
ha
s m
axim
um
or
minim
um
va
lue.
Sta
te t
he m
ax
imu
m o
r m
inim
um
va
lue.
Wh
at
are t
he d
om
ain
an
d r
an
ge o
f th
e f
un
cti
on
?
7. y =
5x
2 -
2x +
2
8. y =
-x
2 +
5x -
10
9. y =
3
−
2 x
2 +
4x -
9
Gra
ph
ea
ch
fu
ncti
on
.
10. f(
x)
= -
x2 +
3
11. f(
x)
= -
2x
2 +
8x -
3
12. f(
x)
= 2
x2 +
8x +
1
xO
f(x)
xO
f(x)
xO
f(x)
13. B
ASEB
ALL A
pla
yer
hit
s a b
ase
ball
in
to t
he o
utf
ield
. T
he e
qu
ati
on
h =
-0.0
05x
2 +
x +
3
giv
es
the p
ath
of
the b
all
, w
here
h i
s th
e h
eig
ht
an
d x
is
the h
ori
zon
tal
dis
tan
ce t
he b
all
tr
avels
.
a
. W
hat
is t
he e
qu
ati
on
of
the a
xis
of
sym
metr
y?
x =
10
0
b
. W
hat
is t
he m
axim
um
heig
ht
reach
ed
by t
he b
ase
ball
? 53 f
t
c.
An
ou
tfie
lder
catc
hes
the b
all
th
ree f
eet
above t
he g
rou
nd
. H
ow
far
has
the b
all
travele
d h
ori
zon
tall
y w
hen
th
e o
utf
ield
er
catc
hes
it?
200 f
t
9-1
x
y O
min
; (0
.2,
1.8
);
D: {x|
all r
eals},
R: {y|
y ≥
1.8}
max;
(2.5
, -
3.7
5);
D
: {x|
all r
eals},
R: {y|
y ≤
-3.7
5}
min
; (-
1 1
−
3 , -
11 2
−
3 );
D: {x|
all r
eals},
R: {y|
y ≥
-11 2
−
3 }
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-1
Ch
ap
ter
9
9
Gle
ncoe A
lgeb
ra 1
1.O
LY
MP
ICS
O
lym
pic
s w
ere
held
in
1896 a
nd
have b
een
held
every
fou
r years
(e
xce
pt
1916,
1940,
an
d 1
944).
Th
e w
inn
ing h
eig
ht
y i
n m
en
’s p
ole
vau
lt a
t an
y n
um
ber
Oly
mp
iad
x c
an
be a
pp
roxim
ate
d b
y t
he e
qu
ati
on
y =
0.3
7x
2 +
4.3
x +
126.
Com
ple
te t
he
table
to e
stim
ate
th
e p
ole
vau
lt h
eig
hts
in
each
of
the O
lym
pic
Gam
es.
Rou
nd
you
r an
swers
to t
he n
eare
st t
en
th.
So
urce:
Nat
ional
Secu
rity
Age
ncy
2. P
HY
SIC
S M
rs.
Cap
well
’s p
hysi
cs c
lass
in
vest
igate
s w
hat
hap
pen
s w
hen
a b
all
is
giv
en
an
in
itia
l p
ush
, ro
lls
up
, an
d t
hen
back
dow
n a
n i
ncl
ined
pla
ne.
Th
e c
lass
fi
nd
s th
at
y =
-x
2 +
6x a
ccu
rate
ly
pre
dic
ts t
he b
all
’s p
osi
tion
y a
fter
roll
ing
x s
eco
nd
s. O
n t
he g
rap
h o
f th
e e
qu
ati
on
,
wh
at
wou
ld b
e t
he y
valu
e w
hen
x =
4?
8
3. A
RC
HIT
EC
TU
RE
A
hote
l’s
main
en
tran
ce i
s in
th
e s
hap
e o
f a p
ara
boli
c arc
h.
Th
e e
qu
ati
on
y =
-x
2 +
10
x m
od
els
the a
rch
heig
ht
y f
or
an
y d
ista
nce
x f
rom
on
e s
ide o
f th
e a
rch
. U
se a
gra
ph
to
dete
rmin
e i
ts m
axim
um
heig
ht.
25 f
t
4.S
OFT
BA
LL
O
lym
pic
soft
ball
gold
m
ed
ali
st M
ich
ele
Sm
ith
pit
ches
a
curv
eball
wit
h a
sp
eed
of
64 f
eet
per
seco
nd
. If
sh
e t
hro
ws
the b
all
str
aig
ht
up
ward
at
this
sp
eed
, th
e b
all
’s h
eig
ht
h
(in
feet)
aft
er
t se
con
ds
is g
iven
by
h =
-16
t2 +
64
t. F
ind
th
e c
oord
inate
s of
the v
ert
ex o
f th
e g
rap
h o
f th
e b
all
’s
heig
ht
an
d i
nte
rpre
t it
s m
ean
ing.
(2,
64);
Aft
er
2 s
eco
nd
s,
the b
all
reach
es i
ts h
igh
est
po
int,
64 f
t ab
ove t
he g
rou
nd
.
5. G
EO
METR
Y T
ed
dy i
s bu
ild
ing t
he
rect
an
gu
lar
deck
sh
ow
n b
elo
w.
x+
6
x-
2
a. W
rite
th
e e
qu
ati
on
rep
rese
nti
ng t
he
are
a o
f th
e d
eck
. y =
(x -
2)(
x +
6)
or
y =
x2 +
4x -
12
b. W
hat
is t
he e
qu
ati
on
of
the a
xis
of
sym
metr
y?
x =
-2
c. G
rap
h t
he e
qu
ati
on
an
d l
abel
its
vert
ex.
y
xO
-4
-6
-8
-10
-12
-14
-16
-2
1-3
-4
-5
( -2,-
16)
9-1
Wo
rd
Pro
ble
m P
racti
ce
Gra
ph
ing
Qu
ad
rati
c F
un
cti
on
s
Year
Oly
mp
iad
(x)
Heig
ht
(y i
nch
es)
1896
1
1900
2
1924
7
1936
10
1964
15
2008
26
130.7
136.1
174.2
206.0
273.8
487.9
Answers (Lesson 9-1)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A4 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
10
Gle
ncoe A
lgeb
ra 1
A c
ub
ic f
un
cti
on
is
a p
oly
nom
ial
wri
tten
in
th
e f
orm
of f(x) =
ax
3 +
bx
2 +
cx +
n,
wh
ere
a ≠ 0
. C
ubic
fu
nct
ion
s d
o n
ot
have a
bso
lute
min
imu
m a
nd
maxim
um
valu
es
lik
e
qu
ad
rati
c fu
nct
ion
s d
o,
bu
t th
ey c
an
have a
lo
ca
l m
inim
um
an
d a
lo
ca
l m
ax
imu
m p
oin
t.
Pare
nt
Function:
f(x) =
x3
Dom
ain
: all
real num
bers
Range:
all
real num
bers
x
f(x
)
U
se a
ta
ble
of
va
lues t
o g
ra
ph
y =
x3 +
3x
2 -
1.
Th
en
use t
he g
ra
ph
to e
sti
ma
te t
he l
oca
tio
ns o
f th
e l
oca
l m
inim
um
an
d l
oca
l m
ax
imu
m p
oin
ts.
x–3
–2
–1
01
y–1
21
–1
2
Gra
ph
th
e o
rdere
d p
air
s, a
nd
con
nect
th
em
to c
reate
a s
mooth
cu
rve.
Th
e “
S”
shap
e e
xte
nd
s to
in
fin
ity i
n t
he p
osi
tive y
dir
ect
ion
an
d t
o
negati
ve i
nfi
nit
y i
n t
he n
egati
ve y
dir
ect
ion
.
Th
e l
oca
l m
inim
um
is
loca
ted
at
(0,
–1).
Th
e l
oca
l m
axim
um
is
loca
ted
at
(–2,
2).
Exerc
ises
Use a
ta
ble
of
va
lues t
o g
ra
ph
ea
ch
eq
ua
tio
n.
Th
en
use t
he g
ra
ph
to
esti
ma
te t
he
loca
tio
ns o
f th
e l
oca
l m
inim
um
an
d l
oca
l m
ax
imu
m p
oin
ts.
1. y =
0.5x
3 +
x2 - 1
2. y =
-2x
3 -
3x
2 - 1
3. y =
x3 +
3x
2 +
x - 4
x
y
x
y
x
y
l
ocal
maxim
um
:
lo
cal
maxim
um
:
lo
cal
maxim
um
: (-
1.3
, -
0.4
);
(0,
-1);
(
-1.8
, -
1.9
);
local
min
imu
m:
l
ocal
min
imu
m:
l
ocal
min
imu
m:
(0,
1)
(-
1,
-2)
(-
0.2
, -
4.1
)
9-1
En
rich
men
t
Gra
ph
ing
Cu
bic
Fu
ncti
on
s
Exam
ple
x
y
(-2
, 2
)
(0,-
1)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-2
Ch
ap
ter
9
11
Gle
ncoe A
lgeb
ra 1
So
lve b
y G
rap
hin
g
Qu
ad
rati
c E
qu
ati
on
an e
quation o
f th
e f
orm
ax
2 +
bx +
c =
0,
where
a ≠
0
Th
e s
olu
tion
s of
a q
uad
rati
c equ
ati
on
are
call
ed
th
e r
oo
ts o
f th
e e
qu
ati
on
. T
he r
oots
of
a q
uad
rati
c equ
ati
on
can
be f
ou
nd
by g
rap
hin
g t
he r
ela
ted
qu
ad
rati
c fu
nct
ion
f(x) =
ax
2 +
bx +
c a
nd
fin
din
g t
he x
-in
terc
ep
ts o
r zero
s o
f th
e f
un
ctio
n.
S
olv
e x
2 +
4x
+ 3
= 0
by
gra
ph
ing
.
Gra
ph
th
e r
ela
ted
fu
nct
ion
f(x
) =
x2 +
4x +
3.
Th
e e
qu
ati
on
of
the a
xis
of
sym
metr
y i
s
x =
-
4 −
2(1
) or -
2.
Th
e v
ert
ex i
s at
(-2, -
1).
Gra
ph
th
e v
ert
ex a
nd
severa
l oth
er
poin
ts o
n
eit
her
sid
e o
f th
e a
xis
of
sym
metr
y.
xO
f(x)
To s
olv
e x
2 +
4x +
3 =
0,
you
need
to k
now
w
here
th
e v
alu
e o
f f(x) =
0.
Th
is o
ccu
rs a
t th
e x
-in
terc
ep
ts, -
3 a
nd
-1.
Th
e s
olu
tion
s are
-3 a
nd
-1.
S
olv
e x
2 -
6x
+ 9
= 0
by
gra
ph
ing
.
Gra
ph
th
e r
ela
ted
fu
nct
ion
f(x
) =
x2 -
6x +
9.
Th
e e
qu
ati
on
of
the a
xis
of
sym
metr
y i
s
x =
6 −
2(1
) or
3.
Th
e v
ert
ex i
s at
(3,
0).
Gra
ph
the v
ert
ex a
nd
severa
l oth
er
poin
ts o
n e
ith
er
sid
e o
f th
e a
xis
of
sym
metr
y.
x
f(x) O
To s
olv
e x
2 -
6x +
9 =
0,
you
need
to k
now
w
here
th
e v
alu
e o
f f(x) =
0.
Th
e v
ert
ex o
f th
e
para
bola
is
the x
-in
terc
ep
t. T
hu
s, t
he o
nly
so
luti
on
is
3.
Exerc
ises
So
lve e
ach
eq
ua
tio
n b
y g
ra
ph
ing
.
1. x
2 +
7x +
12
= 0
2. x
2 -
x -
12 =
0
3. x
2 -
4x +
5 =
0
xO
f(x)
xO4
-4
-8
-12
8-4
-8
f(x)
4
xO
f(x)
-
3,
-4
4,
-3
no
real
roo
ts
9-2
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
Exam
ple
1Exam
ple
2
Answers (Lesson 9-1 and Lesson 9-2)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A5 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
12
Gle
ncoe A
lgeb
ra 1
9-2
Estim
ate S
olu
tio
ns
Th
e r
oots
of
a q
uad
rati
c equ
ati
on
may n
ot
be i
nte
gers
. If
exact
ro
ots
can
not
be f
ou
nd
, th
ey c
an
be e
stim
ate
d b
y f
ind
ing t
he c
on
secu
tive i
nte
gers
betw
een
w
hic
h t
he r
oots
lie
.
S
olv
e x
2 +
6x
+ 6
= 0
by
gra
ph
ing
. If
in
teg
ra
l ro
ots
ca
nn
ot
be f
ou
nd
,
esti
ma
te t
he r
oo
ts b
y s
tati
ng
th
e c
on
secu
tiv
e i
nte
gers b
etw
een
wh
ich
th
e r
oo
ts l
ie.
Gra
ph
th
e r
ela
ted
fu
nct
ion
f(x
) =
x2 +
6x +
6.
xf(
x)
-5
1
-4
-2
-3
-3
-2
-2
-1
1
N
oti
ce t
hat
the v
alu
e o
f th
e f
un
ctio
n c
han
ges
x
f(x)
O
from
negati
ve t
o p
osi
tive b
etw
een
th
e x
-valu
es
of -
5 a
nd
-4 a
nd
betw
een
-2 a
nd
-1.
Th
e x
-in
terc
ep
ts o
f th
e g
rap
h a
re b
etw
een
-5 a
nd
-4 a
nd
betw
een
-2 a
nd
-1.
So o
ne r
oot
is b
etw
een
-5 a
nd
-4,
an
d t
he o
ther
root
is b
etw
een
-2 a
nd
-1.
Exerc
ises
So
lve e
ach
eq
ua
tio
n b
y g
ra
ph
ing
. If
in
teg
ra
l ro
ots
ca
nn
ot
be f
ou
nd
, esti
ma
te t
he
ro
ots
to
th
e n
ea
rest
ten
th.
1. x
2 +
7x +
9 =
0
2. x
2 -
x -
4 =
0
3. x
2 -
4x +
6 =
0
xO
f(x)
xO
f(x)
xO
f(x)
-
6 <
x <
-5,
-2 <
x <
-1,
no
real
roo
ts
-2 <
x <
-1
2 <
x <
3
4. x
2 -
4x -
1 =
0
5. 4x
2 -
12x +
3 =
0
6. x
2 -
2x -
4 =
0
xO
f(x)
xO
f(x)
xO
f(x)
-
1 <
x <
0,
0 <
x <
1,
-2 <
x <
-1,
4 <
x <
5
2 <
x <
3
3 <
x <
4
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
Exam
ple
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-2
Ch
ap
ter
9
13
Gle
ncoe A
lgeb
ra 1
So
lve e
ach
eq
ua
tio
n b
y g
ra
ph
ing
.
1. x
2 -
2x +
3 =
0 Ø
2. c2
+ 6
c +
8 =
0 -
4, -
2
xO
f(x)
cO
f(c)
3. a
2 -
2a
= -
1 1
4. n
2 -
7n
= -
10 2,
5
aO
f(a)
nO
f(n
)
So
lve e
ach
eq
ua
tio
n b
y g
ra
ph
ing
. If
in
teg
ra
l ro
ots
ca
nn
ot
be f
ou
nd
,
esti
ma
te t
he r
oo
ts t
o t
he n
ea
rest
ten
th.
5. p
2 +
4p
+ 2
= 0
6. x
2 +
x -
3 =
0
pO
f(p)
xO
f(x)
-
3.4
, -
0.6
-
2.3
, 1.3
7. d
2 +
6d
= -
3
8. h
2 +
1 =
4h
dO
f(d
)
hO
f(h
)
-
5.5
, -
0.6
0
.3,
3.8
9-2
Sk
ills
Pra
ctic
e
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
Answers (Lesson 9-2)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A6 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
14
Gle
ncoe A
lgeb
ra 1
So
lve e
ach
eq
ua
tio
n b
y g
ra
ph
ing
.
1. x
2 -
5x +
6 =
0 2,
3
2. w
2 +
6w
+ 9
= 0
-
3
3. b
2 -
3b +
4 =
0 Ø
xO
f(x)
wO
f(w)
bOf(
b)
So
lve e
ach
eq
ua
tio
n b
y g
ra
ph
ing
. If
in
teg
ra
l ro
ots
ca
nn
ot
be f
ou
nd
, esti
ma
te t
he
ro
ots
to
th
e n
ea
rest
ten
th.
4. p
2 +
4p
= 3
5. 2m
2 +
5 =
10m
6. 2v
2 +
8v =
-7
pO
f(p)
mO
f(m)
vO
f(v)
-
4.6
, 0.6
0
.6,
4.4
-
2.7
, -
1.3
7. N
UM
BER
TH
EO
RY
T
wo n
um
bers
have a
su
m o
f 2
an
d a
pro
du
ct o
f -
8.
Th
e q
uad
rati
c equ
ati
on
-n
2 +
2n
+ 8
= 0
can
be u
sed
to d
ete
rmin
e
the t
wo n
um
bers
.
a.
Gra
ph
th
e r
ela
ted
fu
nct
ion
f(n
) =
-n
2 +
2n
+ 8
an
d
dete
rmin
e i
ts x
-in
terc
ep
ts. -
2,
4
b.
Wh
at
are
th
e t
wo n
um
bers
? -
2 a
nd
4
8. D
ESIG
N A
footb
rid
ge i
s su
spen
ded
fro
m a
para
boli
c
sup
port
. T
he f
un
ctio
n h
(x)
= -
1
−
25 x
2 +
9 r
ep
rese
nts
the h
eig
ht
in f
eet
of
the s
up
port
above t
he w
alk
way,
wh
ere
x =
0 r
ep
rese
nts
th
e m
idp
oin
t of
the b
rid
ge.
9. G
rap
h t
he f
un
ctio
n a
nd
dete
rmin
e i
ts x
-in
terc
ep
ts. -
15,
15
10. W
hat
is t
he l
en
gth
of
the w
alk
way b
etw
een
th
e t
wo s
up
port
s?30 f
t
xO
61
2
12 6
-6
-1
2
-1
2-
6
h(x)
9-2
Practi
ce
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
nO
f(n)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-2
Ch
ap
ter
9
15
Gle
ncoe A
lgeb
ra 1
1.FA
RM
ING
In
ord
er
for
Ray t
o d
eci
de
how
mu
ch f
ert
iliz
er
to a
pp
ly t
o h
is c
orn
cr
op
th
is y
ear,
he r
evie
ws
reco
rds
from
p
revio
us
years
. H
e f
ind
s th
at
his
cro
p
yie
ld y
dep
en
ds
on
th
e a
mou
nt
of
fert
iliz
er
he a
pp
lies
to h
is f
ield
s x
acc
ord
ing t
o t
he e
qu
ati
on
y =
-x
2 +
4x +
12.
Gra
ph
th
e f
un
ctio
n,
an
d f
ind
th
e p
oin
t at
wh
ich
Ray g
ets
th
e
hig
hest
yie
ld p
oss
ible
.
y
xO1416 12 10 8 6 4 2
45
32
1
( 2,1
6)
2. LIG
HT
A
yzh
a a
nd
Jere
my h
old
a
flash
ligh
t so
th
at
the l
igh
t fa
lls
on
a
pie
ce o
f gra
ph
pap
er
in t
he s
hap
e o
f a
para
bola
. A
yzh
a a
nd
Jere
my s
ketc
h t
he
shap
e o
f th
e p
ara
bola
an
d f
ind
th
at
the
equ
ati
on
y =
x2 -
3x -
10 m
atc
hes
the
shap
e o
f th
e l
igh
t beam
. D
ete
rmin
e t
he
roots
of
the f
un
ctio
n.
-
2 a
nd
5
3. FR
AM
ING
A
rect
an
gu
lar
ph
oto
gra
ph
is
7 i
nch
es
lon
g a
nd
6 i
nch
es
wid
e.
Th
e
ph
oto
gra
ph
is
fram
ed
usi
ng m
ate
rial x
inch
es
wid
e.
If t
he a
rea o
f th
e f
ram
e a
nd
p
hoto
gra
ph
com
bin
ed
is
156 s
qu
are
in
ches,
wh
at
is t
he w
idth
of
the f
ram
ing
mate
rial?
3 i
n.
Ph
oto
gra
ph
Fram
e
6 in
.
7 in
.
x
x
4. W
RA
PP
ING
PA
PE
R C
an
a r
ect
an
gu
lar
pie
ce o
f w
rap
pin
g p
ap
er
wit
h a
n a
rea o
f 81 s
qu
are
in
ches
have a
peri
mete
r of
60
inch
es?
(Hint:
Let
len
gth
= 3
0 –
w.)
E
xp
lain
. S
olv
ing
th
e e
qu
ati
on
(3
0 -
w)w
= 8
1 g
ives w
= 3
or
27.
A 3
in
. b
y 2
7 i
n.
sh
eet
of
pap
er
wo
uld
wo
rk.
5. E
NG
INE
ER
ING
T
he s
hap
e o
f a s
ate
llit
e
dis
h i
s oft
en
para
boli
c beca
use
of
the
refl
ect
ive q
uali
ties
of
para
bola
s. S
up
pose
a p
art
icu
lar
sate
llit
e d
ish
is
mod
ele
d b
y
the f
oll
ow
ing e
qu
ati
on
.
0.5x
2 =
2 +
y
a. A
pp
roxim
ate
th
e s
olu
tion
by
gra
ph
ing.
-
2 a
nd
2
y
xO34 2 1
-2
-3
-4
43
21
-2
-3
-4
AB
b. O
n t
he c
oord
inate
pla
ne a
bove,
tran
slate
th
e p
ara
bola
so t
hat
there
is
on
ly o
ne r
oot.
Label
this
cu
rve A
.
See s
tud
en
ts’
wo
rk.
c. T
ran
slate
th
e p
ara
bola
so t
hat
there
are
no r
oots
. L
abel
this
cu
rve B
.
See s
tud
en
ts’
wo
rk.
9-2
Wo
rd
Pro
ble
m P
racti
ce
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y G
rap
hin
g
Answers (Lesson 9-2)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A7 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
16
Gle
ncoe A
lgeb
ra 1
Para
bo
las T
hro
ug
h T
hre
e G
iven
Po
ints
If y
ou
kn
ow
tw
o p
oin
ts o
n a
str
aig
ht
lin
e,
you
can
fin
d t
he e
qu
ati
on
of
the l
ine.
To f
ind
th
e e
qu
ati
on
of
a p
ara
bola
, you
need
th
ree p
oin
ts o
n
the c
urv
e.
Here
is
how
to a
pp
roxim
ate
an
equ
ati
on
of
the p
ara
bola
th
rou
gh
th
e
poin
ts (
0, -
2),
(3,
0),
an
d (
5,
2).
Use
th
e g
en
era
l equ
ati
on
y =
ax
2 +
bx +
c.
By s
ubst
itu
tin
g t
he g
iven
valu
es
for
x a
nd
y,
you
get
thre
e e
qu
ati
on
s.
(0, -
2):
-
2 =
c
(3,
0):
0 =
9a
+ 3
b +
c
(5,
2):
2 =
25a
+ 5
b +
c
Fir
st,
subst
itu
te -
2 f
or
c in
th
e s
eco
nd
an
d t
hir
d e
qu
ati
on
s.
Th
en
solv
e t
hose
tw
o e
qu
ati
on
s as
you
wou
ld a
ny s
yst
em
of
two e
qu
ati
on
s.
Mu
ltip
ly t
he s
eco
nd
equ
ati
on
by 5
an
d t
he t
hir
d e
qu
ati
on
by -
3.
0 =
9a
+ 3
b -
2
Multip
ly b
y 5
. 0 =
45a
+ 1
5b -
10
0 =
25
a +
5b
- 2
M
ultip
ly b
y -
3.
-6 =
-
75a
- 1
5b
+
6
-6 =
-
30
a
-
4
a =
1
−
15
To f
ind
b,
subst
itu
te
1 −
15 f
or
a i
n e
ith
er
the s
eco
nd
or
thir
d e
qu
ati
on
.
0 =
9 (
1 −
15 ) +
3b -
2
b =
7 −
15
Th
e e
qu
ati
on
of
a p
ara
bola
th
rou
gh
th
e t
hre
e p
oin
ts i
s
y =
1 −
15 x
2 +
7 −
15 x
- 2
.
Fin
d t
he e
qu
ati
on
of
a p
ara
bo
la t
hro
ug
h e
ach
set
of
three p
oin
ts.
1. (1
, 5),
(0,
6),
(2,
3)
2. (-
5,
0),
(0,
0),
(8,
100)
y
= -
1 −
2 x
2 -
1 −
2 x
+ 6
y =
25 −
26 x
2 +
125 −
26 x
3. (4
, -
4),
(0,
1),
(3, -
2)
4. (1
, 3),
(6,
0),
(0,
0)
y
= -
1 −
4 x
2 -
1 −
4 x
+ 1
y
= -
3 −
5 x
2 +
18 −
5 x
5. (2
, 2),
(5, -
3),
(0, -
1)
6. (0
, 4),
(4,
0),
(-
4,
4)
y
= -
19 −
30 x
2 +
83 −
30 x
- 1
y
= -
1 −
8 x
2 -
1 −
2 x
+ 4
9-2
En
rich
men
t
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-2
Ch
ap
ter
9
17
Gle
ncoe A
lgeb
ra 1
9-3
Tran
sla
tio
ns
A t
ra
nsla
tio
n i
s a c
han
ge i
n t
he p
osi
tion
of
a f
igu
re e
ith
er
up
, d
ow
n,
or
dia
gon
al.
Wh
en
a c
on
stan
t c
is a
dd
ed
to o
r su
btr
act
ed
fro
m t
he p
are
nt
fun
ctio
n,
the
resu
ltin
g f
un
ctio
n f
(x) ±
c i
s a t
ran
slati
on
of
the g
rap
h u
p o
r d
ow
n.
The g
raph o
f f(
x) =
x2 +
c t
ransla
tes t
he g
raph o
f f(
x) =
x2 v
ert
ically
.
If c
> 0
, th
e g
raph o
f f(
x) =
x2 is t
ransla
ted |
c|
units u
p.
If c
< 0
, th
e g
raph o
f f(
x) =
x2 is t
ransla
ted |
c|u
nits d
ow
n.
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Tra
nsfo
rmati
on
s o
f Q
uad
rati
c F
un
cti
on
s
f(x)
x
c<
0
c=
0c>
0
D
escrib
e h
ow
th
e g
ra
ph
of
ea
ch
fu
ncti
on
is r
ela
ted
to
th
e g
ra
ph
of
f(x) =
x2.
Exam
ple
a. g
(x) =
x2 +
4
Th
e f
un
ctio
n c
an
be w
ritt
en
as
f(x) =
x2 +
c.
Th
e v
alu
e o
f c
is 4
, an
d 4
> 0
. T
here
fore
, th
e
gra
ph
of
g(x
) =
x2 +
4 i
s a t
ran
slati
on
of
the
gra
ph
of
f(x) =
x2 u
p 4
un
its.
g(x
)
f(x)
x
b. h
(x) =
x2 -
3
Th
e f
un
ctio
n c
an
be w
ritt
en
as
f(x) =
x2 +
c.
Th
e v
alu
e o
f c
is –
3,
an
d –
3 <
0.
Th
ere
fore
, th
e g
rap
h o
f g(x
) =
x2 –
3 i
s a
tran
slati
on
of
the g
rap
h o
f f(
x) =
x2 d
ow
n
3 u
nit
s.y
h(x
)
f(x)
x
Exerc
ises
Describ
e h
ow
th
e g
ra
ph
of
ea
ch
fu
ncti
on
is r
ela
ted
to
th
e g
ra
ph
of f(x
) =
x2.
1. g(x
) =
x2 +
1
2.
h(x
) =
x2 –
6
3.
g(x
) =
x2 –
1
T
ran
sla
tio
n o
f
Tra
nsla
tio
n o
f
T
ran
sla
tio
n o
f f(
x) =
x2 u
p 1
un
it.
f
(x) =
x2 d
ow
n 6
un
its.
f(
x) =
x2 d
ow
n 1
un
it.
4. h
(x) =
20
+ x
2
5.
g(x
) =
–2 +
x2
6.
h(x
) = - 1 −
2 +
x2
T
ran
sla
tio
n o
f
T
ran
sla
tio
n o
f
T
ran
sla
tio
n o
f f(
x) =
x2 u
p 2
0 u
nit
s.
f(
x) =
x2 d
ow
n 2
un
its.
f(
x) =
x2 d
ow
n 1
−
2 u
nit
.
7. g(x
) =
x2 +
8 −
9
8. h
(x) =
x2 –
0.3
9.
g(x
) =
x2 –
4
T
ran
sla
tio
n o
f
T
ran
sla
tio
n o
f
T
ran
sla
tio
n o
f
f(x) =
x2 u
p 8
−
9 u
nit
.
f(x) =
x2 d
ow
n 0
.3 u
nit
.
f(x) =
x2 d
ow
n 4
un
its.
Answers (Lesson 9-2 and Lesson 9-3)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A8 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
18
Gle
ncoe A
lgeb
ra 1
Dilatio
ns a
nd
Refle
ctio
ns
A d
ila
tio
n i
s a t
ran
sform
ati
on
th
at
mak
es
the g
rap
h
narr
ow
er
or
wid
er
than
th
e p
are
nt
gra
ph
. A
refl
ecti
on
fli
ps
a f
igu
re o
ver
the x
- or
y-a
xis
.
The g
raph o
f f(
x)
= a
x2 s
tretc
hes o
r vert
ically
com
pre
sses t
he g
raph o
f f(
x)
= x
2.
If |
a|
> 1
, th
e g
raph o
f f(
x)
= x
2 is s
tretc
hed v
ert
ically
.
If 0
< |
a|
< 1
, th
e g
raph o
f f(
x)
= x
2 is c
om
pre
ssed v
ert
ically
.
The g
raph o
f th
e f
unction –
f(x)
flip
s t
he g
raph o
f f(
x)
= x
2 a
cro
ss t
he x
-axis
.
The g
raph o
f th
e f
unction f
(–x)
flip
s t
he g
raph o
f f(
x)
= x
2 a
cro
ss t
he y
-axis
.
D
escrib
e h
ow
th
e g
ra
ph
of
ea
ch
fu
ncti
on
is r
ela
ted
to
th
e g
ra
ph
of
f(x) =
x2.
9-3
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Tra
nsfo
rmati
on
s o
f Q
uad
rati
c F
un
cti
on
s
x
0<a<
1
a=
1
a>
1
x
y
f(x)=
x2
f(x)=
-x2
Exam
ple
a. g
(x) =
2x
2
Th
e f
un
ctio
n c
an
be w
ritt
en
as
f(x)
= a
x2
wh
ere
a =
2.
Beca
use
|a
| >
1,
the g
rap
h o
f y =
2x
2 i
s th
e g
rap
h
of
y =
x2 t
hat
is
stre
tch
ed
vert
icall
y.
b. h
(x) =
- 1
−
2 x
2 -
3
Th
e n
egati
ve s
ign
cau
ses
a r
efl
ect
ion
acr
oss
the x
-axis
. T
hen
a d
ilati
on
occ
urs
in
wh
ich
a =
1
−
2 a
nd
a t
ran
slati
on
in w
hic
h c
= –
3.
So t
he
gra
ph
of
y =
- 1
−
2 x
2 -
3
is r
efl
ect
ed
acr
oss
th
e
x-a
xis
, d
ilate
d w
ider
than
th
e g
rap
h o
f
f(x)
= x
2,
an
d t
ran
slate
d
dow
n 3
un
its.
Exerc
ises
Describ
e h
ow
th
e g
ra
ph
of
ea
ch
fu
ncti
on
is r
ela
ted
to
th
e g
ra
ph
of f(x
) =
x2.
1. h
(x)
= -
5x
2
2. g(x
) =
-x
2 +
1
3. g(x
) =
- 1
−
4 x
2 -
1
C
om
pre
ssio
n o
f
Tra
nsla
tio
n o
f y =
x2
D
ilati
on
of
y =
x2
y =
x2 n
arr
ow
er
than
re
flecte
d o
ver
the
w
ider
than
th
e g
rap
h o
f
the g
rap
h o
f f(
x) =
x2
x-a
xis
an
d u
p 1
un
it.
f(
x) =
x2 r
efl
ecte
d o
ver
refl
ecte
d o
ver
the
th
e x
-axis
tra
nsla
ted
x-a
xis
.
d
ow
n 1
un
it.
x
y
f(x)=
x2
f(x)=
2x2
x
f(x)
h(x
)
y
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-3
Ch
ap
ter
9
19
Gle
ncoe A
lgeb
ra 1
Describ
e h
ow
th
e g
ra
ph
of
ea
ch
fu
ncti
on
is r
ela
ted
to
th
e g
ra
ph
of f(x
) =
x2.
1. g(x
) =
x2 +
2
2. h
(x)
= -
1 +
x2
3. g(x
) =
x2 -
8
T
ran
sla
tio
n o
f y =
x2
Tra
nsla
tio
n o
f y =
x2
Tra
nsla
tio
n o
f y =
x2
up
2 u
nit
s
do
wn
1 u
nit
d
ow
n 8
un
its
4. h
(x)
= 7
x2
5. g(x
) =
1
−
5 x
2
6. h
(x)
= -
6x
2
C
om
pre
ssio
n o
f y =
x2
Dilati
on
of
y =
x2 w
ider
C
om
pre
ssio
n o
f
narr
ow
er
than
th
e g
rap
h
than
th
e g
rap
h o
f y
= x
2 n
arr
ow
er
than
o
f f(
x) =
x2
f(x
) =
x2
th
e g
rap
h o
f f(
x) =
x2
refl
ecte
d o
ver
the
x-a
xis
7. g(x
) =
-x
2 +
3
8. h
(x)
= 5
- 1
−
2 x
2
9. g(x
) =
4x
2 +
1
R
efl
ecti
on
of
y =
x2
Dilati
on
of
y =
x2 w
ider
Co
mp
ressio
n o
f o
ver
the x
-axis
th
an
th
e g
rap
h o
f f(
x) =
x2
y =
x2 n
arr
ow
er
tran
sla
ted
up
3 u
nit
s
refl
ecte
d o
ver
the
th
an
th
e g
rap
h o
f
x-a
xis
, an
d t
ran
sla
ted
f(x) =
x2 t
ran
sla
ted
u
p 5
un
its
u
p 1
un
it
Ma
tch
ea
ch
eq
ua
tio
n t
o i
ts g
ra
ph
.
10. y =
2x
2 -
2
11. y =
1
−
2 x
2 -
2
12. y =
- 1
−
2 x
2 +
2
13. y =
-2x
2 +
2
9-3
Sk
ills
Pra
ctic
e
Tra
nsfo
rmati
on
s o
f Q
uad
rati
c F
un
cti
on
s
C
B D A
x
y
x
y
x
y
x
yC
.
A.
D.
C.
Answers (Lesson 9-3)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A9 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
20
Gle
ncoe A
lgeb
ra 1
Describ
e h
ow
th
e g
ra
ph
of
ea
ch
fu
ncti
on
is r
ela
ted
to
th
e g
ra
ph
of f(x) =
x2.
1. g(x
) =
10 +
x2
2. h
(x) =
- 2
−
5 +
x2
3. g(x
) =
9 -
x2
T
ran
sla
tio
n o
f y =
x2
Tra
nsla
tio
n o
f y =
x2
Refl
ecti
on
of
u
p 1
0 u
nit
s.
do
wn
2 −
5 u
nit
. y
= x
2 a
cro
ss t
he
x-a
xis
tra
nsla
ted
up
9
un
its.
4. h
(x) =
2x
2 +
2
5. g(x
) =
- 3
−
4 x
2 -
1 −
2
6. h
(x) =
4 - 3
x2
C
om
pre
ssio
n o
f y =
x2
Dilati
on
of
y =
x2 w
ider
Co
mp
ressio
n o
f y =
x2
narr
ow
er
than
th
e
th
an
th
e g
rap
h o
f
narr
ow
er
than
th
e
gra
ph
of
f(x) =
x2
f(x
) =
x2,
refl
ecte
d o
ver
gra
ph
of
f(x) =
x2,
tran
sla
ted
up
2 u
nit
s.
th
e x
-axis
, tr
an
sla
ted
r
efl
ecte
d o
ver
the
do
wn
1 −
2 u
nit
. x
-axis
tra
nsla
ted
up
4 u
nit
s.
Ma
tch
ea
ch
eq
ua
tio
n t
o i
ts g
ra
ph
.
A.
x
y
B.
x
y
C.
x
y
7. y = -
3x
2 - 1
8. y =
- 1
−
3 x
2 +
1
9. y =
3x
2 + 1
9-3
Practi
ce
Tra
nsfo
rmati
on
s o
f Q
uad
rati
c F
un
cti
on
s
AC
B
Lis
t th
e f
un
cti
on
s i
n o
rd
er f
ro
m t
he m
ost
verti
ca
lly
str
etc
hed
to
th
e l
ea
st
verti
ca
lly
str
etc
hed
gra
ph
.
10. f(
x) =
3x
2,
g(x
) =
1 −
2 x
2,
h(x
) = -
2x
2
11. f(
x) =
1 −
2 x
2,
g(x
) =
- 1
−
6 , h
(x) =
4x
2
f
(x),
g(x
), h
(x)
h(x
), f
(x),
g(x
)
12. PA
RA
CH
UTIN
G T
wo p
ara
chu
tist
s ju
mp
fro
m t
wo d
iffe
ren
t p
lan
es
as
part
of
an
aeri
al
show
. T
he h
eig
ht
h1 o
f th
e f
irst
para
chu
tist
in
feet
aft
er
t se
con
ds
is m
od
ele
d b
y t
he f
un
ctio
n
h1 = -16t2
+ 5
000.
Th
e h
eig
ht
h2 o
f th
e s
eco
nd
para
chu
tist
in
feet
aft
er
t se
con
ds
is m
od
ele
d
by t
he f
un
ctio
n h
2 = -16t2
+ 4
000.
h =
t2
a.
Wh
at
is t
he p
are
nt
fun
ctio
n o
f th
e t
wo f
un
ctio
ns
giv
en
?
b.
Desc
ribe t
he t
ran
sform
ati
on
s n
eed
ed
to o
bta
in t
he g
rap
h o
f h
1 f
rom
th
e p
are
nt
fun
ctio
n.
Co
mp
ressio
n o
f y =
x2 n
arr
ow
er
than
th
e g
rap
h o
f f(
x) =
x2,
refl
ecte
d o
ver
the x
-axis
, tr
an
sla
ted
up
5000 u
nit
s.
c.
Wh
ich
para
chu
tist
wil
l re
ach
th
e g
rou
nd
fir
st?
the s
eco
nd
para
ch
uti
st
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-3
Ch
ap
ter
9
21
Gle
ncoe A
lgeb
ra 1
1. SPR
ING
S T
he p
ote
nti
al
en
erg
y s
tore
d i
n
a s
pri
ng i
s giv
en
by t
he f
un
ctio
n U
s =
1 −
2
kx
2 w
here
k i
s a c
on
stan
t k
now
n a
s th
e
spri
ng c
on
stan
t, a
nd
x i
s th
e d
ista
nce
th
e
spri
ng i
s st
retc
hed
or
com
pre
ssed
fro
m
its
init
ial
posi
tion
. E
xp
lain
how
th
e
gra
ph
of
the f
un
ctio
n f
or
a s
pri
ng w
here
k=
2 n
ew
ton
s/m
ete
r d
iffe
rs f
rom
th
e
gra
ph
of
the f
un
ctio
n f
or
a s
pri
ng w
here
k=
10 n
ew
ton
s/m
ete
r.
Th
e g
rap
h o
f U
s=
1 − 2 (
10)x
2 i
s a
c
om
pre
ssio
n o
f th
e g
rap
h.
2. C
YLIN
DER
S T
he v
olu
me o
f a c
yli
nd
er
is
giv
en
by t
he e
qu
ati
on
V=πr
2 ,
wh
ere
ris
th
e r
ad
ius
an
d
is t
he l
en
gth
. A
post
er
com
pan
y w
an
ts t
o i
ncr
ease
th
e v
olu
me o
f it
s 1-f
oot
lon
g s
hip
pin
g t
ube b
y 2
-cu
bic
fe
et
wit
hou
t in
creasi
ng t
he l
en
gth
. E
xp
lain
how
th
e g
rap
h o
f th
e o
rigin
al
tube d
iffe
rs f
rom
th
e g
rap
h o
f th
e n
ew
ly
red
esi
gn
ed
tu
be.
T
he g
rap
h o
f th
e n
ew
tu
be i
s a
tr
an
sla
tio
n d
ow
n 2
un
its.
3. PH
YSIC
S A
ball
is
dro
pp
ed
fro
m a
heig
ht
of
20 f
eet.
Th
e f
un
ctio
n h=-
16t2+
20
mod
els
th
e h
eig
ht
of
the b
all
in
feet
aft
er
t se
con
ds.
Gra
ph
th
e f
un
ctio
n a
nd
co
mp
are
th
is g
rap
h t
o t
he g
rap
h o
f it
s p
are
nt
fun
ctio
n.
Height (ft)3 069
12
15
18
21
24
Tim
e (s
)
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
9-3
Wo
rd
Pro
ble
m P
racti
ce
Tra
nsfo
rmati
on
s o
f Q
uad
rati
c F
un
cti
on
s
4. A
CC
ELER
ATIO
N T
he d
ista
nce
d i
n f
eet
a c
ar
acc
ele
rati
ng a
t 6 f
t/s2
tra
vels
aft
er
t se
con
ds
is m
od
ele
d b
y t
he f
un
ctio
n
d=
3t2
. S
up
pose
th
at
at
the s
am
e t
ime
the f
irst
car
begin
s acc
ele
rati
ng,
a s
eco
nd
ca
r begin
s acc
ele
rati
ng a
t 4 f
t/s2
exact
ly
100 f
eet
dow
n t
he r
oad
fro
m t
he f
irst
car.
T
he d
ista
nce
tra
vele
d b
y s
eco
nd
car
is
mod
ele
d b
y t
he f
un
ctio
n d=
2t2+
100.
a.
Gra
ph
an
d l
abel
each
fu
nct
ion
on
th
e
sam
e c
oord
inate
pla
ne.
Distance (ft)
100 0
200
300
400
500
600
700
800
Tim
e (s
)
24
68
10
12
14
16d=
2t2+
10
0
d=
3t2
b.
Exp
lain
how
each
gra
ph
is
rela
ted
to
the g
rap
h o
f f(
x) =
x2.
d=
3t2
is a
co
mp
ressio
n o
f d=
t2;
d=
2t2+
100 i
s a
co
mp
ressio
n o
f d=
t2
tran
sla
ted
up
100 u
nit
s (
feet)
.
c.
Aft
er
how
man
y s
eco
nd
s w
ill
the f
irst
ca
r p
ass
th
e s
eco
nd
car?
10 s
eco
nd
s
h =
-16t2
+ 2
0 is a
co
mp
ressio
n
of
h =
t2 r
efl
ecte
d o
ver
the x
-axis
an
d t
ran
sla
ted
up
20 u
nit
s (
feet)
.
Answers (Lesson 9-3)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A10 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
22
Gle
ncoe A
lgeb
ra 1
Tra
nsla
tin
g Q
uad
rati
c G
rap
hs
Wh
en
a f
igu
re i
s m
oved
to a
new
posi
tion
wit
hou
t
un
derg
oin
g a
ny r
ota
tion
, th
en
th
e f
igu
re i
s sa
id t
o
have b
een
tra
nsla
ted
to t
hat
posi
tion
.
Th
e g
rap
h o
f a q
uad
rati
c equ
ati
on
in
th
e f
orm
y =
(x -
b)2
+ c
is
a t
ran
slati
on
of
the g
rap
h o
f y =
x2.
Sta
rt w
ith
y =
x2.
Sli
de t
o t
he r
igh
t 4 u
nit
s.
y =
(x -
4)2
Th
en
sli
de u
p 3
un
its.
y =
(x -
4)2
+ 3
Th
ese e
qu
ati
on
s h
av
e t
he f
orm
y =
(x -
b)2
. G
ra
ph
ea
ch
eq
ua
tio
n.
1. y =
(x -
1)2
2. y =
(x -
3)2
3. y =
(x +
2)2
x
y O
x
y O
x
y O
Th
ese e
qu
ati
on
s h
av
e t
he f
orm
y =
(x -
b)2
+ c
. G
ra
ph
ea
ch
eq
ua
tio
n.
4. y =
(x -
2)2
+ 1
5. y =
(x -
1)2
+ 2
6. y =
(x +
1)2
- 2
x
y O
x
y O
x
y O
x
y O
9-3
En
rich
men
t
Lesson X-4
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-4
Ch
ap
ter
9
23
Gle
ncoe A
lgeb
ra 1
Co
mp
lete
th
e S
qu
are
P
erf
ect
squ
are
tri
nom
ials
can
be s
olv
ed
qu
ick
ly b
y t
ak
ing t
he
squ
are
root
of
both
sid
es
of
the e
qu
ati
on
. A
qu
ad
rati
c equ
ati
on
th
at
is n
ot
in p
erf
ect
squ
are
form
can
be m
ad
e i
nto
a p
erf
ect
squ
are
by a
meth
od
call
ed
co
mp
leti
ng
th
e s
qu
are
.
Co
mp
leti
ng
th
e S
qu
are
To c
om
ple
te t
he s
quare
for
any q
uadra
tic e
quation o
f th
e f
orm
x2 +
bx:
Ste
p 1
Fin
d o
ne-h
alf o
f b
, th
e c
oeff
icie
nt
of
x.
Ste
p 2
Square
the r
esult in S
tep 1
.
Ste
p 3
Add t
he r
esult o
f S
tep 2
to x
2 +
bx.
x2
+ b
x +
( b −
2 ) 2
= (x
+ b
−
2 ) 2
F
ind
th
e v
alu
e o
f c t
ha
t m
ak
es x
2 +
2x +
c a
perfe
ct
sq
ua
re t
rin
om
ial.
Ste
p 1
F
ind
1
−
2 o
f 2.
2
−
2 =
1
Ste
p 2
S
qu
are
th
e r
esu
lt o
f S
tep
1.
12
= 1
Ste
p 3
A
dd
th
e r
esu
lt o
f S
tep
2 t
o x
2 +
2x.
x2 +
2x +
1.
Th
us,
c =
1.
Noti
ce t
hat
x2
+ 2
x +
1 e
qu
als
(x +
1)2
Exerc
ises
Fin
d t
he v
alu
e o
f c t
ha
t m
ak
es e
ach
trin
om
ial
a p
erfe
ct
sq
ua
re.
1. x
2 +
10
x +
c 25
2. x
2 +
14
x +
c 49
3. x
2 -
4x +
c 4
4. x
2 -
8x +
c 36
5. x
2 +
5x +
c 2
5 −
4
6. x
2 +
9x +
c 8
1 −
4
7. x
2 -
3x +
c 9 −
4
8. x
2 -
15
x +
c 2
25 −
4
9. x
2 +
28
x +
c 196
10. x
+ 2
2x +
c 121
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y C
om
ple
tin
g t
he S
qu
are
Exam
ple
9-4
Answers (Lesson 9-3 and Lesson 9-4)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A11 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
24
Gle
ncoe A
lgeb
ra 1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n(continued)
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y C
om
ple
tin
g t
he S
qu
are
So
lve b
y C
om
ple
tin
g t
he S
qu
are
S
ince
few
qu
ad
rati
c exp
ress
ion
s are
perf
ect
sq
uare
tri
nom
ials
, th
e m
eth
od
of
co
mp
leti
ng
th
e s
qu
are
can
be u
sed
to s
olv
e s
om
e
qu
ad
rati
c equ
ati
on
s. U
se t
he f
oll
ow
ing s
tep
s to
com
ple
te t
he s
qu
are
for
a q
uad
rati
c exp
ress
ion
of
the f
orm
ax
2 +
bx.
S
olv
e x
2 +
6x
+ 3
= 1
0 b
y c
om
ple
tin
g t
he s
qu
are.
x2 +
6x +
3 =
10
O
rigin
al equation
x
2 +
6x +
3 -
3 =
10 -
3
Subtr
act
3 f
rom
each s
ide.
x2 +
6x =
7
Sim
plif
y.
x2 +
6x +
9 =
7 +
9
Sin
ce ( 6
−
2 ) 2
= 9
, add 9
to e
ach s
ide.
(x
+ 3
)2 =
16
Facto
r x
2 +
6x +
9.
x +
3 =
±4
Take t
he s
quare
root
of
each s
ide.
x =
-3 +
4
Sim
plif
y.
x =
-3 +
4 or
x =
-3 -
4
=
1
= -
7
Th
e s
olu
tion
set
is {-
7,
1}.
Exerc
ises
So
lve e
ach
eq
ua
tio
n b
y c
om
ple
tin
g t
he s
qu
are.
Ro
un
d t
o t
he n
ea
rest
ten
th i
f
necessa
ry
.
1. x
2 -
4x +
3 =
0
2. x
2 +
10x =
-9
3. x
2 -
8x -
9 =
0
1
, 3
-1, -
9
-1,
9
4. x
2 -
6x =
16
5. x
2 -
4x -
5 =
0
6. x
2 -
12x =
9
-
2,
8
-1,
5
-0.7
, 12.7
7. x
2 +
8x =
20
8. x
2 =
2x +
1
9. x
2 +
20x +
11 =
-8
-
10,
2
-0.4
, 2.4
-
19, -
1
10. x
2 -
1 =
5x
11. x
2 =
22x +
23
12. x
2 -
8x =
-7
-
0.2
, 5.2
-
1,
23
1,
7
13. x
2 +
10x =
24
14. x
2 -
18x =
19
15. x
2 +
16x =
-16
-
12,
2
-1,
19
-14.9
, -
1.1
16. 4x
2 =
24
+ 4x
17. 2x
2 +
4x +
2 =
8
18. 4x
2 =
40x +
44
-
2,
3
-3,
1
-1,
11
Ste
p 1
Fin
d b
−
2 .
Ste
p 2
Fin
d ( b
−
2 ) 2
.
Ste
p 3
Add ( b
−
2 ) 2
to a
x2 +
bx.
9-4 Exam
ple
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-4
Ch
ap
ter
9
25
Gle
ncoe A
lgeb
ra 1
Sk
ills
Pra
ctic
e
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y C
om
ple
tin
g t
he S
qu
are
Fin
d t
he v
alu
e o
f c t
ha
t m
ak
es e
ach
trin
om
ial
a p
erfe
ct
sq
ua
re.
1. x
2 +
6x +
c 9
2. x
2 +
4x +
c 4
3. x
2 -
14x +
c 49
4. x
2 -
2x +
c 1
5. x
2 -
18x +
c 81
6. x
2 +
20x +
c 100
7. x
2 +
5x +
c 6.2
5
8. x
2 -
70x +
c 1225
9. x
2 -
11x +
c 30.2
5
10. x
2 +
9x +
c 20.2
5
So
lve e
ach
eq
ua
tio
n b
y c
om
ple
tin
g t
he s
qu
are.
Ro
un
d t
o t
he n
ea
rest
ten
th i
f n
ecessa
ry
.
11. x
2 +
4x -
12 =
0 2, -
6
12. x
2 -
8x +
15 =
0 3,
5
13. x
2 +
6x =
7 -
7,
1
14. x
2 -
2x =
15 -
3,
5
15. x
2 -
14x +
30
= 6
2,
12
16. x
2 +
12x +
21
= 1
0 -
11, -
1
17. x
2 -
4x +
1 =
0 0.3
, 3.7
18. x
2 -
6x +
4 =
0 0.8
, 5.2
19. x
2 -
8x +
10 =
0 1.6
, 6.4
20. x
2 -
2x =
5 -
1.4
, 3.4
21. 2x
2 +
20x =
-2 -
9.9
, -
0.1
22. 0.5x
2 +
8x =
-7 -
15.1
, -
0.9
9-4
Answers (Lesson 9-4)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A12 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
26
Gle
ncoe A
lgeb
ra 1
Practi
ce
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y C
om
ple
tin
g t
he S
qu
are
Fin
d t
he v
alu
e o
f c t
ha
t m
ak
es e
ach
trin
om
ial
a p
erfe
ct
sq
ua
re.
1. x
2 -
24x +
c 144
2. x
2 +
28x +
c 196
3. x
2 +
40x +
c 400
4. x
2 +
3x +
c 9
−
4
5. x
2 -
9x +
c 8
1
−
4
6. x
2 -
x +
c 1
−
4
So
lve e
ach
eq
ua
tio
n b
y c
om
ple
tin
g t
he s
qu
are.
Ro
un
d t
o t
he n
ea
rest
ten
th i
f
necessa
ry
.
7. x
2 -
14x +
24
= 0
8. x
2 +
12x =
13
9. x
2 -
30x +
56 =
-25
2
, 12
-13,
1
3,
27
10. x
2 +
8x +
9 =
0
11. x
2 -
10x +
6 =
-7
12. x
2 +
18x +
50 =
9
-
6.6
, -
1.4
1
.5,
8.5
-
15.3
, -
2.7
13. 3x
2 +
15x -
3 =
0
14. 4x
2 -
72 =
24x
15. 0.9x
2 +
5.4x -
4 =
0
-
5.2
, 0.2
-
2.2
, 8.2
-
6 2
−
3 ,
2
−
3
16. 0.4x
2 +
0.8x =
0.2
17. 1
−
2 x
2 -
x -
10
= 0
1
8.
1 −
4 x
2 +
x -
2 =
0
-
2.2
, 0.2
-
4,
5
-7.1
, 1.1
19. N
UM
BER
TH
EO
RY
T
he p
rod
uct
of
two c
on
secu
tive e
ven
in
tegers
is
728.
Fin
d
the i
nte
gers
.
20. B
USIN
ESS
Jaim
e o
wn
s a b
usi
ness
mak
ing d
eco
rati
ve b
oxes
to s
tore
jew
elr
y,
mem
en
tos,
an
d o
ther
valu
able
s. T
he f
un
ctio
n y
= x
2 +
50x +
1800 m
od
els
th
e p
rofi
t y t
hat
Jaim
e
has
mad
e i
n m
on
th x
for
the f
irst
tw
o y
ears
of
his
bu
sin
ess
.
a. W
rite
an
equ
ati
on
rep
rese
nti
ng t
he m
on
th i
n w
hic
h J
aim
e’s
pro
fit
is $
2400.
x
2 +
50x +
1800 =
2400
b. U
se c
om
ple
tin
g t
he s
qu
are
to f
ind
ou
t in
wh
ich
mon
th J
aim
e’s
pro
fit
is $
2400.
t
he t
en
th m
on
th
21. PH
YSIC
S F
rom
a h
eig
ht
of
256 f
eet
above a
lak
e o
n a
cli
ff,
Mik
aela
th
row
s a r
ock
ou
t over
the l
ak
e.
Th
e h
eig
ht H
of
the r
ock
t s
eco
nd
s aft
er
Mik
aela
th
row
s it
is
rep
rese
nte
d
by t
he e
qu
ati
on
H =
-16t2
+ 3
2t +
256.
To t
he n
eare
st t
en
th o
f a s
eco
nd
, h
ow
lon
g d
oes
it t
ak
e t
he r
ock
to r
each
th
e l
ak
e b
elo
w?
(Hint:
Rep
lace
H w
ith
0.)
5.1
s
9-4 2
6, 28
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-4
Ch
ap
ter
9
27
Gle
ncoe A
lgeb
ra 1
9-4
Wo
rd
Pro
ble
m P
racti
ce
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y C
om
ple
tin
g t
he S
qu
are
1.IN
TER
IOR
DESIG
N M
od
ula
r ca
rpeti
ng i
s in
stall
ed
in
sm
all
pie
ces
rath
er
than
as
a
larg
e r
oll
so t
hat
on
ly a
few
pie
ces
need
to
be r
ep
lace
d i
f a s
mall
are
a i
s d
am
aged
. S
up
pose
th
e r
oom
sh
ow
n i
n t
he d
iagra
m
belo
w i
s bein
g f
itte
d w
ith
mod
ula
r ca
rpeti
ng.
Com
ple
te t
he s
qu
are
to
dete
rmin
e t
he n
um
ber
of
1 f
t by 1
ft
squ
are
s of
carp
eti
ng n
eed
ed
to f
inis
h t
he
room
. F
ill
in t
he m
issi
ng t
erm
s in
th
e
corr
esp
on
din
g e
qu
ati
on
belo
w.
25;
5
x
xxxxx
xx
xx
x2
x
2 +
10x +
_____ =
(x +
_____)2
2. FA
LLIN
G O
BJEC
TS
K
eis
ha t
hro
ws
a
rock
dow
n a
n o
ld w
ell
. T
he d
ista
nce
d
(in
feet)
th
e r
ock
fall
s aft
er t
seco
nd
s ca
n b
e r
ep
rese
nte
d b
y t
he e
qu
ati
on
d
= 1
6t2
+ 6
4t.
If
the w
ate
r in
th
e w
ell
is
80 f
eet
belo
w g
rou
nd
, h
ow
man
y s
eco
nd
s w
ill
it t
ak
e f
or
the r
ock
to h
it t
he w
ate
r?
1 s
eco
nd
3. M
AR
S O
n M
ars
, th
e g
ravit
y a
ctin
g o
n
an
obje
ct i
s le
ss t
han
th
at
on
Eart
h.
On
E
art
h,
a g
olf
ball
hit
wit
h a
n i
nit
ial
up
ward
velo
city
of
26 m
ete
rs p
er
seco
nd
w
ill
reach
a m
axim
um
heig
ht
of
abou
t 34.5
mete
rs.
Th
e h
eig
ht h
of
an
obje
ct o
n
Mars
th
at
leaves
the g
rou
nd
wit
h a
n
init
ial
velo
city
of
26 m
ete
rs p
er
seco
nd
is
giv
en
by t
he e
qu
ati
on
h =
-1.9t2
+ 2
6t.
F
ind
th
e m
axim
um
heig
ht
if t
he s
am
e
golf
ball
is
hit
on
Mars
. R
ou
nd
you
r an
swer
to t
he n
eare
st t
en
th.
88.9
m
4.FR
OG
S A
fro
g s
itti
ng o
n a
stu
mp
3 f
eet
hig
h h
op
s off
an
d l
an
ds
on
th
e g
rou
nd
. D
uri
ng i
ts l
eap
, it
s h
eig
ht h
(in
feet)
is
giv
en
by h=-
0.5d
2+
2d+
3,
wh
ere
d i
s th
e d
ista
nce
fro
m t
he b
ase
of
the s
tum
p.
How
far
is t
he f
rog f
rom
th
e b
ase
of
the
stu
mp
wh
en
it
lan
ded
on
th
e g
rou
nd
?
2 +
√ "
"
10 o
r ab
ou
t 5.1
6 f
t
5. G
AR
DEN
ING
P
eg i
s p
lan
nin
g a
re
ctan
gu
lar
vegeta
ble
gard
en
usi
ng 2
50
feet
of
fen
cin
g m
ate
rial.
Sh
e o
nly
need
s to
fen
ce t
hre
e s
ides
of
the g
ard
en
sin
ce
on
e s
ide b
ord
ers
an
exis
tin
g f
en
ce.
x
a.L
et x=
th
e w
idth
of
the r
ect
an
gle
. W
rite
an
exp
ress
ion
to r
ep
rese
nt
the a
rea o
f th
e g
ard
en
if
she u
ses
all
th
e f
en
cin
g
mate
rial.
x(2
50 -
2x)
b. F
ind
th
e v
ert
ex o
f th
e e
qu
ati
on
an
d
iden
tify
it
as
a m
axim
um
or
a m
inim
um
.
(62.5
, 7812.5
); m
axim
um
c.In
terp
ret
the v
ert
ex o
f th
e e
qu
ati
on
in
te
rms
of
the s
itu
ati
on
.
If t
he s
ho
rt s
ide i
s 6
2.5
ft
an
d t
he
oth
er
sid
e 1
25 f
t, t
he g
ard
en
will
be t
he l
arg
est
po
ssib
le,
wit
h a
n
are
a o
f 7812.5
ft2
.
Answers (Lesson 9-4)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A13 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
28
Gle
ncoe A
lgeb
ra 1
Com
ple
tin
g t
he s
qu
are
is
a u
sefu
l to
ol
for
fact
ori
ng a
nd
solv
ing q
uad
rati
c
exp
ress
ion
s. Y
ou
can
uti
lize a
sim
ilar
tech
niq
ue t
o f
act
or
sim
ple
qu
arti
c p
oly
no
mia
ls o
f
the f
orm
x4 +
c.
Fa
cto
r t
he q
ua
rti
c p
oly
no
mia
l x
4 +
64.
Ste
p 1
F
ind
th
e v
alu
e o
f th
e m
idd
le t
erm
need
ed
to c
om
ple
te t
he s
qu
are
.
Th
is v
alu
e i
s ( 2
√ ""
64 ) (
x2) ,
or
16x
2.
Ste
p 2
R
ew
rite
th
e o
rigin
al
poly
nom
ial
in f
act
ora
ble
form
:
(x4
+ 1
6x
2 + ( 1
6 −
2 ) 2
) - 1
6x
2.
Ste
p 3
F
act
or
the p
oly
nom
ials
: ( x
2 +
8) 2
- (
4x)2
.
Ste
p 4
R
ew
rite
usi
ng t
he d
iffe
ren
ce o
f tw
o s
qu
are
s:
(x2 +
8 +
4x)
(x2 +
8 -
4x)
Th
e f
act
ore
d f
orm
of
x2 +
64 i
s ( x
2 + 4
x +
8)
( x2 - 4
x +
8) .
Th
is c
ou
ld t
hen
be f
act
ore
d
furt
her,
if
need
ed
, to
fin
d t
he s
olu
tion
s to
a q
uart
ic e
qu
ati
on
.
Exerc
ises
Fa
cto
r e
ach
qu
arti
c p
oly
no
mia
l.
1. x
4 +
4
2. x
4 +
32
4
(
x2 +
2x +
2)(
x2 -
2x +
2)
(x2
+ 6
x +
18)(
x2 -
6x
+ 1
8)
3. x
4 +
2500
4. x
4 +
9604
(
x2 +
10x
+ 5
0)(
x2 -
10x
+ 5
0)
(x2
+ 1
4x
+ 9
8)(
x2 -
14x
+ 9
8)
5. x
4 +
1024
6. x
4 +
5184
(x2 +
8x
+ 3
2)(
x2 -
8x
+ 3
2)
(x2
+ 1
2x
+ 7
2)(
x2 -
12x
+ 7
2)
7. x
4 +
484
8. x
4 +
9
(
x2 +
x
√ "
44 +
22)(
x2 -
x
√ "
44 +
22)
(x2
+ x
√ " 6
+ 9
)(x2
- x
√ " 6
+ 9
)
9. x
4 +
144
10.
x8 + 1
6,3
84
(
x2 +
2x
√ " 6
+ 1
2)(
x2 -
2x
√ " 6
+ 1
2)
(x4
+ 1
6x2
+ 1
28)(
x4 -
16x2
+ 1
28)
11.
Fact
or
x4 +
c t
o c
om
e u
p w
ith
a g
en
era
l ru
le f
or
fact
ori
ng q
uart
ic p
oly
nom
ials
.
(
x2 +
x √
""
2 √ "
c +
√ "
c )
+ ( x
2 +
x √
""
2 √ "
c +
√ "
c )
9-4
En
rich
men
t
Facto
rin
g Q
uart
ic P
oly
no
mia
ls
Exam
ple
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-5
Ch
ap
ter
9
29
Gle
ncoe A
lgeb
ra 1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
So
lvin
g Q
uad
ratic E
qu
atio
ns
by
Usi
ng
th
e Q
uad
ratic F
orm
ula
Qu
ad
rati
c F
orm
ula
T
o s
olv
e t
he s
tan
dard
form
of
the q
uad
rati
c equ
ati
on
, a
x 2 +
bx +
c =
0,
use
th
e Q
ua
dra
tic F
orm
ula
.
Qu
ad
rati
c F
orm
ula
the f
orm
ula
x =
-b
± √ """"
b2 -
4ac
−
2a
th
at
giv
es the s
olu
tions o
f ax
2 +
bx +
c =
0, w
here
a ≠
0
S
olv
e x
2 +
2x
= 3
by
usin
g t
he Q
ua
dra
tic F
orm
ula
.
Rew
rite
th
e e
qu
ati
on
in
sta
nd
ard
form
.
x
2 +
2x =
3
Origin
al equation
x
2 +
2x -
3 =
3 -
3
Subtr
act
3 f
rom
each s
ide.
x
2 +
2x -
3 =
0
Sim
plif
y.
Now
let
a =
1,
b =
2,
an
d c
= -
3 i
n t
he
Qu
ad
rati
c F
orm
ula
.
x = -
b ±
b2 -
4a
c −
2a
−
2a
= -
2 ±
√ """"""
(2)2
- 4
(1)(-
3)
−−
2(1
)
=
-2 +
√ ""
16 −
2
x =
-
2 +
4 −
2
or
x =
-
2 +
4 −
2
=
1
= -
3
Th
e s
olu
tion
set
is {-
3,
1}.
S
olv
e x
2 -
6x
- 2
= 0
by
usin
g t
he Q
ua
dra
tic F
orm
ula
. R
ou
nd
to
the n
ea
rest
ten
th i
f n
ecessa
ry
.
For
this
equ
ati
on
a =
1,
b =
-6,
an
d c
= -
2.
x =
-b
± √ """"
b 2 -
4a
c
−
2a
=
6
± √ """""""
(-6) 2
- 4
(1)(-
2)
−−
2(1
)
=
6
± √ ""
44 −
2
x = 6 ±
√ ""
44 −
2
or
x =
6
- √ ""
44 −
2
x ≈
6.3
≈
-0.3
Th
e s
olu
tion
set
is {-
0.3
, 6.3
}.
Exerc
ises
So
lve e
ach
eq
ua
tio
n b
y u
sin
g t
he Q
ua
dra
tic F
orm
ula
. R
ou
nd
to
th
e
nea
rest
ten
th i
f n
ecessa
ry
.
1. x
2 -
3x +
2 =
0 1,
2
2. x
2 -
8x =
-16 4
3. 16x
2 -
8x =
-1 1
−
4
4. x
2 +
5x =
6
-6,
1
5. 3x
2 +
2x =
8
-2,
4
−
3
6. 8
x2 -
8x -
5 =
0
-0.4
, 1.4
7. -
4x
2 +
19x =
21 7
−
4 ,
3
8. 2x
2 +
6x =
5
-3.7
, 0.7
9. 48x
2 +
22
x -
15
= 0
-
5
−
6 ,
3
−
8
10. 8
x2 -
4x =
24
- 3
−
2 ,
2
11. 2x
2 +
5x =
8
-3.6
, 1.1
12. 8
x2 +
9x -
4 =
0
-1.5
, 0.3
13. 2x
2 +
9x +
4 =
0
-4,
- 1
−
2
14. 8
x2 +
17x +
2 =
0
-2,
- 1
−
8
9-5
Exam
ple
1Exam
ple
2
Answers (Lesson 9-4 and Lesson 9-5)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A14 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
30
Gle
ncoe A
lgeb
ra 1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n(c
on
tin
ued
)
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y U
sin
g t
he Q
uad
rati
c F
orm
ula
Th
e D
iscrim
inan
t
In t
he Q
uad
rati
c F
orm
ula
, x =
-b
± √ ####
b 2 -
4a
c
−
2a
, t
he e
xp
ress
ion
un
der
the r
ad
ical
sign
, b
2 -
4a
c,
is c
all
ed
th
e d
iscrim
ina
nt.
Th
e d
iscr
imin
an
t ca
n b
e u
sed
to
dete
rmin
e t
he n
um
ber
of
real
roots
for
a q
uad
rati
c equ
ati
on
.
Case 1
: b
2 -
4ac <
0no r
eal ro
ots
Case 2
: b
2 -
4ac =
0one r
eal ro
ot
Case 3
: b
2 -
4ac >
0tw
o r
eal ro
ots
S
tate
th
e v
alu
e o
f th
e d
iscrim
ina
nt
for e
ach
eq
ua
tio
n.
Th
en
dete
rm
ine t
he n
um
ber o
f rea
l so
luti
on
s o
f th
e e
qu
ati
on
.
a.
12x
2 +
5x =
4
Wri
te t
he e
qu
ati
on
in
sta
nd
ard
form
.
12
x2 +
5x =
4
Origin
al equation
12x
2 +
5x -
4 =
4 -
4
Subtr
act
4 f
rom
each s
ide.
12x
2 +
5x -
4 =
0
Sim
plif
y.
Now
fin
d t
he d
iscr
imin
an
t.
b2 -
4a
c =
(5)2
- 4
(12)(-
4)
=
217
Sin
ce t
he d
iscr
imin
an
t is
posi
tive,
the
equ
ati
on
has
two r
eal
roots
.
b.
2x
2 +
3x
= -
4
2
x2 +
3x =
-4
Origin
al equation
2x
2 +
3x +
4 =
-4 +
4
Add 4
to e
ach s
ide.
2x
2 +
3x +
4 =
0
Sim
plif
y.
Fin
d t
he d
iscr
imin
an
t.
b2 -
4a
c
= (
3)2
- 4
(2)(
4)
=
-23
Sin
ce t
he d
iscr
imin
an
t is
negati
ve,
the
equ
ati
on
has
no r
eal
roots
.
Exerc
ises
Sta
te t
he v
alu
e o
f th
e d
iscrim
ina
nt
for e
ach
eq
ua
tio
n.
Th
en
dete
rm
ine t
he n
um
ber
of
rea
l so
luti
on
s o
f th
e e
qu
ati
on
.
1. 3
x2 +
2x -
3 =
0
2. 3x
2 -
7x -
8 =
0
3. 2
x2 -
10x -
9 =
0
40,
2 r
eal
roo
ts
145,
2 r
eal
so
luti
on
s
172,
2 r
eal
so
luti
on
s
4. 4
x2 =
x +
4
5. 3x
2 -
13x =
10
6. 6
x2 -
10x +
10 =
0
65,
2 r
eal
so
luti
on
s
289,
2 r
eal
so
luti
on
s
7. 2
x2 -
20
= -
x
8. 6x
2 =
-11x -
40
9. 9 -
18x +
9x
2 =
0
161,
2 r
eal
so
luti
on
s
-
839,
no
real
so
luti
on
s
0,
1 r
eal
so
luti
on
10. 12x
2 +
9 =
-6x
11. 9x
2 =
81
12. 16
x2 +
16
x +
4 =
0
-
396,
no
real
so
luti
on
s
2916,
2 r
eal
so
luti
on
s
0,
1 r
eal
so
luti
on
s
13. 8x
2 +
9x =
2
14. 4x
2 -
4x +
4 =
3
15. 3
x2 -
18x =
- 1
4
145,
2 r
eal
so
luti
on
s
0,
1 r
eal
so
luti
on
156,
2 r
eal
so
luti
on
s
9-5 Exam
ple
-140, n
o r
eal so
luti
on
s
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-5
Ch
ap
ter
9
31
Gle
ncoe A
lgeb
ra 1
Sk
ills
Pra
ctic
e
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y U
sin
g t
he Q
uad
rati
c F
orm
ula
So
lve e
ach
eq
ua
tio
n b
y u
sin
g t
he Q
ua
dra
tic F
orm
ula
. R
ou
nd
to
th
e n
ea
rest
ten
th
if n
ecessa
ry
.
1. x
2 -
49
= 0
-
7,
7
2. x
2 -
x -
20 =
0 -
4,
5
3. x
2 -
5x -
36
= 0
-
4,
9
4. x
2 +
11
x +
30 =
0 -
6, -
5
5. x
2 -
7x =
-3 0.5
, 6.5
6. x
2 +
4x =
-1 -
3.7
, -
0.3
7. x
2 -
9x +
22 =
0
8. x
2 +
6x +
3 =
0 -
5.4
, -
0.6
9. 2
x2 +
5x -
7 =
0 - 3
1 −
2 ,
1
10. 2
x2 -
3x =
-1 1
−
2 ,
1
11. 2x
2 +
5x +
4 =
0
12. 2
x2 +
7x =
9 -
4 1
−
2 ,
1
13. 3x
2 +
2x -
3 =
0 -
1.4
, 0.7
14. 3
x2 -
7x -
6 =
0 - 2
−
3 ,
3
Sta
te t
he v
alu
e o
f th
e d
iscrim
ina
nt
for e
ach
eq
ua
tio
n.
Th
en
dete
rm
ine t
he n
um
ber
of
rea
l so
luti
on
s o
f th
e e
qu
ati
on
.
15. x
2 +
4x +
3 =
0
16. x
2 +
2x +
1 =
0
4;
2 r
eal
so
luti
on
s
0;
1 r
eal
so
luti
on
17. x
2 -
4x +
10
= 0
18. x
2 -
6x +
7 =
0
-
24;
no
real
so
luti
on
s
8;
2 r
eal
so
luti
on
s
19. x
2 -
2x -
7 =
0
20. x
2 -
10
x +
25 =
0
32;
2 r
eal
so
luti
on
s
0;
1 r
eal
so
luti
on
21. 2x
2 +
5x -
8 =
0
22. 2
x2 +
6x +
12 =
0
89;
2 r
eal
so
luti
on
s
-
60;
no
real
so
luti
on
s
23. 2x
2 -
4x +
10 =
0
24. 3
x2 +
7x +
3 =
0
-
64;
no
real
so
luti
on
s
13;
2 r
eal
so
luti
on
s
9-5
Answers (Lesson 9-5)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A15 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
32
Gle
ncoe A
lgeb
ra 1
9-5
Practi
ce
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y U
sin
g t
he Q
uad
rati
c F
orm
ula
So
lve e
ach
eq
ua
tio
n b
y u
sin
g t
he Q
ua
dra
tic F
orm
ula
. R
ou
nd
to
th
e n
ea
rest
ten
th
if n
ecessa
ry
.
1. x
2 +
2x -
3 =
0 -
3,
1
2. x
2 +
8x +
7 =
0 -
7, -
1
3. x
2 -
4x +
6 =
0
4. x
2 -
6x +
7 =
0 1.6
, 4.4
5. 2x
2 +
9x -
5 =
0 -
5, 1
−
2
6. 2x
2 +
12x +
10 =
0 -
5, -
1
7. 2
x2 -
9x =
-12
8. 2
x2 -
5x =
12 -
1 1
−
2 ,
4
9. 3
x2 +
x =
4 -
1 1
−
3 ,
1
10. 3
x2 -
1 =
-8
x -
2.8
, 0.1
11. 4
x2 +
7x =
15 -
3,
1 1
−
4
12. 1.6
x2 +
2x +
2.5
= 0
13. 4.5
x2 +
4x -
1.5
= 0
14.
1 −
2 x
2 +
2x +
3 −
2 =
0
15. 3
x2 -
3 −
4 x
= 1
−
2
-
1.2
, 0.3
-
3, -
1
-0.3
, 0.6
Sta
te t
he v
alu
e o
f th
e d
iscrim
ina
nt
for e
ach
eq
ua
tio
n.
Th
en
dete
rm
ine t
he n
um
ber
of
rea
l so
luti
on
s o
f th
e e
qu
ati
on
.
16. x
2 +
8x +
16 =
0
17. x
2 +
3x +
12
= 0
18. 2
x2 +
12x =
-7
0
; 1 r
eal
so
luti
on
-
39;
no
real
so
luti
on
s
88;
2 r
eal
so
luti
on
s
19. 2
x2 +
15
x =
-30
20. 4x
2 +
9 =
12
x
21. 3
x2 -
2x =
3.5
-
15;
no
real
so
luti
on
0
; 1 r
eal
so
luti
on
s
46;
2 r
eal
so
luti
on
s
22. 2.5
x2 +
3x -
0.5
= 0
23.
3 −
4 x
2 -
3x =
-4
24.
1 −
4 x
2 =
-x -
1
1
4;
2 r
eal
so
luti
on
s
-3;
no
real
so
luti
on
s
0;
1 r
eal
so
luti
on
25. C
ON
STR
UC
TIO
N A
roofe
r to
sses
a p
iece
of
roofi
ng t
ile f
rom
a r
oof
on
to t
he g
rou
nd
30
feet
belo
w.
He t
oss
es
the t
ile w
ith
an
in
itia
l d
ow
nw
ard
velo
city
of
10 f
eet
per
seco
nd
.
a. W
rite
an
equ
ati
on
to f
ind
how
lon
g i
t ta
kes
the t
ile t
o h
it t
he g
rou
nd
. U
se t
he m
od
el
for
vert
ical
moti
on
, H
= -
16t2
+ v
t +
h,
wh
ere
H i
s th
e h
eig
ht
of
an
obje
ct a
fter
t se
con
ds,
v i
s th
e i
nit
ial
velo
city
, an
d h
is
the i
nit
ial
heig
ht.
(H
int:
Sin
ce t
he o
bje
ct i
s th
row
n
dow
n,
the i
nit
ial
velo
city
is
negati
ve.)
H
= -
16
t2 -
10t +
30
b.
How
lon
g d
oes
it t
ak
e t
he t
ile t
o h
it t
he g
rou
nd
? ab
ou
t 1.1
s
26. PH
YSIC
S L
up
e t
oss
es
a b
all
up
to Q
uyen
, w
ait
ing a
t a t
hir
d-s
tory
win
dow
, w
ith
an
in
itia
l velo
city
of
30 f
eet
per
seco
nd
. S
he r
ele
ase
s th
e b
all
fro
m a
heig
ht
of
6 f
eet.
Th
e
equ
ati
on
h =
-16
t2 +
30t +
6 r
ep
rese
nts
th
e h
eig
ht
h o
f th
e b
all
aft
er
t se
con
ds.
If
the
ball
mu
st r
each
a h
eig
ht
of
25 f
eet
for
Qu
yen
to c
atc
h i
t, d
oes
the b
all
reach
Qu
yen
? E
xp
lain
. (H
int:
Su
bst
itu
te 2
5 f
or
h a
nd
use
th
e d
iscr
imin
an
t.)
No
; th
e d
iscri
min
an
t,
-316,
is n
eg
ati
ve,
so
th
ere
is n
o s
olu
tio
n.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-5
Ch
ap
ter
9
33
Gle
ncoe A
lgeb
ra 1
Wo
rd
Pro
ble
m P
racti
ce
So
lvin
g Q
uad
rati
c E
qu
ati
on
s b
y U
sin
g t
he Q
uad
rati
c F
orm
ula
1. B
USIN
ESS
T
an
ya r
un
s a c
ate
rin
g
bu
sin
ess
. B
ase
d o
n h
er
reco
rds,
her
week
ly p
rofi
t ca
n b
e a
pp
roxim
ate
d b
y t
he
fun
ctio
n f
(x) =
x2 +
2x -
37,
wh
ere
x i
s th
e n
um
ber
of
meals
sh
e c
ate
rs.
If f
(x)
is
negati
ve,
it m
ean
s th
at
the b
usi
ness
has
lost
mon
ey.
Wh
at
is t
he l
east
nu
mber
of
meals
th
at
Tan
ya n
eed
s to
cate
r in
ord
er
to h
ave a
pro
fit?
6 m
eals
2. A
ER
ON
AU
TIC
S A
t li
ftoff
, th
e s
pace
sh
utt
le D
isco
ver
y h
as
a c
on
stan
t acc
ele
rati
on
of
16.4
feet
per
seco
nd
sq
uare
d a
nd
an
in
itia
l velo
city
of
1341
feet
per
seco
nd
du
e t
o t
he r
ota
tion
of
Eart
h.
Th
e d
ista
nce
Dis
cover
y h
as
travele
d t
seco
nd
s aft
er
lift
off
is
giv
en
by
the e
qu
ati
on
d(t
) =
1341
t +
8.2
t2.
How
lo
ng a
fter
lift
off
has
Dis
cover
y t
ravele
d
40,0
00 f
eet?
Rou
nd
you
r an
swer
to t
he
neare
st t
en
th.
25.8
seco
nd
s
3. A
RC
HIT
EC
TU
RE
Th
e G
old
en
Rati
o
ap
pears
in
th
e
desi
gn
of
the G
reek
P
art
hen
on
beca
use
th
e w
idth
an
d
heig
ht
of
the f
aça
de a
re r
ela
ted
by t
he
equ
ati
on
W +
H −
W
= W
−
H .
If t
he h
eig
ht
of
a
mod
el
of
the P
art
hen
on
is
16 i
nch
es,
w
hat
is i
ts w
idth
? R
ou
nd
you
r an
swer
to
the n
eare
st t
en
th.
25.9
in
.
4. C
RA
FTS
M
ad
ely
n c
ut
a 6
0-i
nch
pip
e
clean
er
into
tw
o u
nequ
al
pie
ces,
an
d t
hen
sh
e u
sed
each
pie
ce t
o m
ak
e a
squ
are
. T
he s
um
of
the a
reas
of
the s
qu
are
s w
as
117 s
qu
are
in
ches.
Let
x =
th
e l
en
gth
of
on
e p
iece
. W
rite
an
d s
olv
e a
n e
qu
ati
on
to
rep
rese
nt
the s
itu
ati
on
an
d f
ind
th
e
len
gth
s of
the t
wo o
rigin
al
pie
ces.
(
60 -
x
−
4
) 2
+ ( x
−
4 ) 2
= 1
17
2
4 i
n.
an
d 3
6 i
n.
5. SIT
E D
ESIG
N T
he t
ow
n o
f S
mall
port
p
lan
s to
bu
ild
a n
ew
wate
r tr
eatm
en
t p
lan
t on
a r
ect
an
gu
lar
pie
ce o
f la
nd
75
yard
s w
ide a
nd
200 y
ard
s lo
ng.
Th
e
bu
ild
ings
an
d f
aci
liti
es
need
to c
over
an
are
a o
f 10,0
00 s
qu
are
yard
s. T
he t
ow
n’s
zon
ing b
oard
wan
ts t
he s
ite d
esi
gn
er
to
all
ow
as
mu
ch r
oom
as
poss
ible
betw
een
each
ed
ge o
f th
e s
ite a
nd
th
e b
uil
din
gs
an
d f
aci
liti
es.
Let
x r
ep
rese
nt
the w
idth
of
the b
ord
er.
Bu
ildin
gs
and
Fac
iliti
es
Bo
rder
75
yd
20
0 y
d
x x
xx
a
. U
se a
n e
qu
ati
on
sim
ilar
to A
= �
× w
to
rep
rese
nt
the s
itu
ati
on
.
10,0
00 =
(200 -
2x)(
75 -
2x)
b
. W
rite
th
e e
qu
ati
on
in
sta
nd
ard
qu
ad
rati
c fo
rm.
4x
2 -
550x +
5000 =
0;
9.8
an
d
127.7
c.
Wh
at
shou
ld b
e t
he w
idth
of
the
bord
er?
Rou
nd
you
r an
swer
to t
he
neare
st t
en
th.
9.7
9 y
d
9-5
Answers (Lesson 9-5)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A16 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
34
Gle
ncoe A
lgeb
ra 1
En
rich
men
t
Go
lden
Recta
ng
les
A g
old
en
recta
ng
le h
as
the p
rop
ert
y t
hat
its
sid
es
sati
sfy
the f
oll
ow
ing p
rop
ort
ion
. a
+ b −
a
= a
−
b
Tw
o q
uad
rati
c equ
ati
on
s ca
n b
e w
ritt
en
fro
m t
he p
rop
ort
ion
. T
hese
are
som
eti
mes
call
ed
go
lden
qu
ad
ra
tic e
qu
ati
on
s.
1. In
th
e p
rop
ort
ion
, le
t a
= 1
. U
se
2. S
olv
e t
he e
qu
ati
on
in
Exerc
ise 1
for
b.
cross
-mu
ltip
lica
tion
to w
rite
aqu
ad
rati
c equ
ati
on
. b
= -
1 +
√ "
5 −
2
b
2 +
b -
1 =
0
3. In
th
e p
rop
ort
ion
, le
t b =
1.
Wri
te
4. S
olv
e t
he e
qu
ati
on
in
Exerc
ise 3
for
a.
a q
uad
rati
c equ
ati
on
in
a.
a
2 -
a -
1 =
0
a =
1 +
√ "
5 −
2
5.
Exp
lain
wh
y 1
−
2 (
√ #
5 +
1)
an
d 1
−
2 (
√ #
5 -
1)
are
call
ed
gold
en
rati
os.
Th
ey a
re t
he r
ati
os o
f th
e s
ides i
n a
go
lden
recta
ng
le.
Th
e f
irst
is t
he r
ati
o
of
the l
on
g s
ide t
o t
he s
ho
rt s
ide;
the s
eco
nd
is s
ho
rt s
ide:
lon
g s
ide.
An
oth
er
pro
pert
y o
f gold
en
re
ctan
gle
s is
th
at
a s
qu
are
dra
wn
in
sid
e a
gold
en
rect
an
gle
cre
ate
s an
oth
er,
sm
all
er
gold
en
rect
an
gle
.
In t
he d
esi
gn
at
the r
igh
t, o
pp
osi
te
vert
ices
of
each
squ
are
have b
een
co
nn
ect
ed
wit
h q
uart
ers
of
circ
les.
For
exam
ple
, th
e a
rc f
rom
poin
t B
to
poin
t C
is
create
d b
y p
utt
ing t
he
poin
t of
a c
om
pass
at
poin
t A
. T
he
rad
ius
of
the a
rc i
s th
e l
en
gth
BA
.
6. O
n a
sep
ara
te s
heet
of
pap
er,
dra
w a
larg
er
vers
ion
of
the d
esi
gn
. S
tart
wit
h a
gold
en
re
ctan
gle
wit
h a
lon
g s
ide o
f 10 i
nch
es.
T
he s
ho
rt s
ide s
ho
uld
be a
bo
ut
6 3
−
16 i
nch
es.
b
a
9-5
C
BA
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-6
Ch
ap
ter
9
35
Gle
ncoe A
lgeb
ra 1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Exp
on
en
tial
Fu
ncti
on
s
Gra
ph
Exp
on
en
tial
Fu
nct
ion
sE
xp
on
en
tial
Fu
ncti
on
a f
unction d
efined b
y a
n e
quation o
f th
e f
orm
y =
ab
x,
where
a ≠
0,
b >
0,
and b
≠ 1
You
can
use
valu
es
of
x t
o f
ind
ord
ere
d p
air
s th
at
sati
sfy a
n e
xp
on
en
tial
fun
ctio
n.
Th
en
you
ca
n u
se t
he o
rdere
d p
air
s to
gra
ph
th
e f
un
ctio
n.
G
ra
ph
y =
3x
. F
ind
th
e
y-i
nte
rcep
t a
nd
sta
te t
he d
om
ain
an
d
ra
ng
e.
xy
-2
1 −
9
-1
1 −
3
01
13
29
Th
e y
-in
terc
ep
t is
1.
Th
e d
om
ain
is
all
real
nu
mbers
, an
d t
he
ran
ge i
s all
posi
tive n
um
bers
.x
y
O
G
ra
ph
y =
( 1 −
4 ) x
. U
se t
he
gra
ph
to
ap
pro
xim
ate
th
e v
alu
e o
f ( 1
−
4 ) -
0.5
.
Th
e v
alu
e o
f ( 1
−
4 ) -
0.5
is
abou
t 2.
x
y
O2
8
Exerc
ises
1. G
rap
h y
= 0
.3x.
Fin
d t
he y
-in
terc
ep
t. T
hen
use
th
e g
rap
h t
o
ap
pro
xim
ate
th
e v
alu
e o
f 0.3-
1.5.
Use
a c
alc
ula
tor
to c
on
firm
th
e v
alu
e.
1;
ab
ou
t 6
Gra
ph
ea
ch
fu
ncti
on
. F
ind
th
e y
-in
tercep
t a
nd
sta
te t
he d
om
ain
an
d r
an
ge.
2. y =
3x +
1 2
3. y =
( 1 −
3 ) x
+ 1
2
4. y =
( 1 −
2 ) x
- 2
-
1
x
y
O1
2
x
y
O
x
y
O1
2
9-6
Exam
ple
1Exam
ple
2
xy
-2
16
-1
4
01
1 1
−
4
2 1
−
16
x
y
O1
2
Answers (Lesson 9-5 and Lesson 9-6)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A17 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
36
Gle
ncoe A
lgeb
ra 1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n (c
on
tin
ued
)
Exp
on
en
tial
Fu
ncti
on
s
Iden
tify
Exp
on
en
tial
Beh
avio
r It
is
som
eti
mes
use
ful
to k
now
if
a s
et
of
data
is
exp
on
en
tial.
On
e w
ay t
o t
ell
is
to o
bse
rve t
he s
hap
e o
f th
e g
rap
h.
An
oth
er
way i
s to
obse
rve
the p
att
ern
in
th
e s
et
of
data
.
D
ete
rm
ine w
heth
er t
he s
et
of
da
ta s
ho
wn
belo
w d
isp
lay
s
ex
po
nen
tia
l b
eh
av
ior.
Writ
e yes o
r no
. E
xp
lain
wh
y o
r w
hy
no
t.
Meth
od
1: L
ook
for
a P
att
ern
Th
e d
om
ain
valu
es
incr
ease
by r
egu
lar
inte
rvals
of
2,
wh
ile t
he r
an
ge v
alu
es
have
a c
om
mon
fact
or
of
1
−
2 .
Sin
ce t
he d
om
ain
valu
es
incr
ease
by r
egu
lar
inte
rvals
an
d
the r
an
ge v
alu
es
have a
com
mon
fact
or,
th
e d
ata
are
pro
bably
exp
on
en
tial.
Meth
od
2: G
rap
h t
he D
ata
T
he g
rap
h s
how
s ra
pid
ly d
ecr
easi
ng
valu
es
of
y a
s x
incr
ease
s. T
his
is
chara
cteri
stic
of
exp
on
en
tial
beh
avio
r.
Exerc
ises
Dete
rm
ine w
heth
er t
he s
et
of
da
ta s
ho
wn
belo
w d
isp
lay
s e
xp
on
en
tia
l b
eh
av
ior.
Writ
e yes o
r no
. E
xp
lain
wh
y o
r w
hy
no
t.
1.
x0
12
3
y5
10
15
20
2.
x0
12
3
y3
92
781
N
o;
the d
om
ain
valu
es a
re a
t re
gu
lar
Yes;
the d
om
ain
valu
es a
re a
t
inte
rvals
, an
d t
he r
an
ge v
alu
es h
ave
reg
ula
r in
terv
als
, an
d t
he r
an
ge
a c
om
mo
n d
iffe
ren
ce 5
.
valu
es h
ave a
co
mm
on
facto
r 3.
3.
x-
11
35
y32
16
84
4.
x-
10
12
3
y3
33
33
Y
es;
the d
om
ain
valu
es a
re a
t
No
; th
e d
om
ain
valu
es a
re a
t
reg
ula
r in
terv
als
, an
d t
he r
an
ge
reg
ula
r in
terv
als
, b
ut
the r
an
ge
valu
es h
ave a
co
mm
on
facto
r 1
−
2 .
valu
es d
o n
ot
ch
an
ge.
5.
x-
50
51
0
y1
0.5
0.2
50.1
25
6.
x0
12
34
y 1
−
3
1
−
9
1
−
27
1
−
81
1
−
243
Y
es;
the d
om
ain
valu
es a
re a
t
Yes;
the d
om
ain
valu
es a
re a
t
reg
ula
r in
terv
als
, an
d t
he r
an
ge
reg
ula
r in
terv
als
, an
d t
he r
an
ge
valu
es h
ave a
co
mm
on
facto
r 0.5
. v
alu
es
ha
ve
a c
om
mo
n f
ac
tor
1
−
3 .
9-6 Exam
ple
x0
24
68
10
y64
32
16
84
2
x
y
O2
8
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-6
Ch
ap
ter
9
37
Gle
ncoe A
lgeb
ra 1
Sk
ills
Pra
ctic
e
Exp
on
en
tial
Fu
ncti
on
sG
ra
ph
ea
ch
fu
ncti
on
. F
ind
th
e y
-in
tercep
t, a
nd
sta
te t
he d
om
ain
an
d r
an
ge.
Th
en
use t
he g
ra
ph
to
dete
rm
ine t
he a
pp
ro
xim
ate
va
lue o
f th
e g
iven
ex
pressio
n.
Use
a c
alc
ula
tor t
o c
on
firm
th
e v
alu
e.
1. y =
2x;
22
.3
2. y =
( 1
−
3 ) x
; ( 1
−
3 ) -
1.6
1
; D
= a
ll r
eals
, R
= {
y | y
> 0
};
4.9
1
; D
= a
ll r
eals
, R
= {
y |
y >
0};
5.8
Gra
ph
ea
ch
fu
ncti
on
. F
ind
th
e y
-in
tercep
t, a
nd
sta
te t
he d
om
ain
an
d r
an
ge.
3. y =
3(2
x)
4. y =
3x +
2
3
; D
= a
ll r
eals
, R
= {
y | y
> 0
}
3;
D =
all r
eals
, R
= {
y | y
> 2
}
Dete
rm
ine w
heth
er t
he s
et
of
da
ta s
ho
wn
belo
w d
isp
lay
s e
xp
on
en
tia
l b
eh
av
ior.
Writ
e yes o
r no.
Ex
pla
in w
hy
or w
hy
no
t.
5.
x-
3-
2-
10
y9
12
15
18
6.
x0
51
015
y20
10
52.5
N
o;
the d
om
ain
valu
es a
re a
t
Yes;
the d
om
ain
valu
es a
re a
t
reg
ula
r in
terv
als
an
d t
he r
an
ge
reg
ula
r in
terv
als
an
d t
he r
an
ge
valu
es h
ave a
co
mm
on
v
alu
es h
ave a
co
mm
on
facto
r 0.5
.
dif
fere
nce 3
.
7.
x4
812
16
y20
40
80
160
8.
x50
30
10
-10
y90
70
50
30
Y
es;
the d
om
ain
valu
es a
re a
t
No
; th
e d
om
ain
valu
es a
re a
t
reg
ula
r in
terv
als
an
d t
he r
an
ge
reg
ula
r in
terv
als
an
d t
he r
an
ge
valu
es h
ave a
co
mm
on
facto
r 2.
valu
es h
ave a
co
mm
on
dif
fere
nce 2
0.
x
y
Ox
y
O
x
y
Ox
y
O
9-6
Answers (Lesson 9-6)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A18 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
38
Gle
ncoe A
lgeb
ra 1
9-6
Practi
ce
Exp
on
en
tial
Fu
ncti
on
sG
ra
ph
ea
ch
fu
ncti
on
. F
ind
th
e y
-in
tercep
t a
nd
sta
te t
he d
om
ain
an
d r
an
ge.
Th
en
use t
he g
ra
ph
to
dete
rm
ine t
he a
pp
ro
xim
ate
va
lue o
f th
e g
iven
ex
pressio
n.
Use
a c
alc
ula
tor t
o c
on
firm
th
e v
alu
e.
1. y =
( 1 −
10 ) x
; ( 1
−
10 ) -
0.5
2. y =
3x;
31
.9
3. y =
( 1 −
4 ) x
; ( 1
−
4 ) -
1.4
Gra
ph
ea
ch
fu
ncti
on
. F
ind
th
e y
-in
tercep
t, a
nd
sta
te t
he d
om
ain
an
d r
an
ge.
4. y =
4(2
x) +
1
5. y =
2(2
x -
1)
6. y =
0.5
(3x -
3)
Dete
rm
ine w
heth
er t
he s
et
of
da
ta s
ho
wn
belo
w d
isp
lay
s e
xp
on
en
tia
l b
eh
av
ior.
Writ
e yes o
r no
. E
xp
lain
wh
y o
r w
hy
no
t.
7.
x2
58
11
y48
120
30
7.5
8.
x21
18
15
12
y30
23
16
9
Y
es;
the d
om
ain
valu
es a
re a
t
No
; th
e d
om
ain
valu
es a
re a
t re
gu
lar
inte
rvals
an
d t
he r
an
ge
reg
ula
r in
terv
als
an
d t
he r
an
ge
valu
es h
ave a
co
mm
on
v
alu
es h
ave a
co
mm
on
fa
cto
r 0.2
5.
dif
fere
nce 7
.
9. LEA
RN
ING
M
s. K
lem
pere
r to
ld h
er
En
gli
sh c
lass
th
at
each
week
stu
den
ts t
en
d t
o f
org
et
on
e s
ixth
of
the v
oca
bu
lary
word
s th
ey l
earn
ed
th
e p
revio
us
week
. S
up
pose
a s
tud
en
t le
arn
s 60 w
ord
s. T
he n
um
ber
of
word
s re
mem
bere
d c
an
be d
esc
ribed
by t
he f
un
ctio
n
W
(x) =
60 ( 5
−
6 ) x
, w
here
x i
s th
e n
um
ber
of
week
s th
at
pass
. H
ow
man
y w
ord
s w
ill
the
st
ud
en
t re
mem
ber
aft
er
3 w
eek
s? ab
ou
t 35
10. B
IOLO
GY
S
up
pose
a c
ert
ain
cell
rep
rod
uce
s it
self
in
fou
r h
ou
rs.
If a
lab r
ese
arc
her
begin
s w
ith
50 c
ell
s, h
ow
man
y c
ell
s w
ill
there
be a
fter
on
e d
ay,
two d
ays,
an
d t
hre
e
days?
(H
int:
Use
th
e e
xp
on
en
tial
fun
ctio
n y
= 5
0(2
x).
)
3200 c
ells;
204,8
00 c
ells;
13,1
07,2
00 c
ells
x
y
Ox
y
O
x
y
O
x
y
Ox
y
Ox
y
O
1;
D =
{all
real
nu
mb
ers
};
R =
{ y
| y >
0};
32
1;
D =
{all
real
nu
mb
ers
};
R =
{ y
| y >
0};
8.1
1;
D =
{a
ll r
eal
nu
mb
ers
};
R =
{ y
| y >
0};
7.0
5;
D =
{all
real
nu
mb
ers
};
R =
{ y
| y >
1};
0;
D =
{all
real
nu
mb
ers
};
R =
{ y
| y >-
2}
-1;
D =
{a
ll r
eal
nu
mb
ers
};
R =
{ y
| y >
1.5
}
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-6
Ch
ap
ter
9
39
Gle
ncoe A
lgeb
ra 1
1. W
ASTE
S
up
pose
th
e w
ast
e g
en
era
ted
by n
on
recy
cled
pap
er
an
d c
ard
board
p
rod
uct
s is
ap
pro
xim
ate
d b
y t
he
foll
ow
ing f
un
ctio
n.
y=
1000(2
)0.3
x
Sk
etc
h t
he e
xp
on
en
tial
fun
ctio
n o
n t
he
coord
inate
gri
d b
elo
w. y
xO
950
650
350
1850
1550
2450
2150
1250
43
21
-2-1
-3
-4
2. M
ON
EY
T
aty
an
a’s
gra
nd
fath
er
gave h
er
on
e p
en
ny o
n t
he d
ay s
he w
as
born
. H
e
pla
ns
to d
ou
ble
th
e a
mou
nt
he g
ives
her
every
day.
Est
imate
how
mu
ch s
he w
ill
rece
ive f
rom
her
gra
nd
fath
er
on
th
e 1
2th
d
ay o
f h
er
life
.
ab
ou
t $20
3.PIC
TU
RE F
RA
MES
Sin
ce a
pic
ture
fra
me
incl
ud
es
a b
ord
er,
th
e
pic
ture
mu
st b
e
small
er
in a
rea t
han
th
e e
nti
re f
ram
e.
Th
e
table
sh
ow
s th
e
rela
tion
ship
betw
een
p
ictu
re a
rea a
nd
fra
me
len
gth
for
a p
art
icu
lar
lin
e o
f fr
am
es.
Is
this
an
exp
on
en
tial
rela
tion
ship
? E
xp
lain
.
No
; th
ere
is n
o c
om
mo
n f
acto
r b
etw
een
th
e p
ictu
re a
reas.
4.D
EPR
EC
IATIO
N T
he v
alu
e o
f R
oyce
C
om
pan
y’s
com
pu
ter
equ
ipm
en
t is
d
ecr
easi
ng i
n v
alu
e a
ccord
ing t
o t
he
foll
ow
ing f
un
ctio
n.
y=
4000(0
.87)x
In t
he e
qu
ati
on
, x i
s th
e n
um
ber
of
years
th
at
have e
lap
sed
sin
ce t
he
equ
ipm
en
t w
as
pu
rch
ase
d a
nd
y i
s in
d
oll
ars
. W
hat
was
the v
alu
e 5
years
aft
er
it w
as
pu
rch
ase
d?
Rou
nd
you
r an
swer
to t
he n
eare
st d
oll
ar.
$1994
5. M
ETEO
RO
LO
GY
T
he a
tmosp
heri
c p
ress
ure
(in
mil
libars
) at
a g
iven
alt
itu
de
x,
in m
ete
rs,
can
be a
pp
roxim
ate
d b
y t
he
foll
ow
ing f
un
ctio
n.
Th
e f
un
ctio
n i
s vali
d
for
valu
es
of
x b
etw
een
0 a
nd
10,0
00.
f (x) =
1038(1
.000134)-
x
a
. W
hat
is t
he p
ress
ure
at
sea l
evel?
1038 m
illib
ars
b
. T
he M
cDon
ald
Obse
rvato
ry i
n T
exas
is a
t an
alt
itu
de o
f 2000 m
ete
rs.
Wh
at
is t
he a
pp
roxim
ate
atm
osp
heri
c p
ress
ure
th
ere
?
794 m
illib
ars
c.
As
alt
itu
de i
ncr
ease
s, w
hat
hap
pen
s to
atm
osp
heri
c p
ress
ure
?
It d
ecre
ases.
Wo
rd
Pro
ble
m P
racti
ce
Exp
on
en
tial
Fu
ncti
on
s
9-6
Sid
e
Pic
ture
Len
gth
A
rea
(in
.)
(in
2)
5
6
6
12
7
20
8
30
9
42
Answers (Lesson 9-6)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A19 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
40
Gle
ncoe A
lgeb
ra 1
En
rich
men
t
Rati
on
al
Exp
on
en
ts
You
have
dev
elop
ed t
he
foll
owin
g p
rop
erti
es o
f p
ower
s w
hen
a i
s a p
osit
ive
real
nu
mber
an
d
m a
nd
p a
re i
nte
ger
s.
am �
ap =
am
+ p
(a
b)m
= a
mb
m
a
0 =
1
(am)p
= a
mp
a
m
−
a
p =
am
- p
a-
m =
1
−
am
Exp
onen
ts n
eed
not
be
rest
rict
ed t
o in
teger
s. W
e ca
n d
efin
e ra
tion
al
exp
onen
ts s
o th
at
oper
ati
ons
invol
vin
g t
hem
wil
l be
gov
ern
ed b
y t
he
pro
per
ties
for
in
teger
exp
onen
ts.
( a 1
−
2 ) 2
= a
1 −
2 ·
2 =
a
( a
1 −
3 ) 3
= a
1 −
3 ·
3
( a
1
−
n ) n
= a
1
−
n ·
n =
a
a 1
−
2 s
qu
are
d i
s a
. a
1 −
3 c
ubed
is
a.
a 1
−
n t
o th
e n
pow
er i
s a
.
a 1
−
2 i
s a s
qu
are
roo
t of
a.
a 1
−
3 i
s a c
ube
root
of
a.
a 1
−
n i
s an
nth
roo
t of
a.
a 1
−
2 =
√ #
a
a
1 −
3 =
3 √ #
a
a
1
−
n =
n √ #
a
Now
let
us
inves
tigate
th
e m
ean
ing o
f a
m
−
p .
a m
−
p =
a m
· 1
−
p =
( a
m ) 1
−
p =
p √ ##
a
m
a
m
−
p =
a 1
−
p ·
m =
( a 1
−
p ) m
= (
p √ #
a
) m
Th
eref
ore,
a m
−
p =
p √ ##
a
m o
r (
p √ #
a
) m .
W
rit
e
4 √ "
"
a 3 i
n
ex
po
nen
tia
l fo
rm
.
4 √ #
a
3 =
a 3
−
4
W
rit
e a
2
−
5 i
n r
ad
ica
l fo
rm
.
a 2
−
5 =
5 √ #
a
2
F
ind
a 2
−
3
−
a 1
−
2 .
a 2
−
3
−
a 1
−
2 =
a 2
−
3 -
1 −
2 =
a 4
−
6 -
3 −
6 =
a 1
−
6 o
r 6 √ #
a
Writ
e e
ach
ex
press
ion
in
ra
dic
al
form
.
1. b
2 −
3
3 √ "
b
2
2. 3 c 1
−
2 3
√ "
c
3. (3
c ) 1
−
2
√ "
"
3c
Writ
e e
ach
ex
press
ion
in
ex
po
nen
tia
l fo
rm
.
4. 3 √ #
b 4 b
4
−
3
5. √
##
4 a
3 (4
a 3 ) 1
−
2 =
2 a 3
−
2
6. 2
·
3 √ #
b 2 2 b
2
−
3
Perfo
rm
th
e o
pera
tio
n i
nd
ica
ted
. A
nsw
ers
sho
uld
sh
ow
po
siti
ve e
xp
on
en
ts o
nly
.
7. ( a
3 b 1
−
4 ) 2
a 6 b
1
−
2
8. -
8 a
3 −
4
−
2 a
1 −
2
-4 a
1
−
4
9. ( b
1 −
2
−
b -
2 −
3 ) 3
b 7
−
2
10. √
#
a 3 ·
√ #
a
a
2
11. ( a
2 b -
1 −
3 ) -
1 −
2
b 1
−
6
−
a
12. -
2 a
1 −
3 b
0 ( 5
a 1
−
2 b
- 2
−
3 )
-10 a 5
−
6
−
b 2
−
3
9-6
Exam
ple
1Exam
ple
2
Exam
ple
3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-7
Ch
ap
ter
9
41
Gle
ncoe A
lgeb
ra 1
9-7
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Gro
wth
an
d D
ecay
Exp
on
en
tial
Gro
wth
P
opu
lati
on i
ncr
ease
s an
d g
row
th o
f m
onet
ary
in
ves
tmen
ts a
re
exam
ple
s of
ex
po
nen
tia
l g
ro
wth
. T
his
mea
ns
that
an
in
itia
l am
oun
t in
crea
ses
at
a s
tead
y
rate
over
tim
e.
T
he g
enera
l equation f
or
exponential gro
wth
is y
= a
(1 +
r)t .
• y r
epre
sents
the f
inal am
ount.
Exp
on
en
tial
Gro
wth
• a r
epre
sents
the initia
l am
ount.
• r
repre
sents
the r
ate
of
change e
xpre
ssed a
s a
decim
al.
• t
repre
sents
tim
e.
PO
PU
LA
TIO
N T
he
po
pu
lati
on
of
Jo
hn
son
Cit
y i
n 2
000 w
as
25,0
00. S
ince t
hen
, th
e p
op
ula
tio
n h
as
gro
wn
at
an
av
era
ge r
ate
of
3.2
% e
ach
yea
r.
a.
Writ
e a
n e
qu
ati
on
to
rep
rese
nt
the
po
pu
lati
on
of
Jo
hn
son
Cit
y s
ince 2
000.
Th
e ra
te 3
.2%
can
be
wri
tten
as
0.0
32.
y =
a(1
+ r
)t
y =
25,0
00(1
+ 0
.032)t
y =
25,0
00(1
.032)t
b.
Acco
rd
ing
to
th
e e
qu
ati
on
, w
ha
t w
ill
the p
op
ula
tio
n o
f J
oh
nso
n C
ity
be i
n
the y
ea
r 2
010?
In 2
010,
t w
ill
equ
al
2010
- 2
000 o
r 10.
Su
bst
itu
te 1
0 f
or t
in
th
e eq
uati
on f
rom
p
art
a.
y =
25,0
00(1
.032)1
0t
= 1
0
≈
34,2
56
In 2
010,
the
pop
ula
tion
of
Joh
nso
n C
ity w
ill
be
abou
t 34,2
56.
INV
ES
TM
EN
T T
he
Ga
rcia
s h
av
e $
12,0
00 i
n a
sa
vin
gs
acco
un
t. T
he b
an
k p
ay
s 3.5
% i
nte
rest
o
n s
av
ing
s a
cco
un
ts,
co
mp
ou
nd
ed
m
on
thly
. F
ind
th
e b
ala
nce i
n 3
yea
rs.
Th
e ra
te 3
.5%
can
be
wri
tten
as
0.0
35.
Th
e sp
ecia
l eq
uati
on f
or c
omp
oun
d
inte
rest
is
A=
P(1
+r − n
)nt ,
wh
ere
A
rep
rese
nts
th
e bala
nce
, P
is
the
init
ial
am
oun
t, r
rep
rese
nts
th
e an
nu
al
rate
ex
pre
ssed
as
a d
ecim
al,
n r
epre
sen
ts t
he
nu
mber
of
tim
es t
he
inte
rest
is
com
pou
nd
ed e
ach
yea
r, a
nd
t r
epre
sen
ts
the
nu
mber
of
yea
rs t
he
mon
ey i
s in
ves
ted
.
A=
P(1
+r − n
)nt
A=
12,0
00
(1 +
0.0
35
−12
)36
A≈
12,0
00(1
.00292)3
6
A ≈
13,3
28.0
9
In t
hre
e yea
rs,
the
bala
nce
of
the
acc
oun
t w
ill
be
$13,3
26.4
9.
Exe
rcis
es
1. P
OP
ULA
TIO
N T
he
pop
ula
tion
of
the
Un
ited
2.
INV
ES
TM
EN
T D
eter
min
e th
e S
tate
s h
as
bee
n i
ncr
easi
ng a
t an
aver
age
am
oun
t of
an
in
ves
tmen
t of
$2500 i
f it
an
nu
al
rate
of
0.9
1%
. If
th
e p
opu
lati
on o
f th
e
is i
nves
ted
at
in i
nte
rest
rate
of
5.2
5%
U
nit
ed S
tate
s w
as
abou
t 303,1
46,0
00 i
n t
he
com
pou
nd
ed m
onth
ly f
or 4
yea
rs.
yea
r 2008,
pre
dic
t th
e U
.S.
pop
ula
tion
in
th
e
$3082.7
8yea
r 2012.
ab
ou
t 308,3
85,8
45
3. P
OP
ULA
TIO
N I
t is
est
imate
d t
hat
the
4.
INV
ES
TM
EN
T D
eter
min
e th
e p
opu
lati
on o
f th
e w
orld
is
incr
easi
ng a
t an
am
oun
t of
an
in
ves
tmen
t of
$100,0
00
aver
age
an
nu
al
rate
of
1.3
%.
If t
he
pop
ula
tion
if i
t is
in
ves
ted
at
an
in
tere
st r
ate
of
of
th
e w
orld
was
abou
t 6,6
41,0
00,0
00 i
n t
he
5.2
% c
omp
oun
ded
qu
art
erly
for
yea
r 2008,
pre
dic
t th
e w
orld
pop
ula
tion
in
th
e
12 y
ears
.
yea
r 2015.
ab
ou
t 7,0
84,8
81,7
69
$185,8
88.8
7
Exam
ple
1Exam
ple
2
Answers (Lesson 9-6 and Lesson 9-7)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A20 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
42
Gle
ncoe A
lgeb
ra 1
9-7
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n (c
on
tin
ued
)
Gro
wth
an
d D
ecay
Exp
on
en
tial
Deca
y
Rad
ioact
ive
dec
ay a
nd
dep
reci
ati
on a
re e
xam
ple
s of
ex
po
nen
tia
l d
eca
y.
Th
is m
ean
s th
at
an
in
itia
l am
oun
t d
ecre
ase
s at
a s
tead
y r
ate
over
a p
erio
d o
f ti
me.
The g
enera
l equation f
or
exponential decay is y
= a
(1 -
r)t .
• y r
epre
sents
the f
inal am
ount.
Exp
on
en
tial
Decay
• a r
epre
sents
the initia
l am
ount.
• r
repre
sents
the r
ate
of
decay e
xpre
ssed a
s a
decim
al.
• t
repre
sents
tim
e.
D
EPR
EC
IATIO
N T
he o
rig
ina
l p
ric
e o
f a
tra
cto
r w
as
$45,0
00.
Th
e
va
lue o
f th
e t
ra
cto
r d
ecrea
ses
at
a s
tea
dy
ra
te o
f 12%
per y
ea
r.
a.
Writ
e a
n e
qu
ati
on
to
rep
rese
nt
the v
alu
e o
f th
e t
ra
cto
r s
ince i
t w
as
pu
rch
ase
d.
Th
e ra
te 1
2%
can
be
wri
tten
as
0.1
2.
y =
a(1
- r
)t G
enera
l equation f
or
exponential decay
y =
45,0
00(1
- 0
.12)t
a
= 4
5,0
00 a
nd r
= 0
.12
y =
45,0
00(0
.88)t
Sim
plif
y.
b.
Wh
at
is t
he v
alu
e o
f th
e t
ra
cto
r i
n 5
yea
rs?
y =
45,0
00(0
.88)t
Equation f
or
decay f
rom
part
a
y =
45,0
00(0
.88)5
t
= 5
y ≈
23,7
47.9
4
Use a
calc
ula
tor.
In 5
yea
rs,
the
tract
or w
ill
be
wor
th a
bou
t $23,7
47.9
4.
Exerc
ises
1. PO
PU
LA
TIO
N T
he
pop
ula
tion
of
Bu
lgari
a h
as
bee
n d
ecre
asi
ng a
t an
an
nu
al
rate
of
0.8
9%
. If
th
e p
opu
lati
on o
f B
ulg
ari
a w
as
abou
t 7,4
50,3
49 i
n t
he
yea
r 2005,
pre
dic
t it
s p
opu
lati
on i
n t
he
yea
r 2015.
ab
ou
t 6,8
13,2
04
2. D
EPR
EC
IATIO
N M
r. G
osse
ll i
s a m
ach
inis
t. H
e bou
gh
t so
me
new
mach
iner
y f
or a
bou
t $125,0
00.
He
wan
ts t
o ca
lcu
late
th
e valu
e of
th
e m
ach
iner
y o
ver
th
e n
ext
10 y
ears
for
ta
x p
urp
oses
. If
th
e m
ach
iner
y d
epre
ciate
s at
the
rate
of
15%
per
yea
r, w
hat
is t
he
valu
e of
th
e m
ach
iner
y (
to t
he
nea
rest
$100)
at
the
end
of
10 y
ears
? ab
ou
t $24,6
00
3. A
RC
HA
EO
LO
GY
Th
e h
alf
-lif
e of
a r
ad
ioact
ive
elem
ent
is d
efin
ed a
s th
e ti
me
that
it
tak
es f
or o
ne-
half
a q
uan
tity
of
the
elem
ent
to d
ecay.
Rad
ioact
ive
carb
on-1
4 i
s fo
un
d i
n
all
liv
ing o
rgan
ism
s an
d h
as
a h
alf
-lif
e of
5730 y
ears
. C
onsi
der
a l
ivin
g o
rgan
ism
wit
h a
n
orig
inal
con
cen
trati
on o
f ca
rbon
-14 o
f 100 g
ram
s.
a.
If t
he
organ
ism
liv
ed 5
730 y
ears
ago,
wh
at
is t
he
con
cen
trati
on o
f ca
rbon
-14
tod
ay?
50 g
b.
If t
he
organ
ism
liv
ed 1
1,4
60 y
ears
ago,
det
erm
ine
the
con
cen
trati
on o
f ca
rbon
-14
tod
ay.
25 g
4. D
EP
RE
CIA
TIO
N A
new
car
cost
s $32,0
00.
It i
s ex
pec
ted
to
dep
reci
ate
12%
each
yea
r fo
r 4 y
ears
an
d t
hen
dep
reci
ate
8%
each
yea
r th
erea
fter
. F
ind
th
e valu
e of
th
e ca
r in
6 y
ears
. ab
ou
t $16,2
42.6
3
Exam
ple
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-7
Ch
ap
ter
9
43
Gle
ncoe A
lgeb
ra 1
9-7
Sk
ills
Pra
ctic
e
Gro
wth
an
d D
ecay
1. PO
PU
LA
TIO
N T
he
pop
ula
tion
of
New
Yor
k C
ity i
ncr
ease
d f
rom
8,0
08,2
78 i
n 2
000
to 8
,168,3
88 i
n 2
005.
Th
e an
nu
al
rate
of
pop
ula
tion
in
crea
se f
or t
he
per
iod
was
abou
t 0.4
%.
a.
Wri
te a
n e
qu
ati
on f
or t
he
pop
ula
tion
t y
ears
aft
er 2
000.
P=
8,0
08,2
78(1
.004)t
b.
Use
th
e eq
uati
on t
o p
red
ict
the
pop
ula
tion
of
New
Yor
k C
ity i
n 2
015.
ab
ou
t 8,5
02,4
65
2. SA
VIN
GS
Th
e F
resh
an
d G
reen
Com
pan
y h
as
a s
avin
gs
pla
n f
or i
ts e
mp
loyee
s. I
f an
em
plo
yee
mak
es a
n i
nit
ial
con
trib
uti
on o
f $1000,
the
com
pan
y p
ays
8%
in
tere
st
com
pou
nd
ed q
uart
erly
.
a.
If a
n e
mp
loyee
part
icip
ati
ng i
n t
he
pla
n w
ith
dra
ws
the
bala
nce
of
the
acc
oun
t aft
er
5 y
ears
, h
ow m
uch
wil
l be
in t
he
acc
oun
t? $
1485.9
5
b.
If a
n e
mp
loyee
part
icip
ati
ng i
n t
he
pla
n w
ith
dra
ws
the
bala
nce
of
the
acc
oun
t aft
er
35 y
ears
, h
ow m
uch
wil
l be
in t
he
acc
oun
t? $
15,9
96.4
7
3. H
OU
SIN
G M
r. a
nd
Mrs
. B
oyce
bou
gh
t a h
ouse
for
$96,0
00 i
n 1
995.
Th
e re
al
esta
te
bro
ker
in
dic
ate
d t
hat
hou
ses
in t
hei
r are
a w
ere
ap
pre
ciati
ng a
t an
aver
age
an
nu
al
rate
of
7%
. If
th
e ap
pre
ciati
on r
emain
ed s
tead
y a
t th
is r
ate
, w
hat
was
the
valu
e of
th
e B
oyce
’s h
ome
in 2
009?
ab
ou
t $247,5
39
4. M
AN
UFA
CT
UR
ING
Zel
ler
Ind
ust
ries
bou
gh
t a p
iece
of
wea
vin
g e
qu
ipm
ent
for
$60,0
00.
It i
s ex
pec
ted
to
dep
reci
ate
at
an
aver
age
rate
of
10%
per
yea
r.
a.
Wri
te a
n e
qu
ati
on f
or t
he
valu
e of
th
e p
iece
of
equ
ipm
ent
aft
er t
yea
rs.
E =
60,0
00(0
.90)t
b.
Fin
d t
he
valu
e of
th
e p
iece
of
equ
ipm
ent
aft
er 6
yea
rs.
ab
ou
t $31,8
86
5. FIN
AN
CES
Kyle
saved
$500 f
rom
a s
um
mer
job
. H
e p
lan
s to
sp
end
10%
of
his
savin
gs
each
wee
k o
n v
ari
ous
form
s of
en
tert
ain
men
t. A
t th
is r
ate
, h
ow m
uch
wil
l K
yle
have
left
aft
er 1
5 w
eek
s? $
102.9
5
6. T
RA
NS
PO
RT
AT
ION
Tif
fan
y’s
mot
her
bou
gh
t a c
ar
for
$9000 f
ive
yea
rs a
go.
Sh
e w
an
ts
to s
ell
it t
o T
iffa
ny b
ase
d o
n a
15%
an
nu
al
rate
of
dep
reci
ati
on.
At
this
rate
, h
ow m
uch
w
ill
Tif
fan
y p
ay f
or t
he
car?
ab
ou
t $3993
Answers (Lesson 9-7)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A21 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
44
Gle
ncoe A
lgeb
ra 1
9-7
Practi
ce
Gro
wth
an
d D
ecay
1.C
OM
MU
NIC
AT
ION
S S
por
ts r
ad
io s
tati
ons
nu
mber
ed 2
20 i
n 1
996.
Th
e n
um
ber
of
spor
ts r
ad
io s
tati
ons
has
sin
ce i
ncr
ease
d b
y a
pp
roxim
ate
ly 1
4.3%
per
yea
r.
a.
Wri
te a
n e
qu
ati
on f
or t
he
nu
mber
of
spor
ts r
ad
io s
tati
ons
for
t yea
rs a
fter
1996.
R=
220(1
.143)t
b.
If t
he
tren
d c
onti
nu
es,
pre
dic
t th
e n
um
ber
of
spor
ts r
ad
io s
tati
ons
in t
his
for
mat
for
the
yea
r 2010.
ab
ou
t 1429 s
tati
on
s
2. IN
VE
ST
ME
NT
S D
eter
min
e th
e am
oun
t of
an
in
ves
tmen
t if
$500 i
s in
ves
ted
at
an
in
tere
st r
ate
of
4.2
5%
com
pou
nd
ed q
uart
erly
for
12 y
ears
.
$830.4
1
3. IN
VE
ST
ME
NT
S D
eter
min
e th
e am
oun
t of
an
in
ves
tmen
t if
$300 i
s in
ves
ted
at
an
in
tere
st r
ate
of
6.7
5%
com
pou
nd
ed s
emia
nn
uall
y f
or 2
0 y
ears
.
$1131.7
3
4. H
OU
SIN
G T
he
Gre
ens
bou
gh
t a c
ond
omin
ium
for
$110,0
00 i
n 2
005.
If i
ts v
alu
e ap
pre
ciate
s at
an
aver
age
rate
of
6%
per
yea
r, w
hat
wil
l th
e valu
e be
in 2
010?
ab
ou
t $147,2
05
5. D
EFO
RE
ST
AT
ION
Du
rin
g t
he
1990s,
th
e fo
rest
ed a
rea o
f G
uate
mala
dec
rease
d a
t an
aver
age
rate
of
1.7
%.
a.
If t
he
fore
sted
are
a i
n G
uate
mala
in
1990 w
as
abou
t 34,4
00 s
qu
are
kil
omet
ers,
wri
te
an
equ
ati
on f
or t
he
fore
sted
are
a f
or t
yea
rs a
fter
1990.
C =
34,4
00(0
.983)t
b.
If t
his
tre
nd
con
tin
ues
, p
red
ict
the
fore
sted
are
a i
n 2
015.
ab
ou
t 22,4
07.6
5 k
m2
6. B
US
INE
SS
A p
iece
of
mach
iner
y v
alu
ed a
t $25,0
00 d
epre
ciate
s at
a s
tead
y r
ate
of
10%
yea
rly.
Wh
at
wil
l th
e valu
e of
th
e p
iece
of
mach
iner
y b
e aft
er 7
yea
rs?
ab
ou
t $11,9
57
7. T
RA
NS
PO
RT
AT
ION
A n
ew c
ar
cost
s $18,0
00.
It i
s ex
pec
ted
to
dep
reci
ate
at
an
aver
age
rate
of
12%
per
yea
r. F
ind
th
e valu
e of
th
e ca
r in
8 y
ears
.
ab
ou
t $6473
8. P
OP
ULA
TIO
N T
he
pop
ula
tion
of
Osa
ka,
Jap
an
, d
ecli
ned
at
an
aver
age
an
nu
al
rate
of
0.0
5%
for
th
e fi
ve
yea
rs b
etw
een
1995 a
nd
2000.
If t
he
pop
ula
tion
of
Osa
ka w
as
11,0
13,0
00 i
n 2
000 a
nd
it
con
tin
ues
to
dec
lin
e at
the
sam
e ra
te,
pre
dic
t th
e p
opu
lati
on
in 2
050.
ab
ou
t 10,7
41,0
00
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-7
Ch
ap
ter
9
45
Gle
ncoe A
lgeb
ra 1
9-7
1. D
EPR
EC
IATIO
N T
he
valu
e of
a n
ew
pla
sma t
elev
isio
n d
epre
ciate
s by a
bou
t 7%
each
yea
r. A
eryn
pu
rch
ase
s a 5
0-i
nch
p
lasm
a t
elev
isio
n f
or $
3000.
Wh
at
is i
ts
valu
e aft
er 4
yea
rs?
Rou
nd
you
r an
swer
to
th
e n
eare
st h
un
dre
d.
ab
ou
t $2200
2. M
ON
EY
Han
s op
ens
a s
avin
gs
acc
oun
t by d
epos
itin
g $
1200 i
n a
n a
ccou
nt
that
earn
s 3 p
erce
nt
inte
rest
com
pou
nd
ed
wee
kly
. H
ow m
uch
wil
l h
is i
nves
tmen
t be
wor
th i
n 1
0 y
ears
? A
ssu
me
that
ther
e are
ex
act
ly 5
2 w
eek
s in
a y
ear
an
d r
oun
d
you
r an
swer
to
the
nea
rest
ce
nt.
$1619.6
9
3. H
IGH
ER
ED
UC
AT
ION
Th
e ta
ble
lis
ts
the
aver
age
cost
s of
att
end
ing a
fou
r-yea
r co
lleg
e in
th
e U
nit
ed S
tate
s d
uri
ng t
he
2005–2006 s
choo
l yea
rs.
So
urce:
Colle
ge B
oar
d
Ru
ss’s
pare
nts
in
ves
ted
mon
ey i
n a
sa
vin
gs
acc
oun
t ea
rnin
g a
n a
ver
age
of 4
.5
per
cen
t in
tere
st,
com
pou
nd
ed m
onth
ly.
Aft
er 1
5 y
ears
, th
ey h
ave
exact
ly t
he
righ
t am
oun
t to
cov
er t
he
tuit
ion
, fe
es,
room
an
d b
oard
for
Ru
ss’s
fir
st y
ear
at
a
pu
bli
c co
lleg
e. W
hat
was
thei
r in
itia
l in
ves
tmen
t? R
oun
d y
our
an
swer
to
the
nea
rest
dol
lar.
$6182
4.P
OP
ULA
TIO
N I
n 2
007 t
he
U.S
. C
ensu
s B
ure
au
est
imate
d t
he
pop
ula
tion
of
the
Un
ited
Sta
tes
esti
mate
d a
t 301 m
illi
on.
Th
e an
nu
al
rate
of
gro
wth
is
abou
t 0.8
9%
. A
t th
is r
ate
, w
hat
is t
he
exp
ecte
d
pop
ula
tion
at
the
tim
e of
th
e 2020
cen
sus?
Rou
nd
you
r an
swer
to
the
nea
rest
ten
mil
lion
.
340 m
illio
n
5.M
ED
ICIN
E W
hen
doc
tors
pre
scri
be
med
icati
on,
they
have
to c
onsi
der
th
e ra
te a
t w
hic
h t
he
bod
y f
ilte
rs a
dru
g f
rom
th
e blo
odst
ream
. S
up
pos
e it
tak
es t
he
hu
man
bod
y 6
days
to f
ilte
r ou
t h
alf
of
the
Flu
-B-G
one
vacc
ine.
Th
e am
oun
t of
F
lu-B
-Gon
e vacc
ine
rem
ain
ing i
n t
he
blo
odst
ream
x h
ours
aft
er a
n i
nje
ctio
n i
s giv
en b
y t
he
equ
ati
on y
= y
0(0
.5)
x −
6 ,
wh
ere
y0 i
s th
e in
itia
l am
oun
t. S
up
pos
e a
doc
tor
inje
cts
a p
ati
ent
wit
h 2
0 µ
g
(mic
rogra
ms)
of
Flu
-B-G
one.
a.
How
mu
ch o
f th
e vacc
ine
wil
l re
main
aft
er 1
day?
Rou
nd
you
r an
swer
to
the
nea
rest
ten
th.
17.8
µg
b.
How
mu
ch o
f th
e vacc
ine
wil
l re
main
aft
er 1
2 d
ays?
Rou
nd
you
r an
swer
to
the
nea
rest
ten
th.
5 µ
g
c.
Aft
er h
ow m
an
y d
ays
wil
l th
e am
oun
t of
vacc
ine
be
less
th
an
1 µ
g?
aft
er
26 d
ays
Wo
rd
Pro
ble
m P
racti
ce
Gro
wth
an
d D
ecay
Co
lleg
e
Secto
r
Tu
itio
n a
nd
Fees
Ro
om
an
d
Bo
ard
Four-
year
Public
$5941
$6636
Four-
year
Private
$21,2
35
$7,7
91
Answers (Lesson 9-7)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A22 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
46
Gle
ncoe A
lgeb
ra 1
9-7
You
can
use
th
e fo
rmu
la f
or c
omp
oun
d i
nte
rest
A =
P (1
+ r
−
n ) n
t wh
en n
, th
e n
um
ber
of
tim
es
a y
ear
inte
rest
is
com
pou
nd
ed,
is k
now
n.
A s
pec
ial
typ
e of
com
pou
nd
in
tere
st c
alc
ula
tion
re
gu
larl
y u
sed
in
fin
an
ce i
s co
nti
nu
ou
sly
co
mp
ou
nd
ed
in
terest
, w
her
e n
ap
pro
ach
es
infi
nit
y.
IN
VESTIN
G M
r.
Riv
era
pla
ced
$5000 i
n a
n i
nv
est
men
t a
cco
un
t th
at
ha
s a
n i
nte
rest
ra
te o
f 9.5
% p
er y
ea
r.
Exerc
ises
1. SA
VIN
GS
Mr.
Harr
is s
aves
$20,0
00 i
n a
mon
ey-m
ark
et a
ccou
nt
at
an
in
tere
stra
te o
f 5.2
%.
a.
Det
erm
ine
the
valu
e of
his
in
ves
tmen
t aft
er 1
0 y
ears
if
inte
rest
is
com
pou
nd
ed q
uart
erly
. $33,5
28.0
1
b.
Det
erm
ine
the
valu
e of
his
in
ves
tmen
t aft
er 1
0 y
ears
if
inte
rest
is
com
pou
nd
ed c
onti
nu
ousl
y.
$33,6
40.5
4
2. C
OLLEG
E S
AV
ING
S S
han
non
is
choo
sin
g b
etw
een
tw
o d
iffe
ren
t sa
vin
gs
acc
oun
ts t
o
kee
p h
er c
olle
ge
fun
d i
n.
Th
e fi
rst
acc
oun
t co
mp
oun
ds
inte
rest
sem
ian
nu
all
y a
t a r
ate
of
11.0
%.
Th
e se
con
d a
ccou
nt
com
pou
nd
s in
tere
st c
onti
nu
ousl
y a
t a r
ate
of
10.8
%.
If
Sh
an
non
pla
ns
to k
eep
her
mon
ey i
n t
he
acc
oun
t fo
r 5 y
ears
, w
hic
h a
ccou
nt
shou
ld s
he
choo
se?
Exp
lain
. S
he s
ho
uld
ch
oo
se t
he c
on
tin
uo
us c
om
po
un
din
g a
cco
un
t at
10.8
%.
Aft
er
5 y
ears
, th
e v
alu
e o
f th
e c
on
tin
uo
usly
co
mp
ou
nd
ing
acco
un
t w
ill
be 1
.716P
, w
here
as t
he q
uart
erl
y c
om
po
un
din
g a
cco
un
t w
ill
on
ly h
ave a
valu
e o
f 1.7
08
P.
En
rich
men
t
Co
nti
nu
ou
sly
Co
mp
ou
nd
ing
In
tere
st
Genera
l fo
rmula
: A
= P
ert
A =
curr
ent
am
ount
of
investm
ent
P =
initia
l am
ount
of
investm
ent
e =
natu
ral lo
garith
m,
a c
onsta
nt
appro
xim
ate
ly e
qual to
2.7
1828
r =
annual ra
te o
f in
tere
st, e
xpre
ssed a
s a
decim
al
t =
num
ber
of
years
the m
oney is investe
d
Exam
ple
a.
Ho
w m
uch
mo
ney
wil
l b
e i
n t
he
acco
un
t a
fter 5
yea
rs
if i
nte
rest
is
co
mp
ou
nd
ed
mo
nth
ly?
A
= P
(1 +
r −
n ) n
t C
om
pound inte
rest
equation
= 5
000 (1
+ 0
.095 −
12
) 12
(5) P
= 5
000,
r =
0.0
95,
n =
12,
and t
= 5
= 5
000(1
.0079)6
0
Sim
plif
y.
= 8
025.0
5
Use a
calc
ula
tor.
Th
ere
wil
l be
abou
t $8025.0
5 i
n t
he
acc
oun
t if
in
tere
st i
s co
mp
oun
ded
mon
thly
.
b.
Ho
w m
uch
mo
ney
wil
l b
e i
n t
he
acco
un
t a
fter 5
yea
rs
if i
nte
rest
is
co
mp
ou
nd
ed
co
nti
nu
ou
sly
?
A
= P
ert
Continuous inte
rest
equation
= 5
000(2
.71828)0
.095(5
) P
= 5
000,
e =
2.7
1828,
r =
0.0
95,
and t
= 5
= 5
000(2
.71828)0
.475
Sim
plif
y.
= 8040.07
Use a
calc
ula
tor.
Th
ere
wil
l be
abou
t $8040.0
7 i
n t
he
acc
oun
t if
in
tere
st i
s co
mp
oun
ded
con
tin
uou
sly.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-7
Ch
ap
ter
9
47
Gle
ncoe A
lgeb
ra 1
9-7
Sp
read
sheet
Act
ivit
y
Co
mp
ou
nd
In
tere
st
Exerc
ises
Use
th
e s
prea
dsh
eet
of
acco
un
ts i
nv
olv
ing
co
mp
ou
nd
in
terest
.
1. H
ow a
re t
he
dou
bli
ng t
imes
aff
ecte
d i
f th
e acc
oun
ts c
omp
oun
d i
nte
rest
qu
art
erly
in
stea
d
of m
onth
ly?
Th
e 5
% a
nd
8%
acco
un
ts d
ou
ble
in
th
e s
am
e n
um
ber
of
years
as b
efo
re,
bu
t th
e 1
0%
acco
un
t w
ill
do
ub
le i
n 8
years
.
2. H
ow l
ong w
ill
it t
ak
e ea
ch a
ccou
nt
to r
each
$4000 i
f th
e in
tere
st i
s co
mp
oun
ded
m
onth
ly?
qu
art
erly
? m
on
thly
: 5%
in
28 y
r, 8
% i
n 1
8 y
r, 1
0%
in
14 y
r;
qu
art
erl
y:
5%
in
28 y
r, 8
% i
n 1
8 y
r, 1
0%
in
15 y
r
3. H
ow d
o th
e in
tere
st r
ate
an
d t
he
nu
mber
of
tim
es t
he
inte
rest
is
com
pou
nd
ed a
ffec
t th
e gro
wth
of
an
in
ves
tmen
t? T
he a
cco
un
ts w
ith
hig
her
inte
rest
rate
an
d i
nte
rest
co
mp
ou
nd
ed
mo
re o
ften
gro
w m
ore
qu
ickly
.
Ban
ks
ofte
n u
se s
pre
ad
shee
ts t
o ca
lcu
late
an
d s
tore
fin
an
cial
data
. O
ne
ap
pli
cati
on i
s to
ca
lcu
late
com
pou
nd
in
tere
st o
n a
n a
ccou
nt.
U
se a
sp
rea
dsh
eet
to f
ind
th
e t
ime i
t w
ill
tak
e a
n i
nv
est
men
t o
f $1000 t
o d
ou
ble
. S
up
po
se y
ou
ca
n c
ho
ose
fro
m i
nv
est
men
ts t
ha
t h
av
e a
nn
ua
l in
terest
ra
tes
of
5%
, 8%
, o
r 1
0%
co
mp
ou
nd
ed
mo
nth
ly.
Th
e co
mp
oun
d i
nte
rest
equ
ati
on i
s A
= P
(1 +
r −
n ) n
t , w
her
e P
is
the
pri
nci
pal
or i
nit
ial
inves
tmen
t, A
is
the
fin
al
am
oun
t of
th
e in
ves
tmen
t, r
is
the
an
nu
al
inte
rest
rate
, n
is
the
nu
mber
of
tim
es i
nte
rest
is
paid
, or
com
pou
nd
ed,
each
yea
r, a
nd
t i
s th
e n
um
ber
of
yea
rs.
In t
his
case
, P = 1
000 a
nd
n = 1
2.
Ste
p 1
U
se C
olu
mn
A o
f th
e sp
read
shee
t fo
r th
e yea
rs.
Ste
p 2
C
olu
mn
s B
, C
, an
d D
con
tain
th
e fo
rmu
las
for
the
fin
al
am
oun
ts o
f th
e in
ves
tmen
ts.
For
mat
the
cell
s in
th
ese
colu
mn
s as
curr
ency
so
that
the
am
oun
ts a
re s
how
n i
n
dol
lars
an
d c
ents
. E
ach
for
mu
la
wil
l u
se t
he
valu
es i
n C
olu
mn
A a
s th
e valu
e of
t.
For
exam
ple
, in
cel
l C
3,
the
form
ula
is
1000 *
(1 + (
0.0
8/1
2))^(1
2 * A
3).
Stu
dy t
he
spre
ad
shee
t fo
r th
e ti
mes
wh
en
each
in
ves
tmen
t ex
ceed
s $2000.
At
5%
, th
e $1000 w
ill
dou
ble
in
14 y
ears
, at
8%
it
wil
l d
ouble
in
9 y
ears
, an
d a
t 10%
it
wil
l d
ouble
in
7 y
ears
.
A1 3 4 5 6 7 8 9 10
112
BC
D
12
13
14
15
16
Years
$1,0
51.1
6
$1,1
04.9
4
$1,1
61.4
7
$1,2
20.9
0
$1,2
83.3
6
$1,3
49.0
2
$1,4
18.0
4
$1,4
90.5
9
$1,5
66.8
5
$1,6
47.0
1
$1,7
31.2
7
$1,8
19.8
5
$1,9
12.9
6
$2,0
10.8
3
5%
8%
10%
$1,0
83.0
0
$1,1
72.8
9
$1,2
70.2
4
$1,3
75.6
7
$1,4
89.8
5
$1,6
13.5
0
$1,7
47.4
2
$1,8
92.4
6
$2,0
49.5
3
$2,2
19.6
4
$2,4
03.8
7
$2,6
03.3
9
$2,8
19.4
7
$3,0
53.4
8
$1,1
04.7
1
$1,2
20.3
9
$1,3
48.1
8
$1,4
89.3
5
$1,6
45.3
0
$1,8
17.5
9
$2,0
07.9
2
$2,2
18.1
8
$2,4
50.4
5
$2,7
07.0
4
$2,9
90.5
0
$3,3
03.6
5
$3,6
49.5
8
$4,0
31.7
4
1 2 3 4 5 6 7 8 9
10
11
12
13
14
Sh
eet
1S
heet
2S
heet
3
Exam
ple
Answers (Lesson 9-7)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A23 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
48
Gle
ncoe A
lgeb
ra 1
9-8
Reco
gn
ize G
eo
metr
ic S
eq
uen
ces
A g
eom
etr
ic s
equ
en
ce i
s a s
equ
en
ce i
n w
hic
h
each
term
aft
er
the f
irst
is
fou
nd
by m
ult
iply
ing t
he p
revio
us
term
by a
non
zero
con
stan
t r
call
ed
th
e c
om
mo
n r
ati
o.
Th
e c
om
mon
rati
o c
an
be f
ou
nd
by d
ivid
ing a
ny t
erm
by i
ts
pre
vio
us
term
.
Exerc
ises
Dete
rm
ine w
heth
er e
ach
seq
uen
ce i
s a
rit
hm
eti
c,
geom
etr
ic,
or n
eit
her.
Ex
pla
in.
1. 1,
2,
4,
8,
. .
. 2. 9,
14,
6,
11,
. .
.
3.
2
−
3 ,
1
−
3 ,
1
−
6 ,
1
−
12 ,
. .
. 4. –
2,
5,
12,
19,
. .
.
Fin
d t
he n
ex
t th
ree t
erm
s i
n e
ach
geo
metr
ic s
eq
uen
ce.
5. 648,
–216,
72,
. .
. 6. 25,
–5,
1,
. .
.
7.
1
−
16 ,
1
−
2 ,
4,
. .
. 8. 72,
36,
18,
. .
.
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Geo
metr
ic S
eq
uen
ces a
s E
xp
on
en
tial
Fu
ncti
on
s
Exam
ple
1
Dete
rm
ine w
heth
er t
he
seq
uen
ce i
s a
rit
hm
eti
c,
geom
etr
ic,
or
neit
her:
21,
63,
189,
567,
. .
.
Fin
d t
he r
ati
os
of
the c
on
secu
tive t
erm
s.
If t
he r
ati
os
are
con
stan
t, t
he s
equ
en
ce i
s
geom
etr
ic.
21
63
189
567
63
−
21
=
1
89
−
63
=
5
67
−
189
= 3
Beca
use
th
e r
ati
os
are
con
stan
t, t
he s
equ
en
ce
is g
eom
etr
ic. T
he c
om
mon
rati
o i
s 3.
F
ind
th
e n
ex
t th
ree
term
s i
n t
his
geo
metr
ic s
eq
uen
cew
:
-1215,
405,
-135,
45,
. .
.
Ste
p 1
F
ind
th
e c
om
mon
rati
o.
–1215
405
–135
45
405
−
-1215
=
-135
−
405
=
45
−
-135
=
- 1
−
3
Th
e v
alu
e o
f r i
s -
1
−
3 .
Ste
p 2
M
ult
iply
each
term
by t
he c
om
mon
rati
o t
o f
ind
th
e n
ext
thre
e t
erm
s.
45
–15
5
-
5
−
3
× (-
5
−
3 )
×
(- 5
−
3 )
×
(- 5
−
3 )
Th
e n
ext
thre
e t
erm
s of
the s
equ
en
ce
are
–15,
5,
an
d -
5
−
3 .
Exam
ple
2
geo
metr
ic c
om
mo
n r
ati
o =
2;
neit
her
no
co
mm
on
rati
o o
r d
iffe
ren
ce;
geo
metr
ic co
mm
on
rati
o =
1
−
2 ;
ari
thm
eti
c c
om
mo
n d
iffe
ren
ce =
7;
-24, 8,
-2
2
−
3
32, 256, 2048
-1
−
5 ,
1
−
25 ,
-1
−
125
9, 4
1
−
2 ,
2 1
−
4
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-8
Ch
ap
ter
9
49
Gle
ncoe A
lgeb
ra 1
9-8
Geo
metr
ic S
eq
uen
ces
an
d F
un
ctio
ns
Th
e n
th t
erm
an o
f a g
eom
etr
ic s
equ
en
ce
wit
h f
irst
term
a1 a
nd
com
mon
rati
o r
is
giv
en
by t
he f
oll
ow
ing f
orm
ula
, w
here
n i
s an
y
posi
tive i
nte
ger:
an =
a1 ·
rn
– 1.
Exerc
ises
1. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce –
2,
10,
–50,
. .
. .
Fin
d t
he e
leven
th t
erm
of
this
sequ
en
ce.
2. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 5
12,
128,
32,
. .
. .
Fin
d t
he s
ixth
term
of
this
sequ
en
ce.
3. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 4
−
9 ,
4,
36,
. .
. .
Fin
d t
he e
igh
th t
erm
of
this
sequ
en
ce.
4. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 6
, –54,
486,
. .
. .
Fin
d t
he n
inth
term
of
this
sequ
en
ce.
5. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 1
00,
80,
64,
. .
. .
Fin
d t
he s
even
th t
erm
of
this
sequ
en
ce.
6. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 2
−
5 ,
1
−
10 ,
1
−
40 ,
. .
. .
Fin
d t
he s
ixth
term
of
this
sequ
en
ce.
7. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 3
−
8 ,
- 3
−
2 ,
6,
. .
. .
Fin
d t
he t
en
th t
erm
of
this
sequ
en
ce.
8. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce –
3,
–21,
–147,
. .
. .
Fin
d t
he f
ifth
term
of
this
sequ
en
ce.
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n (c
on
tin
ued
)
Geo
metr
ic S
eq
uen
ces a
s E
xp
on
en
tial
Fu
ncti
on
s
Exam
ple
a
. W
rit
e a
n e
qu
ati
on
fo
r
the n
th t
erm
of
the g
eo
metr
ic s
eq
uen
ce
5,
20,
80,
320,
. .
.
Th
e f
irst
term
of
the s
equ
en
ce i
s 320.
So,
a1 =
320.
Now
fin
d t
he c
om
mon
rati
o.
5
20
80
320
20
−
5
=
80
−
20 =
3
20
−
80
=
4
Th
e c
om
mon
rati
o i
s 4.
So,
r =
4.
an =
a1 ·
rn
– 1
Form
ula
for
nth
term
an =
5 ·
4n
– 1
a1 =
5 a
nd
r =
4
b.
Fin
d t
he s
ev
en
th t
erm
of
this
seq
uen
ce.
Beca
use
we a
re l
ook
ing f
or
the s
even
th t
erm
,
n =
7.
an =
a1 ·
rn
– 1
Form
ula
for
nth
term
a7 =
5 ·
47
– 1
n =
7
=
5 ·
46
Sim
pli
fy.
=
5 ·
4096
46 =
4096
=
20,4
80
Mu
ltip
ly.
Th
e s
even
th t
erm
of
the s
equ
en
ce i
s 20,4
80.
an =
-3 ·
7n
- 1;
-7203
an =
3
−
8 ·
(-
4)n
- 1;
-98,3
04
an =
2
−
5 ·
( 1
−
4 ) n
- 1
;
1
−
2560
an =
100 ·
( 4
−
5 ) n
- 1
; 26 1
34
−
625
an =
6 ·
(-
9)n
- 1;
258, 280, 326
an =
4
−
9 ·
9n
- 1;
2,1
25,7
64
an =
512 ·
( 1
−
4 ) n
- 1
; 1
−
2
an =
-2 ·
(-
5)n
- 1;
-19,5
31, 250
Answers (Lesson 9-8)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A24 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
50
Gle
ncoe A
lgeb
ra 1
9-8
Dete
rm
ine w
heth
er e
ach
seq
uen
ce i
s a
rit
hm
eti
c,
geom
etr
ic,
or n
eit
her.
Ex
pla
in.
1. 7,
13,
19,
25,
…
2. –96,
–48,
–24,
–12,
…
3. 108,
66,
141,
99,
…
4. 3,
9,
81,
6561,
…
5. 7
−
3 ,
14,
84,
504,
…
6. 3
−
8 ,
- 1
−
8
, -
5
−
8 ,
- 9
−
8
, …
Fin
d t
he n
ex
t th
ree t
erm
s i
n e
ach
geo
metr
ic s
eq
uen
ce.
7. 2500,
500,
100,
…
8. 2,
6,
18,
…
9.
–4,
24,
–144,
…
10. 4
−
5 ,
2
−
5 ,
1
−
5 ,
…
11.
–3,
–12,
–48,
…
12. 72,
12,
2,
…
13. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 3
, –
24,
192,
….
Fin
d t
he n
inth
term
of
this
sequ
en
ce.
14. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce
9
−
16 ,
3
−
8 ,
1
−
4 ,
….
Fin
d t
he s
even
th t
erm
of
this
sequ
en
ce.
15. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 1
000,
200,
40,
….
Fin
d t
he f
ifth
term
of
this
sequ
en
ce.
16. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce –
8,
– 2
, -
1
−
2 ,
….
Fin
d t
he e
igh
th t
erm
of
this
sequ
en
ce.
17. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 3
2,
48,
72,
….
Fin
d t
he s
ixth
term
of
this
sequ
en
ce.
18. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce
3
−
100 ,
3
−
10 ,
3,
….
Fin
d t
he n
inth
term
of
this
sequ
en
ce.
Sk
ills
Practi
ce
Geo
metr
ic S
eq
uen
ces a
s E
xp
on
en
tial
Fu
ncti
on
s
Ari
thm
eti
c;
co
mm
on
dif
fere
nce is 6
.
Geo
metr
ic;
the c
om
mo
n
rati
o is 1
−
2 .
Neit
her;
th
ere
is n
o c
om
mo
n
dif
fere
nce o
r ra
tio
.
Neit
her;
th
ere
is n
o c
om
mo
n
dif
fere
nce o
r ra
tio
.
Geo
metr
ic;
co
mm
on
rati
o is 6
.A
rith
meti
c;
co
mm
on
dif
fere
nce is -
1
−
2 .
20, 4,
4
−
5
54, 162, 486
864;
-5184;
31,1
04
-192,
-768,
-3072
1
−
10 ,
1
−
20 ,
1
−
40
1
−
3 ,
1
−
18 ,
1
−
108
3(-
8) n
- 1;
50,3
31,6
48
32
( 3
−
2 ) n
- 1 ;
243
-8
( 1
−
4 ) n
- 1 ;
-
1
−
2048
3
−
100 ·
10
n -
1;
3,0
00,0
00
9
−
16 ( 2
−
3 ) n
-1
; 4
−
81
1000
( 1
−
5 ) n
-1
; 8
−
5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-8
Ch
ap
ter
9
51
Gle
ncoe A
lgeb
ra 1
9-8
Dete
rm
ine w
heth
er e
ach
seq
uen
ce i
s a
rit
hm
eti
c,
geom
etr
ic,
or n
eit
her.
Ex
pla
in.
1. 1,
-5,
-11,
-17,
…
2. 3,
3
−
2 ,
1,
3
−
4 ,
…
3. 108,
36,
12,
4,
…
4.
-2,
4,
-6,
8,
…
Fin
d t
he n
ex
t th
ree t
erm
s i
n e
ach
geo
metr
ic s
eq
uen
ce.
5. 64,
16,
4,
…
6. 2,
-12,
72,
…
7. 3750,
750,
150,
…
8. 4,
28,
196,
…
9. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 8
96,
-448,
224,
… .
Fin
d t
he e
igh
th t
erm
of
this
sequ
en
ce.
10. W
rite
an
equ
ati
on
for
the n
th t
erm
of
the g
eom
etr
ic s
equ
en
ce 3
584,
896,
224,
… .
Fin
d t
he s
ixth
term
of
this
sequ
en
ce.
11. F
ind
th
e s
ixth
term
of
a g
eom
etr
ic s
equ
en
ce f
or
wh
ich
a2 =
288 a
nd
r =
1
−
4 .
12. F
ind
th
e e
igh
th t
erm
of
a g
eom
etr
ic s
equ
en
ce f
or
wh
ich
a3 =
35 a
nd
r =
7.
13. PEN
NIE
S T
hom
as
is s
avin
g p
en
nie
s in
a j
ar.
Th
e f
irst
day h
e s
aves
3 p
en
nie
s, t
he
seco
nd
day 1
2 p
en
nie
s, t
he t
hir
d d
ay 4
8 p
en
nie
s, a
nd
so o
n.
How
man
y p
en
nie
s d
oes
Th
om
as
save o
n t
he e
igh
th d
ay?
Practi
ce
Geo
metr
ic S
eq
uen
ces a
nd
Fu
ncti
on
s
Ari
thm
eti
c;
the c
om
mo
n
dif
fere
nce is -
6.
Neit
her;
th
ere
is n
o c
om
mo
n
dif
fere
nce o
r ra
tio
.
Geo
metr
ic;
the c
om
mo
n
rati
o is 1
−
3 .
Neit
her;
th
ere
is n
o c
om
mo
n
dif
fere
nce o
r ra
tio
.
1;
1
−
4 ;
1
−
16
-432;
2592;
-15,5
52
30;
6;
6
−
5
1372;
9604;
67,2
28
896
−
(-2)n
-1 ,
-7
3584
−
4n
-1 ,
3.5
1.1
25
588,2
45
49,1
52
Answers (Lesson 9-8)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A25 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
52
Gle
ncoe A
lgeb
ra 1
9-8
1. W
OR
LD
PO
PU
LA
TIO
N T
he
CIA
es
tim
ate
s th
e w
orld
pop
ula
tion
is
gro
win
g a
t a r
ate
of
1.1
67%
each
yea
r.
Th
e w
orld
pop
ula
tion
for
2007 w
as
abou
t 6.6
bil
lion
.
a.
Wri
te a
n e
qu
ati
on f
or t
he
wor
ld
pop
ula
tion
aft
er n
yea
rs.
(Hin
t: T
he
com
mon
rati
o is
not
ju
st 0
.01167.)
b.
Wh
at
wil
l th
e es
tim
ate
d w
orld
p
opu
lati
on b
e in
2017?
2. M
USEU
MS
Th
e ta
ble
sh
ows
the
an
nu
al
vis
itor
s to
a m
use
um
in
mil
lion
s. W
rite
an
equ
ati
on f
or t
he
pro
ject
ed n
um
ber
of
vis
itor
s aft
er n
yea
rs.
3. B
AN
KIN
G A
rnol
d h
as
a b
an
k a
ccou
nt
wit
h a
beg
inn
ing b
ala
nce
of
$5000.
He
spen
ds
one-
fift
h o
f th
e bala
nce
each
m
onth
. H
ow m
uch
mon
ey w
ill
be
in t
he
acc
oun
t aft
er 6
mon
ths?
4. PO
PU
LA
TIO
N T
he
table
sh
ows
the
pro
ject
ed p
opu
lati
on o
f th
e U
nit
ed S
tate
s th
rou
gh
2050.
Doe
s th
is t
able
sh
ow a
n
ari
thm
etic
seq
uen
ce,
2 g
eom
etri
c se
qu
ence
or
nei
ther
? E
xp
lain
.
So
urce:
U.S
. C
ensus B
ure
au
5. SA
VIN
GS A
CC
OU
NTS
A b
an
k o
ffer
s a
savin
gs
acc
oun
t w
ith
a 0
.5%
ret
urn
each
m
onth
.
a.
Wri
te a
n e
qu
ati
on f
or t
he
bala
nce
of
a
savin
gs
acc
oun
t aft
er n
mon
ths.
(H
int:
T
he
com
mon
rati
o is
not
ju
st 0
.005.)
b.
Giv
en a
n i
nit
ial
dep
osit
of
$500,
wh
at
wil
l th
e acc
oun
t bala
nce
be
aft
er 1
5
mon
ths?
Wo
rd
Pro
ble
m P
racti
ce
Geo
metr
ic S
eq
uen
ces a
s E
xp
on
en
tial
Fu
ncti
on
s
Year
Vis
ito
rs
(millio
ns)
14
26
39
413 1
−
2
n?
Year
Pro
jecte
d
Po
pu
lati
on
2000
282,1
25,0
00
2010
308,9
36,0
00
2020
335,8
05,0
00
2030
363,5
84,0
00
2040
391,9
46,0
00
2050
419,8
54,0
00
an =
6,6
00,0
00,0
00 ·
1.0
1167 n
- 1
ab
ou
t 7.4
billio
n
an =
4 ·
( 3
−
2 ) n
- 1
$1310.7
2
Neit
her,
th
ere
is n
o c
om
mo
n r
ati
oo
r d
iffe
ren
ce.
an =
p ·
1.0
05
n
$538.8
4
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-8
Ch
ap
ter
9
53
Gle
ncoe A
lgeb
ra 1
9-8
Pay i
t Fo
rward
Th
e id
ea b
ehin
d “
pay i
t fo
rward
” is
th
at
on t
he
firs
t d
ay,
one
per
son
doe
s a g
ood
dee
d f
or
thre
e d
iffe
ren
t p
eop
le.
Th
en,
on t
he
seco
nd
day,
thos
e th
ree
peo
ple
wil
l ea
ch p
erfo
rm g
ood
d
eed
s fo
r 3 m
ore
peo
ple
, so
th
at
on D
ay 2
, th
ere
are
3 ×
3 o
r 9 g
ood
dee
ds
bei
ng d
one.
C
onti
nu
e th
is p
roce
ss t
o fi
ll i
n t
he
chart
. A
tre
e d
iagra
m w
ill
hel
p y
ou f
ill
in t
he
chart
.
1. G
rap
h t
he
data
you
fou
nd
in
th
e ch
art
as
ord
ered
pair
s an
d c
onn
ect
wit
h a
sm
ooth
cu
rve.
2. W
hat
typ
e of
fu
nct
ion
is
you
r gra
ph
fro
m p
roble
m 1
? W
rite
an
equ
ati
on t
hat
can
be
use
d t
o d
eter
min
e th
e n
um
ber
of
goo
d d
eed
s on
an
y g
iven
day,
x.
exp
on
en
tial;
y =
3x
3. H
ow m
an
y g
ood
dee
ds
wil
l be
per
form
ed o
n D
ay 2
1?
10,4
60,3
53,2
03
Th
e fo
rmu
la,
3 n
+1 -
3
−
2
, ca
n b
e u
sed
to
det
erm
ine
the
to
ta
l n
um
ber
of
goo
d d
eed
s th
at
have
bee
n p
erfo
rmed
, w
her
e n
rep
rese
nts
th
e d
ay.
For
exam
ple
, on
Day 2
, th
ere
have
bee
n 3
+ 9
or 1
2 g
ood
dee
ds
per
form
ed.
Usi
ng t
he
form
ula
, you
get
, 3
2+
1 -
3
−
2
or
3 3 -
3
−
2
+ 2
7 -
3
−
2
or
a
tota
l of
12 g
ood
dee
ds
per
form
ed.
4. U
se t
his
for
mu
la t
o d
eter
min
e th
e ap
pro
xim
ate
nu
mber
of
goo
d d
eed
s th
at
have
bee
n
per
form
ed t
hro
ugh
Day 2
1.
15,6
90,5
29,8
00
5. L
ook
up
th
e w
orld
pop
ula
tion
. H
ow d
oes
you
r n
um
ber
fro
m E
xer
cise
4 c
omp
are
to
the
wor
ld p
opu
lati
on?
Th
e w
orl
d p
op
ula
tio
n i
s a
pp
roxim
ate
ly 6
,560,5
26,0
46.
Th
e n
um
ber
of
go
od
deed
s p
erf
orm
ed
is g
reate
r th
an
th
e e
nti
re w
orl
d p
op
ula
tio
n.
Pers
on
1
Da
y 2
Da
y 1
New
Pers
on
2
New
Pers
on
1
New
Pers
on
3
y
xO
240
210
180
120
150
90
60
30
43
27
86
51
Day
# o
f
Deed
s
01
13
29
327
481
5243
En
ric
hm
en
t
Answers (Lesson 9-8)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A26 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
54
Gle
ncoe A
lgeb
ra 1
9-9
Iden
tify
Fu
nct
ion
s L
inear
fun
ctio
ns,
qu
ad
rati
cfu
nct
ion
s, a
nd
exp
on
en
tial
fun
ctio
ns
can
all
be
use
d t
o m
od
el
data
. T
he g
en
era
l fo
rms
of
the
equ
ati
on
s are
lis
ted
belo
w.
You
can
als
o i
den
tify
data
as
lin
ear,
qu
ad
rati
c, o
r exp
on
en
tial
base
d o
n p
att
ern
s of
beh
avio
r of
their
y-v
alu
es.
G
ra
ph
th
e s
et
of
ord
ered
p
air
s {
(–3,
2),
(–2,
–1),
(–1,
–2),
(0,
–1),
(1
, 2)}
. D
ete
rm
ine w
heth
er t
he o
rd
ered
p
air
s r
ep
resen
t a
lin
ea
r f
un
cti
on
, a
q
ua
dra
tic f
un
cti
on
, o
r a
n e
xp
on
en
tia
l
fun
cti
on
.
Th
e o
rdere
d p
air
s ap
pear
to r
ep
rese
nt
a
qu
ad
rati
c fu
nct
ion
.
L
oo
k f
or a
pa
ttern
in
th
e
tab
le t
o d
ete
rm
ine w
hic
h m
od
el
best
describ
es t
he d
ata
.
Sta
rt b
y c
om
pari
ng t
he f
irst
dif
fere
nce
s.
Th
e f
irst
dif
fere
nce
s are
not
all
equ
al.
Th
e
table
does
not
rep
rese
nt
a l
inear
fun
ctio
n.
Fin
d t
he s
eco
nd
dif
fere
nce
s an
d c
om
pare
.
Th
e t
able
does
not
rep
rese
nt
a q
uad
rati
c
fun
ctio
n.
Fin
d t
he r
ati
os
of
the y
-valu
es.
Th
e r
ati
os
are
equ
al.
Th
ere
fore
, th
e t
able
ca
n b
e m
od
ele
d b
y a
n e
xp
on
en
tial
fun
ctio
n.
Exam
ple
1
x
y
Exam
ple
2
x–2
–1
01
2
y4
21
0.5
0.2
5
-2
-
1
-
0.5
-0.2
5
+ 1
+ 0
.5
+ 0
.25
4
2
1
0.5
0.2
5
× 0
.5
×
0.5
× 0
.5
× 0
.5
4
2
1
0.5
0.2
5
-2
-1
-
0.5
-
0.2
5
Exerc
ises
Gra
ph
ea
ch
set
of
ord
ered
pa
irs.
Dete
rm
ine w
heth
er t
he o
rd
ered
pa
irs r
ep
resen
t a
lin
ea
r f
un
cti
on
, a
qu
ad
ra
tic f
un
cti
on
, o
r a
n e
xp
on
en
tia
l fu
ncti
on
.
1. (0
, –1),
(1, 1),
(2, 3),
(3, 5)
2. (–
3, –1),
(–2, –4),
(–1, –5),
(0, –4),
(1, –1)
x
y
x
y
Lo
ok
fo
r a
pa
ttern
in
ea
ch
ta
ble
to
dete
rm
ine w
hic
h m
od
el
best
describ
es t
he d
ata
.
3.
x–2
–1
01
2
y6
54
32
4.
x–2
–1
01
2
y6.2
52.5
10.4
0.1
6
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
An
aly
zin
g F
un
ctio
ns
with
Su
ccess
ive D
iffe
ren
ces
an
d R
atio
s
Lin
ear
Function
y
= m
x +
b
Quadra
tic F
unction
y
= a
x2 +
bx
+ c
Exponential F
unction
y
= a
bx
lin
ear
qu
ad
rati
c
lin
ear
exp
on
en
tial
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-9
Ch
ap
ter
9
55
Gle
ncoe A
lgeb
ra 1
9-9
Wri
te E
qu
ati
on
s O
nce
you
fin
d t
he m
od
el
that
best
desc
ribes
the d
ata
, you
can
wri
te
an
equ
ati
on
for
the f
un
ctio
n.
Basic
Fo
rms
Lin
ear
Function
y
= m
x +
b
Quadra
tic F
unction
y
= a
x2
Exponential F
unction
y
= a
bx
D
ete
rm
ine w
hic
h m
od
el
best
describ
es t
he d
ata
. T
hen
writ
e a
n
eq
ua
tio
n f
or t
he f
un
cti
on
th
at
mo
dels
th
e d
ata
.
x 0
1 2
34
y3
61
224
48
Ste
p 1
D
ete
rmin
e w
heth
er
the d
ata
is
mod
ele
d b
y a
lin
ear,
qu
ad
rati
c, o
r exp
on
en
tial
fun
ctio
n.
Fir
st d
iffe
ren
ces:
3
6
12
24
48
+
3
+
6
+
12
+
24
Seco
nd
dif
fere
nce
s:
3
6
12
24
+
3
+ 6
+ 1
2
y-v
alu
e r
ati
os:
3
6
12
24
48
×
2
× 2
× 2
× 2
Th
e r
ati
os
of
succ
ess
ive y
-valu
es
are
equ
al.
Th
ere
fore
, th
e t
able
of
valu
es
can
be m
od
ele
d b
y a
n e
xp
on
en
tial
fun
ctio
n.
Ste
p 2
W
rite
an
equ
ati
on
for
the f
un
ctio
n t
hat
mod
els
th
e d
ata
. T
he
equ
ati
on
has
the f
orm
y =
ab
x.
Th
e y
-valu
e r
ati
o i
s 2,
so t
his
is
the
valu
e o
f th
e b
ase
.
y =
ab
x
Equation f
or
exponential fu
nction
3 =
a(2
)o
x =
0,
y =
5,
and b
= 2
3 =
a
Sim
plif
y.
An
equ
ati
on
th
at
mod
els
th
e d
ata
is
y =
3.2
x.
To c
heck
th
e r
esu
lts,
you
can
veri
fy t
hat
the
oth
er
ord
ere
d p
air
s sa
tisf
y t
he f
un
ctio
n.
Exerc
ises
Lo
ok
fo
r a
pa
ttern
in
ea
ch
ta
ble
of
va
lues t
o d
ete
rm
ine w
hic
h m
od
el
best
describ
es t
he d
ata
. T
hen
writ
e a
n e
qu
ati
on
fo
r t
he f
un
cti
on
th
at
mo
dels
th
e d
ata
.
1.
x–2
–1
01
2
y12
30
31
2
qu
ad
rati
c;
y=
3x
2
2.
x–1
01
23
y–2
14
710
lin
ear;
y =
3x
+ 1
3.
x–1
01
23
y0.7
53
12
48
192
exp
on
en
tial;
y =
3.4
x
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n(c
on
tin
ued
)
An
aly
zin
g F
un
cti
on
s w
ith
Su
ccessiv
e D
iffe
ren
ces a
nd
Rati
os
Exam
ple
Answers (Lesson 9-9)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he M
cG
raw
-Hill C
om
pan
ies,
Inc.
Chapter 9 A27 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
56
Gle
ncoe A
lgeb
ra 1
9-9
Gra
ph
ea
ch
set
of
ord
ered
pa
irs.
Dete
rm
ine w
heth
er t
he o
rd
ered
pa
irs r
ep
resen
t
a linear f
un
cti
on
, a
quadratic f
un
cti
on
, o
r a
n exponential
fun
cti
on
.
1. (2
, 3),
(1,
1),
(0, –1),
(–1, –3),
(–3, –5)
2.
(–1,
0.5
), (
0,
1),
(1,
2),
(2,
4)
x
y
l
inear
x
y
exp
on
en
tial
3. (–
2,
4),
(–1,
1),
(0,
0),
(1,
1),
(2,
4)
4.
(–3,
5),
(–2,
2),
(–1,
1),
(0,
2),
(1,
5)
x
y
q
uad
rati
c
x
y
q
uad
rati
c
Lo
ok
fo
r a
pa
ttern
in
ea
ch
ta
ble
of
va
lues t
o d
ete
rm
ine w
hic
h m
od
el
best
describ
es t
he d
ata
. T
hen
writ
e a
n e
qu
ati
on
fo
r t
he f
un
cti
on
th
at
mo
dels
th
e d
ata
.
5.
x–3
–2
–1
01
2
y–32
16
84
21
exp
on
en
tial;
y =
4 ·
(0.5
)x
6.
x–1
01
23
y7
3–1
–5
–9
lin
ear;
y =
-4x
+ 3
7.
x–3
–2
–1
01
y–27
–12
–3
0–3
q
uad
rati
c;
y =
-3
x2
8.
x0
12
34
y0.5
1.5
4.5
13.5
40.5
exp
on
en
tial;
y =
1 −
2 ·
3x
9.
x–2
–1
01
2
y–8
–4
04
8
lin
ear;
y =
4x
Sk
ills
Practi
ce
An
aly
zin
g F
un
cti
on
s w
ith
Su
ccessiv
e D
iffe
ren
ces a
nd
Rati
os
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-9
Ch
ap
ter
9
57
Gle
ncoe A
lgeb
ra 1
9-9
Practi
ce
An
aly
zin
g F
un
cti
on
s w
ith
Su
ccessiv
e D
iffe
ren
ces a
nd
Rati
os
Gra
ph
ea
ch
set
of
ord
ered
pa
irs.
Dete
rm
ine w
heth
er t
he o
rd
ered
pa
irs r
ep
resen
t
a linear f
un
cti
on
, a
quadratic f
un
cti
on
, o
r a
n exponential
fun
cti
on
.
1. (4
, 0.5
), (
3,
1.5
), (
2,
2.5
), (
1,
3.5
), (
0,
4.5
) 2. (–1
, 1
−
9 ) ,
(0,
1 −
3 ) ,
(1,
1),
(2,
3)
x
y
lin
ear
x
y
exp
on
en
tial
3. (–
4,
4),
(–2,
1),
(0,
0),
(2,
1),
(4,
4)
4.
(–4,
2),
(–2,
1),
(0,
0),
(2, –1),
(4, –2)
x
y
q
uad
rati
c
x
y
l
inear
Lo
ok
fo
r a
pa
ttern
in
ea
ch
ta
ble
of
va
lues t
o d
ete
rm
ine w
hic
h m
od
el
best
describ
es t
he d
ata
. T
hen
writ
e a
n e
qu
ati
on
fo
r t
he f
un
cti
on
th
at
mo
dels
th
e d
ata
.
5.
x –
3 –
11
35
y–5
–2
1
4 7
lin
ear;
y =
3 −
2 x
- 1
−
2
6.
x–2
–1
01
2
y0.0
20.2
220
200
exp
on
en
tial;
y =
2! 1
0x
7.
x–1
01
23
y6
06
24
54
q
uad
rati
c;
y =
6x
2
8.
x–2
–1
01
2
y18
90
–9
–18
lin
ear;
y =
-9
x
9. IN
SEC
TS
T
he l
oca
l zoo k
eep
s tr
ack
of
the n
um
ber
of
dra
gon
flie
s bre
ed
ing i
n t
heir
in
sect
exh
ibit
each
day.
Day
12
34
5
Dra
go
nfl
ies
918
36
72
144
a.
Dete
rmin
e w
hic
h f
un
ctio
n b
est
mod
els
th
e d
ata
. exp
on
en
tial
b
. W
rite
an
equ
ati
on
for
the f
un
ctio
n t
hat
mod
els
th
e d
ata
. y =
9 !
2x
c.
Use
you
r equ
ati
on
to d
ete
rmin
e t
he n
um
ber
of
dra
gon
flie
s th
at
wil
l be b
reed
ing a
fter
9 d
ays.
4608 d
rag
on
flie
s
Answers (Lesson 9-9)
Co
pyrig
ht ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pan
ies, In
c.
Chapter 9 A28 Glencoe Algebra 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
9
58
Gle
ncoe A
lgeb
ra 1
9-9
Wo
rd
Pro
ble
m P
racti
ce
An
aly
zin
g F
un
cti
on
s w
ith
Su
ccessiv
e D
iffe
ren
ces a
nd
Rati
os
1. W
EA
TH
ER
T
he S
an
Mate
o w
eath
er
stati
on
reco
rds
the a
mou
nt
of
rain
fall
si
nce
th
e b
egin
nin
g o
f a t
hu
nd
ers
torm
. D
ata
for
a s
torm
is
reco
rded
as
a s
eri
es
of
ord
ere
d p
air
s sh
ow
n b
elo
w,
wh
ere
th
e
x v
alu
e i
s th
e t
ime i
n m
inu
tes
sin
ce t
he
start
of
the s
torm
, an
d t
he y
valu
e i
s th
e
am
ou
nt
of
rain
in
in
ches
that
has
fall
en
si
nce
th
e s
tart
of
the s
torm
.
(2,
0.3
), (
4,
0.6
), (
6,
0.9
), (
8,
1.2
), (
10,
1.5
)
Gra
ph
th
e o
rdere
d p
air
s. D
ete
rmin
e
wh
eth
er
the o
rdere
d p
air
s re
pre
sen
t a
lin
ear
fun
ctio
n,
a q
ua
dra
tic
fun
ctio
n,
or
an
exp
on
enti
al
fun
ctio
n.
Total Rainfall (in.)
0.2 0
0.4
0.6
0.8
1.0
1.2
1.4
1.6
Tim
e (m
in)
24
68
10
12
2. IN
VESTIN
G T
he v
alu
e o
f a c
ert
ain
parc
el
of
lan
d h
as
been
in
creasi
ng i
n v
alu
e e
ver
sin
ce i
t w
as
pu
rch
ase
d.
Th
e t
able
sh
ow
s th
e v
alu
e o
f th
e l
an
d p
arc
el
over
tim
e.
Year
Sin
ce
Pu
rch
asin
g0
12
34
Lan
d V
alu
e
(th
ou
san
ds $
)$1.0
5$2.1
0$4.2
0$8.4
0$16.8
0
Look
for
a p
att
ern
in
th
e t
able
of
valu
es
to d
ete
rmin
e w
hic
h m
od
el
best
desc
ribes
the d
ata
. T
hen
wri
te a
n e
qu
ati
on
for
the
fun
ctio
n t
hat
mod
els
th
e d
ata
.
3. B
OA
TS
T
he v
alu
e o
f a b
oat
typ
icall
y
dep
reci
ate
s over
tim
e.
Th
e t
able
sh
ow
s th
e v
alu
e o
f a b
oat
over
a p
eri
od
of
tim
e.
Years
01
23
4
Bo
at
Valu
e (
$)
8250
6930
5821.2
04889.8
14107.4
4
Wri
te a
n e
qu
ati
on
for
the f
un
ctio
n t
hat
mod
els
th
e d
ata
. T
hen
use
th
e e
qu
ati
on
to
dete
rmin
e h
ow
mu
ch t
he b
oat
is w
ort
h
aft
er
9 y
ears
.
4. N
UC
LEA
R W
ASTE
R
ad
ioact
ive m
ate
rial
slow
ly d
eca
ys
over
tim
e.
Th
e a
mou
nt
of
tim
e n
eed
ed
for
an
am
ou
nt
of
rad
ioact
ive
mate
rial
to d
eca
y t
o h
alf
its
in
itia
l qu
an
tity
is
kn
ow
n a
s it
s h
alf
-lif
e.
Con
sid
er
a 2
0-g
ram
sam
ple
of
a
rad
ioact
ive i
soto
pe.
Half
-Liv
es
Ela
psed
01
23
4
Am
ou
nt
of
Iso
top
e
Rem
ain
ing
(g
ram
s)
20
10
52.5
1.2
5
a.
Is r
ad
ioact
ive d
eca
y a
lin
ear
deca
y,
qu
ad
rati
c d
eca
y,
or
an
exp
on
enti
al
deca
y?
b.
Wri
te a
n e
qu
ati
on
to d
ete
rmin
e h
ow
m
an
y g
ram
s y o
f a r
ad
ioact
ive i
soto
pe
wil
l be r
em
ain
ing a
fter
x h
alf
-liv
es.
c.
How
man
y g
ram
s of
the i
soto
pe w
ill
rem
ain
aft
er
11 h
alf
-liv
es?
d.
Plu
ton
ium
-238 i
s on
e o
f th
e m
ost
d
an
gero
us
wast
e p
rod
uct
s of
nu
clear
pow
er
pla
nts
. If
th
e h
alf
-lif
e o
f p
luto
niu
m-2
38 i
s 87.7
years
, h
ow
lo
ng w
ou
ld i
t ta
ke f
or
a 2
0-g
ram
sa
mp
le o
f p
luto
niu
m-2
38 t
o d
eca
y
to 0
.078 g
ram
s?
exp
on
en
tial
y =
20
(
0.5
)x
exp
on
en
tial;
y =
1.0
5
2n
701.6
years
y =
8250
(0.8
4)x
; $1717.7
8
lin
ear
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 9-9
Ch
ap
ter
9
59
Gle
ncoe A
lgeb
ra 1
Sie
rpin
ski
Tri
an
gle
Sie
rpin
ski
Tri
an
gle
is
an
exam
ple
of
a f
ract
al
that
chan
ges
exp
on
en
tiall
y.
Sta
rt w
ith
an
equ
ilate
ral
tria
ngle
an
d f
ind
th
e m
idp
oin
ts o
f each
sid
e.
Th
en
con
nect
th
e m
idp
oin
ts t
o
form
a s
mall
er
tria
ngle
. R
em
ove t
his
sm
all
er
tria
ngle
fro
m t
he l
arg
er
on
e.
Rep
eat
the p
roce
ss t
o c
reate
th
e n
ext
tria
ngle
in
th
e s
equ
en
ce.
Fin
d t
he m
idp
oin
ts o
f th
e s
ides
of
the t
hre
e r
em
ain
ing t
rian
gle
s an
d c
on
nect
th
em
to f
orm
sm
all
er
tria
ngle
s to
be r
em
oved
.
1. F
ind
th
e n
ext
tria
ngle
in
th
e s
equ
en
ce.
How
mu
ch h
as
been
cu
t ou
t? W
hat
is t
he a
rea o
f th
e f
ou
rth
fig
ure
in
th
e s
equ
en
ce?
1
−
4 +
3
−
16 +
9
−
64 o
r 3
7
−
64 ;
1
−
4
1 -
37
−
64 o
r 2
7
−
64
2. M
ak
e a
con
ject
ure
as
to w
hat
you
need
to m
ult
iply
th
e p
revio
us
am
ou
nt
cut
by t
o f
ind
the n
ew
am
ou
nt
cut.
3
−
4
3. F
ill
in t
he c
hart
to r
ep
rese
nt
the a
mou
nt
cut
an
d t
he a
rea r
em
ain
ing f
or
each
tri
an
gle
in
th
e s
equ
en
ce. fig
ure
12
34
56
am
ou
nt
cu
t0
1
−
4
7
−
16
37
−
64
175
−
256
781
−
1024
are
a r
em
ain
ing
1 3
−
4
9
−
16
27
−
64
81
−
256
243
−
1024
4. W
rite
an
equ
ati
on
to r
ep
rese
nt
the a
rea t
hat
is l
eft
in
th
e n
th t
rian
gle
in
th
e s
equ
en
ce.
( 3
−
4 ) n
5. If
th
is p
roce
ss i
s co
nti
nu
ed
, m
ak
e a
con
ject
ure
as
to t
he r
em
ain
ing a
rea.
Th
e r
em
ain
ing
are
a g
ets
very
clo
se t
o 0
.
fig
ure
2fig
ure
3fig
ure
1
Cu
t o
ut:
1 4
Are
a=
1-
o
r1 4
3 4
Cu
t o
ut:
1 4
3 16
Are
a=
1-
o
r7 16
9 16
7 16
+o
r
9-9
En
ric
hm
en
t
Answers (Lesson 9-9)
ERROR: undefined
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