y 1.1 graphs the rectangular coordinate system: ( x y · the rectangular coordinate system: (x,...
Post on 05-Jul-2020
5 Views
Preview:
TRANSCRIPT
S
ecti
on
1.1
G
rap
hs
5
5
1.1
G
rap
hs
M
uch
of
alg
ebra
is
con
cern
ed w
ith
so
lvin
g e
qu
atio
ns.
M
any
alg
ebra
ic
tech
niq
ues
hav
e b
een
dev
elo
ped
to
pro
vid
e in
sig
hts
in
to v
ario
us
sort
s o
f eq
uat
ion
s,
and
th
ose
tec
hn
iqu
es a
re e
ssen
tial
fo
r u
nd
erst
and
ing
th
e b
asic
id
eas
of
calc
ulu
s an
d
mo
re a
dv
ance
d m
ath
emat
ical
to
pic
s.
In r
ecen
t d
ecad
es p
eop
le h
ave
also
use
d c
alcu
lato
rs a
nd
com
pu
ters
fo
r ca
lcula
tin
g a
pp
rox
imat
e so
luti
ons
to c
erta
in t
yp
es o
f eq
uat
ion
s. S
uch
nu
mer
ical
so
luti
on
met
ho
ds
are
oft
en e
ssen
tial
fo
r so
lvin
g r
eal
pro
ble
ms
in e
ng
inee
rin
g a
nd
sc
ien
ce. I
n t
his
co
urs
e w
e in
ves
tig
ate
pro
ble
ms
bo
th a
lgeb
raic
ally
an
d g
rap
hic
ally
.
It i
s o
ften
use
ful
to c
on
sid
er a
sit
uat
ion
fro
m s
ever
al v
iew
po
ints
: a
dif
fere
nt
dir
ecti
on
can
pro
vid
e in
sig
hts
th
at a
re n
ot
ob
vio
us
fro
m t
he
ori
gin
al p
osi
tio
n.
Th
e g
eom
etri
cal
app
roac
h t
o s
olv
ing
eq
uat
ion
s h
as b
een
mad
e m
ore
p
rod
uct
ive
by
th
e in
ven
tio
n o
f th
e g
rap
hin
g c
alcu
lato
r.
Th
ese
calc
ula
tors
qu
ick
ly
con
stru
ct g
rap
hs
that
are
fai
rly
acc
ura
te a
nd
rel
iab
le. I
n p
re-c
alcu
lato
r y
ears
, th
e g
rap
hic
al a
pp
roac
h w
as m
uch
mo
re t
edio
us.
In
th
e n
ext
two
sec
tio
ns
(1.2
an
d 1
.3)
we
wil
l ex
plo
re v
ario
us
met
ho
ds
to
solv
e eq
uat
ion
s g
rap
hic
ally
. (
It i
s as
sum
ed t
hat
we
alre
ady
kn
ow
th
e al
geb
raic
m
eth
od
s. T
ho
se m
eth
od
s ar
e o
utl
ined
in
sec
tio
ns
2.1
an
d 2
.2 b
ut
we
wil
l n
ot
spen
d
tim
e to
rev
iew
th
em n
ow
.) In
th
is s
ecti
on
, w
e w
ill
dis
cuss
th
e re
ctan
gu
lar
coo
rdin
ate
syst
em, p
rop
erti
es o
f g
rap
hs,
an
d h
ow
to
use
th
e ca
lcu
lato
r to
dis
pla
y a
g
rap
h.
Th
e R
ecta
ng
ula
r C
oo
rdin
ate
Sy
stem
: (x
, y)
-pla
ne
Th
e re
ctan
gu
lar
coord
inat
e sy
stem
is
a sy
stem
fo
r la
bel
ing
poin
ts i
n t
he
pla
ne.
T
he
coo
rdin
ate
pla
ne
is c
on
stru
cted
fro
m t
wo
nu
mb
er l
ines
, th
e x-
axis
an
d
the
y-ax
is. T
he
x-ax
is a
nd
th
e y-
axis
are
per
pen
dic
ula
r to
eac
h o
ther
an
d i
nte
rsec
t at
the
po
int
0 o
n b
oth
lin
es. (
Per
pen
dic
ula
r m
ean
s th
at t
he
two
lin
es m
eet
at 9
0�
ang
les.
) T
he
po
int
of
inte
rsec
tio
n i
s ca
lled
th
e o
rig
in. S
ee F
igu
re 1
bel
ow
.
y-ax
is
III
ori
gin
x-
axis
III
IV
F
igu
re 1
56
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
Th
e p
lan
e is
div
ided
in
to f
ou
r q
uad
ran
ts, la
bel
ed I
, II
, II
I, a
nd I
V a
s in
Fig
ure
1.
Th
ese
inte
rsec
tin
g n
um
ber
lin
es a
llo
w u
s to
rep
rese
nt
each
po
int
in t
he
pla
ne
by
an
o
rder
ed-p
air
wri
tten
as
(x,
y). E
ach
po
int
of
the
pla
ne
is e
ith
er o
n a
n a
xis
or
in o
ne
of
the
qu
adra
nts
. T
his
co
ord
inat
e sy
stem
pro
vid
es a
maj
or
ben
efit
: it
all
ow
s u
s to
so
lve
geo
met
rica
l p
rob
lem
s u
sin
g a
lgeb
ra. T
wo
use
ful
feat
ure
s ar
e th
e fo
rmu
las
to
com
pu
te t
he
dis
tan
ce b
etw
een
tw
o p
oin
ts a
nd
fo
r fi
nd
ing
mid
po
ints
of
lin
e se
gm
ents
.
Dis
tan
ce F
orm
ula
Th
e D
ista
nce
Fo
rmu
la s
tate
s th
at t
he
dis
tance
d b
etw
een
th
e po
ints
��
11,
yx
and
i
s g
iven
by
th
e fo
rmu
la:
�2
2,
yx
�
�
��
�2
21
2
21
yy
xx
d�
��
�
T
he
der
ivat
ion
of
the
dis
tan
ce f
orm
ula
use
s th
e P
yth
ago
rean
Th
eore
m a
nd
is
incl
ud
ed a
s an
ex
erci
se a
t th
e en
d o
f th
e se
ctio
n.
E
xa
mp
le 1
(D
ista
nce
Fo
rmu
la)
Fin
d t
he
dis
tan
ce b
etw
een
th
e p
oin
ts (
28
, 7
6)
and
(5
9, 3
2).
S
olu
tio
n:
To
so
lve
this
pro
ble
m w
e u
se t
he
dis
tan
ce f
orm
ula
an
d l
et (
59
, 3
2)
be
an
d (
28
, 7
6)
be
�1
1,
yx
��
�2
2,
yx
.
� �
��
�2
276
32
28
59
��
��
d
Subst
itute
the
val
ues
into
the
form
ula
.
��
��2
24
43
1�
��
d
�
28
97
�d
�
d =
53
.82
37
86
… (
an i
nfi
nit
e d
ecim
al)
�
No
te t
hat
2
89
7 i
s th
e ex
act
answ
er b
ut
it i
s o
ften
mo
re u
sefu
l to
hav
e an
app
rox
imat
e (r
ou
nd
ed o
ff)
answ
er.
So
, af
ter
rou
nd
ing
to
tw
o d
ecim
al
pla
ces,
th
e dis
tance
bet
wee
n t
he
two
po
ints
is
appro
xim
atel
y 5
3.8
2. �
Mid
po
int
Fo
rmu
la
Th
e m
idp
oin
t o
f a
lin
e se
gm
ent
is t
he
po
int
that
is
hal
f-w
ay b
etw
een
th
e en
dp
oin
ts. T
hat
is,
it
is t
he
on
ly p
oin
t o
n t
he
lin
e se
gm
ent
that
is
the
sam
e d
ista
nce
fr
om
eac
h e
nd
. O
n a
nu
mb
er l
ine,
the
nu
mb
er t
hat
is
hal
fway
bet
wee
n t
wo
nu
mb
ers
is t
he
aver
age
of
the
two
nu
mb
ers.
F
or
inst
ance
, to
fin
d t
he
nu
mb
er h
alfw
ay
bet
wee
n a
and b
, w
e co
mp
ute
th
e val
ue
2
ba�
. F
or
po
ints
in
th
e p
lan
e, t
he
x-
coord
inat
e of
the
mid
poin
t is
hal
fway
bet
wee
n t
he
x-co
ord
inat
es o
f th
e en
dpoin
ts,
and
sim
ilar
ly f
or
the
y-co
ord
inat
e of
the
mid
poin
t. S
o i
f w
e le
t th
e poin
ts �
�1
1,
yx
S
ecti
on
1.1
G
rap
hs
5
7
and
be
the
endpoin
ts o
f th
e li
ne
segm
ent,
then
the
Mid
poin
t F
orm
ula
sta
tes
that
th
e m
idp
oin
t is
:
�2
2,
yx
� �
�� �
��
2,
2
21
21
yy
xx
E
xa
mp
le 2
(M
idp
oin
t F
orm
ula
) F
ind
th
e m
idp
oin
t o
f th
e li
ne
seg
men
t b
etw
een
th
e tw
o p
oin
ts (
4, 9
) an
d (
27
, –
6)
So
luti
on
:
Her
e w
e le
t �
�1
1,
yx
be
the
poin
t (4
, 9)
and �
�2
2,
yx
be
the
poin
t (2
7, –6).
Th
en t
he
mid
po
int
equ
als:
�� �
��
�2
)6
(9
,2
27
4
�� �
23,
231
or
(15.5
, 1.5
) �
To b
e su
re t
hat
�
23
,231
�� i
s th
e m
idpoin
t, w
e ca
n c
hec
k t
hat
the
dis
tance
bet
wee
n (
4, 9)
and
�� �
23,
231
is
the
sam
e as
the
dis
tance
bet
wee
n (
27, –6)
and
�� �
23,
231
. T
his
is
left
as
an e
xer
cise
in u
sing t
he
dis
tance
form
ula
. �
Com
puti
ng d
ista
nce
s an
d m
idpoin
ts i
s use
ful
but
the
real
uti
lity
of
the
rect
angula
r co
ord
inat
e sy
stem
can
be
seen
when
we
look a
t an
expre
ssio
n l
ike
. T
his
expre
ssio
n i
nvolv
es a
sin
gle
var
iable
x, w
hic
h r
epre
sents
an u
nknow
n
num
ber
. I
t ca
n b
e vie
wed
as
a co
mm
and, “c
han
ge
the
num
ber
x b
y m
ult
iply
ing b
y
3 a
nd t
hen
subtr
acti
ng 4
.” T
hin
kin
g g
eom
etri
call
y o
n a
sin
gle
num
ber
lin
e, w
e m
ay
vie
w t
his
as
the
com
man
d “
move
x to
a s
pot
3 t
imes
as
far
from
the
ori
gin
and t
hen
sh
ift
that
poin
t to
the
left
4 u
nit
s.”
43�x
The
beh
avio
r of
such
an e
xpre
ssio
n i
s re
pre
sente
d m
ore
cle
arly
when
we
intr
oduce
a n
ew v
aria
ble
y a
nd c
onsi
der
the
equat
ion
43�
�x
y a
s a
rela
tionsh
ip
bet
wee
n t
wo v
aria
ble
s. T
he
rela
tionsh
ip c
an b
e des
crib
ed u
sing a
tab
le a
s in
Tab
le
1. N
ote
th
at t
he
table
does
no
t co
nta
in a
ll o
f th
e poss
ible
val
ues
fo
r x
sin
ce t
her
e ar
e in
finit
ely m
any.
58
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
X Y
0
–4
1
–1
2
2
3
5
4
8
Tab
le 1
Sin
ce w
e hav
e both
an x
-val
ue
and
y-v
alue
for
each
entr
y i
n t
he
table
, w
e ca
n p
lot
each
row
of
the
table
as
a poin
t in
the
pla
ne.
In
this
exam
ple
we
would
put
dots
at
the
poin
ts (
0, –4),
(1, –1),
(2, 2),
(3, 5),
and (
4, 8).
B
y r
epre
senti
ng a
ll
poss
ible
entr
ies
from
the
table
in s
uch
a f
ashio
n, w
e obta
in t
he
“gra
ph”
of
. W
e ca
n n
ot
dra
w e
ver
y d
ot
separ
atel
y s
ince
ther
e ar
e in
finit
ely m
any,
but
we
should
plo
t en
ough o
f th
em t
o s
ee t
he
pat
tern
. E
ach o
f th
e so
luti
ons
(or
poss
ible
entr
ies
in t
he
table
) is
incl
uded
on t
he
gra
ph a
nd t
he
gra
ph r
epre
sents
the
set
of
all
solu
tions
to t
he
equat
ion. I
n t
his
cas
e, t
he
gra
ph i
s a
stra
ight
line.
43�
�x
y
A g
raph g
ives
a “
pic
ture
” of
the
beh
avio
r of
an e
quat
ion a
nd r
epre
sents
the
set
of
all
solu
tions
to t
he
equat
ion. R
emem
ber
that
the
gra
ph s
ket
ch i
s a
finit
e re
pre
senta
tion o
f an
infi
nit
e re
lati
onsh
ip. J
ust
as
the
table
is
lim
ited
in t
he
entr
ies
that
it
can
dis
pla
y, th
e g
rap
h i
s li
mit
ed t
oo
. A
t an
y t
ime,
we
see
on
ly a
sm
all
port
ion o
f th
e gra
ph o
f an
equat
ion i
n t
wo v
aria
ble
s. S
ince
this
is
the
case
, w
e nee
d
to b
e su
re t
hat
we
are
loo
kin
g a
t a
“pic
ture
” w
hic
h b
est
des
crib
es t
he
beh
avio
r an
d
dis
pla
ys
the
import
ant
feat
ure
s of
the
gra
ph. T
his
is
par
ticu
larl
y t
rue
when
usi
ng a
gra
phin
g c
alcu
lato
r.
Gra
ph
s a
nd
th
e G
rap
hin
g C
alc
ula
tor
O
n C
D:
� �
Bas
ics
of
Gra
ph
ing
Get
tin
g Y
ou
r F
un
ctio
n I
nto
th
e P
rop
er
Fo
rmat
fo
r
Gra
ph
ing
Ente
ring e
quat
ions
into
the
TI–
83 c
alcu
lato
r is
done
by u
sing t
he
Equat
ion
Win
dow
. T
his
fea
ture
all
ow
s you t
o s
tore
var
ious
equat
ions
in t
he
mem
ory
of
the
calc
ula
tor.
G
raphs
of
those
equat
ions
are
dis
pla
yed
in t
he
vie
win
g w
indow
. T
he
stan
dar
d v
iew
ing w
indow
of
the
TI–
83 d
ispla
ys
the
val
ues
of
x fr
om
–1
0 t
o 1
0, an
d
the
val
ues
of
y fr
om
–10 t
o 1
0, ea
ch w
ith a
sca
le o
f 1. (
The
scal
e re
fers
to t
he
dis
tance
bet
wee
n t
he
tic
mar
ks
on t
he
axes
.) T
o v
iew
the
dim
ensi
ons
of
the
curr
ent
vie
win
g w
indow
, pre
ss t
he
[WIN
DO
W]
key
. T
o g
et a
good p
ictu
re o
f th
e gra
ph,
we
wil
l dev
elop s
om
e st
rate
gie
s fo
r se
lect
ing a
win
dow
that
show
s al
l th
e im
port
ant
feat
ure
s of
the
gra
ph. T
he
trac
e, z
oom
, an
d t
able
fea
ture
s w
ill
be
use
d t
o h
elp
det
erm
ine
good v
iew
ing w
indow
s. T
o u
se t
he
calc
ula
tor
effe
ctiv
ely, w
e nee
d t
o b
e ab
le t
o a
cces
s an
d u
nder
stan
d a
ll o
f th
ese
feat
ure
s.
A s
um
mar
y o
f th
e ca
lcula
tor
feat
ure
s is
pro
vid
ed i
n T
able
2. P
leas
e note
:
Addit
ional
hel
p o
n u
sing t
he
calc
ula
tor
can b
e fo
und o
n t
he
incl
uded
CD
-RO
M.
S
ecti
on
1.1
G
rap
hs
5
9
Fea
ture
C
alcu
lato
r K
ey
Use
s
Equat
ion W
indow
[Y
=]
Use
d f
or
ente
ring e
quat
ions
Gra
ph
s
Gra
phin
g E
quat
ions
[GR
AP
H]
All
ow
s th
e use
r to
vie
w t
he
gra
phs
of
any o
r al
l of
the
equat
ions
ente
red i
n
the
calc
ula
tor.
Vie
win
g W
indow
[W
IND
OW
] D
ispla
ys
the
dim
ensi
ons
of
the
curr
ent
vie
win
g w
indow
. T
hes
e dim
ensi
ons
can
be
adju
sted
by
th
e use
r.
Sta
ndar
d W
indow
[Z
OO
M]:
sel
ect
ZS
tandar
d [
6]
Ret
urn
s th
e ca
lcula
tor
to t
he
stan
dar
d
vie
win
g w
indow
Zo
om
In
[Z
OO
M]:
sel
ect
Zo
om
In
[2
] A
dju
sts
the
vie
win
g w
indow
in 2
way
s:
1.
re-c
ente
rs t
he
win
dow
at
a dif
fere
nt
poin
t in
the
pla
ne;
2.
dec
reas
es t
he
win
dow
val
ues
by
a fa
ctor
of
4.
(Th
is m
akes
th
e p
ictu
re o
f th
e g
rap
h
look b
igger
.)
Zo
om
Ou
t [Z
OO
M]:
sel
ect
Zoo
m O
ut
[3]
Do
es t
he
sam
e as
Zo
om
In
ex
cept
that
it
incr
ease
s th
e w
indow
val
ues
by a
fa
ctor
of
4. (
This
mak
es t
he
pic
ture
of
the
gra
ph l
ook s
mal
ler.
)
Tra
cing t
he
gra
ph
[TR
AC
E]
All
ow
s th
e use
r to
move
along t
he
gra
ph
usi
ng
th
e �
and �
arr
ow
s. It
dis
pla
ys
the
coord
inat
es o
f th
e hig
hli
ghte
d p
oin
t of
the
gra
ph a
t th
e bott
om
of
the
vie
win
g w
indow
.
Tab
les
T
able
for
equat
ions
[2n
d]
[GR
AP
H]
This
dis
pla
ys
a ta
ble
wit
h e
ntr
ies
rep
rese
nti
ng s
olu
tio
ns
to t
he
sele
cted
eq
uat
ion
(o
r eq
uat
ions)
. T
hes
e ar
e poin
ts o
n t
he
gra
ph. S
croll
thro
ugh t
he
val
ues
usi
ng t
he �
and �
key
s.
Tab
le S
etup
[2n
d]
[WIN
DO
W]
Dis
pla
ys
the
star
ting v
alue
for
the
table
as
wel
l as
the
incr
emen
t fo
r d
eter
min
ing
th
e n
ext
x val
ue.
T
hes
e val
ues
can
be
adju
sted
by t
he
use
r.
T
able
2
60
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
One
of
the
most
dif
ficu
lt a
spec
ts o
f usi
ng a
gra
phin
g c
alcu
lato
r is
fin
din
g a
n
appro
pri
ate
win
dow
in w
hic
h t
o v
iew
the
gra
ph. R
emem
ber
that
we
can s
ee o
nly
a
finit
e port
ion o
f an
infi
nit
e re
lati
onsh
ip s
o i
t is
im
port
ant
to d
evel
op a
n
under
stan
din
g o
f w
hat
the
gra
ph o
f th
e giv
en e
quat
ion s
hould
look l
ike.
T
he
foll
ow
ing
exam
ple
s ex
plo
re h
ow
to
use
th
e fe
ature
s fr
om
Tab
le 2
to
ad
just
th
e w
indow
and t
o f
ind s
olu
tions
to e
quat
ions
in t
wo v
aria
ble
s. In
som
e ca
ses
we
nee
d
to s
ee o
nly
a p
ort
ion o
f th
e gra
ph, su
ch a
s an
inte
rcep
t, b
ut
oth
er t
imes
we
nee
d t
o
see
a co
mp
lete
gra
ph
. A
co
mp
lete
gra
ph
is
a gra
ph t
hat
dis
pla
ys
all
of
the
import
ant
feat
ure
s of
the
giv
en e
quat
ion s
uch
as
pea
ks,
val
leys,
and i
nte
rcep
ts.
Fin
din
g g
ood v
iew
ing w
indow
s is
not
an e
xac
t sc
ience
and t
her
e ar
e m
any d
iffe
rent
win
dow
s th
at s
how
a c
om
ple
te g
raph f
or
a giv
en e
quat
ion.
E
xa
mp
le 3
(G
rap
hin
g E
qu
ati
on
s)
a) G
raph t
he
equat
ion
in t
he
stan
dar
d w
indow
and i
n t
he
win
dow
wit
h x
-val
ues
in t
he
inte
rval
[–15, 10]
and y
-val
ues
in t
he
inte
rval
xx
xy
520
22
3�
��
[–10,
300].
A
dju
st t
he
y-sc
ale
to 3
0.
S
olu
tio
n:
a) T
o v
iew
th
e g
rap
hs
we
must
fir
st e
nte
r th
e eq
uat
ion
in
to t
he
calc
ula
tor
usi
ng t
he
equat
ion w
indow
. O
nce
the
equat
ion i
s en
tere
d, w
e vie
w t
he
gra
ph i
n t
he
stan
dar
d v
iew
ing w
indow
.
W
e now
must
adju
st t
he
win
dow
so t
hat
we
are
usi
ng t
he
giv
en v
alues
. W
e th
en d
ispla
y t
he
gra
ph i
n t
he
new
win
dow
. (W
hat
hap
pen
s w
hen
the
y-sc
ale
is l
eft
at 1
?)
O
n C
D:
�
Ov
erv
iew
Win
do
w
Men
u
N
oti
ce t
hat
this
win
dow
giv
es u
s a
much
bet
ter
“pic
ture
” of
the
gra
ph a
nd i
t al
so s
how
s a
com
ple
te g
raph o
f th
e eq
uat
ion. I
t ca
n b
e har
d t
o b
elie
ve
that
b
oth
of
the
pic
ture
s sh
ow
th
e sa
me
gra
ph
sin
ce t
hey
ap
pea
r to
be
ver
y
dif
fere
nt.
T
his
new
win
dow
does
pro
vid
e a
com
ple
te g
raph b
ecau
se i
t tu
rns
S
ecti
on
1.1
G
rap
hs
6
1
out
that
ther
e ar
e no m
ore
“w
iggle
s” t
hat
occ
ur
outs
ide
the
win
dow
. T
his
know
ledge
com
es w
ith e
xper
ience
wit
h g
raphin
g e
quat
ions
and r
ecogniz
ing
that
cer
tain
fam
ilie
s of
equat
ions
hav
e so
me
feat
ure
s in
com
mon. Y
ou
should
alr
eady b
e fa
mil
iar
wit
h t
he
char
acte
rist
ics
of
the
linea
r fa
mil
y a
nd
the
quad
rati
c fa
mil
y. �
b)
Use
the
gra
ph t
o f
ind t
wo s
olu
tions
to t
he
equat
ion. U
se t
he
table
fea
ture
to f
ind
two m
ore
solu
tions
to t
he
equat
ion. A
lso, fi
nd y
when
x i
s 5 a
nd w
hen
x i
s 62.3
.
b)
We
wil
l u
se t
he
trac
e fe
atu
re o
f th
e ca
lcula
tor
to f
ind
tw
o s
olu
tio
ns
to t
he
equat
ion. O
nce
we
ente
r th
e tr
ace
mode,
the
calc
ula
tor
dis
pla
ys
the
solu
tio
ns
at t
he
bo
tto
m o
f th
e ca
lcu
lato
r.
O
n C
D:
� T
raci
ng
th
e F
un
ctio
n:
Th
e
Tra
ce K
ey
T
her
e ar
e m
any c
hoic
es f
or
solu
tions.
U
sing t
he
arro
ws
we
find o
ne
at
appro
xim
atel
y (
1.7
55, 81.2
16)
and a
noth
er o
ne
at a
ppro
xim
atel
y
(–4.0
96, 177.6
10).
R
emem
ber
that
, in
most
cas
es, th
e ca
lcula
tor
wil
l giv
e only
appro
xim
ate
val
ues
. U
sing t
he
table
fea
ture
we
can d
ispla
y m
ore
solu
tions
at o
ne
tim
e. H
ere
we
see
a li
st o
f se
ven
solu
tions.
T
his
als
o i
llust
rate
s th
e ta
ble
set
up p
roce
dure
.
T
o f
ind t
he
y-val
ues
for
the
giv
en v
alues
of
x, w
e ca
n u
se t
he
trac
e or
table
fe
atu
res.
T
o u
se t
he
trac
e fe
atu
re, en
ter
the
trac
e m
od
e an
d t
yp
e th
e v
alu
e fo
r x.
In
this
cas
e w
e en
ter
5. O
nce
we
pre
ss [
Ente
r] t
he
mac
hin
e dis
pla
ys
the
coo
rdin
ates
fo
r th
e poin
t w
ith
x-c
oord
inat
e 5.
62
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
S
o y
= 7
75 w
hen
x =
5.
To f
ind t
he
val
ue
for
y w
hen
x =
62.3
, w
e ca
n u
se t
he
table
fea
ture
. (
If y
ou
try
to
use
the
trac
e m
eth
od
yo
u w
ill
get
an
err
or
un
less
yo
u a
dju
st t
he
win
dow
.) E
nte
r th
e ta
ble
set
up a
nd s
et t
he
val
ue
for
TblS
tart
= 6
2 a
nd
�Tbl=
0.1
. T
his
pro
duce
s th
e fo
llow
ing t
able
.
F
rom
this
we
can s
ee t
hat
when
x =
62.3
, y
is a
ppro
xim
atel
y 5
61546. N
ote
th
at t
he
table
can
dis
pla
y o
nly
6 c
har
acte
rs i
n t
he
y co
lum
n. �
E
xa
mp
le 4
(C
om
ple
te G
rap
hs)
Fin
d t
he
com
ple
te g
raph o
f th
e eq
uat
ion
. 10
004
.332
.1
3.008
.2
34
5�
��
��
xx
xx
y
So
luti
on
:
We
firs
t en
ter
the
equat
ion i
n t
he
equat
ion w
indow
and l
ook a
t th
e gra
ph i
n
the
stan
dar
d w
indow
. T
he
stan
dar
d w
indow
is
a good p
lace
to s
tart
when
tr
yin
g t
o f
ind a
com
ple
te g
raph.
O
n C
D:
� G
rap
hin
g
Po
lyn
om
ials
N
oti
ce t
hat
we
seem
to b
e m
issi
ng a
pie
ce o
f th
e gra
ph w
hen
x i
s bet
wee
n 0
an
d 5
. S
o t
his
pic
ture
is
not
a co
mple
te g
raph. W
e nee
d t
o f
ind a
win
dow
th
at s
how
s al
l of
the
import
ant
feat
ure
s of
the
gra
ph a
nd s
till
giv
es t
he
gra
ph
S
ecti
on
1.1
G
rap
hs
6
3
a nic
e sh
ape.
S
ince
som
e of
the
gra
ph i
s outs
ide
the
curr
ent
win
dow
, w
e sh
ould
zoom
out.
A
fter
we
zoom
out
we
hav
e th
e fo
llow
ing p
ictu
re.
O
n C
D:
� Z
oo
m M
enu
Ov
erv
iew
W
e ca
n n
ow
see
more
of
the
gra
ph
but
we
also
dis
cov
er t
hat
we
are
mis
sin
g
anoth
er p
iece
. W
e sh
ould
adju
st t
he
min
imu
m v
alues
for
y to
see
th
e re
st o
f th
e gra
ph. T
o d
o t
his
, tr
ace
the
gra
ph a
nd w
atch
the
val
ues
for
y. W
e th
en
reac
h t
he
smal
lest
val
ue
for
y w
hen
x i
s ab
out
25.5
.
N
ow
we
use
this
info
rmat
ion f
rom
the
firs
t tw
o p
ictu
res
of
the
gra
ph t
o
adju
st o
ur
win
dow
and g
et a
com
ple
te g
raph.
To c
hec
k t
hat
this
is
a co
mple
te g
raph w
e ca
n u
se t
he
trac
e fe
ature
to s
ee
that
the
val
ues
fo
r y t
hat
are
outs
ide
the
win
dow
get
lar
ger
as
we
move
to t
he
right
and s
mal
ler
as w
e m
ove
to t
he
left
. T
his
indic
ates
that
we
hav
e, m
ore
th
an l
ikel
y, fo
und a
com
ple
te g
raph. H
ow
ever
, th
e to
p p
ort
ion o
f th
e gra
ph
her
e is
indis
tinguis
hab
le f
rom
the
x-ax
is s
ince
ther
e is
a b
ig d
iffe
rence
b
etw
een
th
e Y
min
an
d Y
max
val
ues
an
d t
he
Ym
ax i
s cl
ose
to
th
e x-
axis
.
So
met
imes
a b
ette
r re
pre
sen
tati
on
of
a co
mp
lete
gra
ph
can
be
mad
e b
y a
han
d-d
raw
n s
ket
ch t
han
fro
m a
cal
cula
tor
pic
ture
. T
he
sket
ch m
ay n
ot
be
dra
wn t
o s
cale
but
it m
ight
pro
vid
e a
clea
rer
dis
pla
y o
f th
e gra
phs
import
ant
feat
ure
s. �
64
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
Exam
ple
4 s
how
s th
at f
indin
g a
good w
indow
is
an a
rt f
orm
more
than
it
is a
sc
ience
. F
or
each
pro
ble
m t
hat
you e
nco
unte
r, y
ou w
ill
nee
d t
o f
irst
hav
e an
idea
of
the
gen
eral
shap
e of
the
gra
ph. T
his
wil
l co
me
wit
h p
ract
ice
and
ex
per
ien
ce w
ith
al
geb
ra. E
ver
yone
should
know
that
quad
rati
c eq
uat
ions
all
hav
e a
sim
ilar
type
of
gra
ph. T
he
sam
e is
tru
e fo
r li
nea
r eq
uat
ions.
M
ore
com
pli
cate
d p
oly
nom
ial
gra
phs
hav
e a
gen
eral
shap
e w
hic
h i
s det
erm
ined
by t
he
deg
ree
(or
hig
hes
t pow
er)
of
the
poly
nom
ial.
R
atio
nal
, ex
ponen
tial
, an
d l
ogar
ithm
ic e
quat
ions
all
hav
e dis
tinct
ive
shap
es. D
evel
opin
g t
he
pro
per
ties
of
thes
e fa
mil
ies
of
funct
ions
is l
eft
for
futu
re
cou
rses
.
Th
ere
are
a fe
w i
ssu
es a
bo
ut
gra
phin
g c
alcu
lato
rs a
nd
gra
ph
s th
at w
ere
men
tioned
ear
lier
and s
hould
be
rest
ated
.
Fir
st:
Th
e ca
lcu
lato
r p
rov
ides
ap
pro
xim
ate
so
luti
on
s to
eq
ua
tio
ns.
It
pro
vid
es t
he
an
swer
s a
s d
ecim
als
an
d h
as
a l
imit
ed a
ccu
racy
fo
r th
e
nu
mb
ers
it c
an
dis
pla
y.
Sec
on
d:
Yo
u s
ho
uld
alw
ay
s tr
y t
o f
ind
a c
om
ple
te g
rap
h o
f th
e eq
ua
tio
n
un
less
th
e co
nte
xt
of
the
pro
ble
m s
ug
ges
ts o
ther
wis
e. W
e w
ill
use
calc
ula
tors
to
hel
p u
s so
lve
ma
ny
pro
ble
ms
an
d w
e n
eed
to
kn
ow
wh
en
it i
s a
pp
rop
ria
te t
o f
ocu
s o
n a
pa
rtic
ula
r p
ort
ion
of
the
gra
ph
ra
ther
tha
n t
he
enti
re g
rap
h.
E
xer
cise
s 1
.1
Fo
r p
rob
lem
s 1
– 6
, fi
nd
th
e d
ista
nce
bet
wee
n t
he
two
po
ints
an
d t
he
mid
po
int
of
the
lin
e se
gm
ent
join
ing
th
em.
1.
(–5, 6)
and (
6, –5)
2.
(–8, 2)
and (
8, –2)
3.
(7, 8)
and (
7, 14)
4.
(23, 9)
and (
–4, 9)
5.
(13, –21)
and (
27, 59)
6.
(x, y)
and (
1, 2)
7.
Tw
o p
oin
ts �
�0
,1x
and �
�0
,2x
on t
he
x-ax
is.
8.
Tw
o p
oin
ts �
�1
,0
y a
nd �
�2
,0
y o
n t
he
y-ax
is.
Fo
r p
rob
lem
s 9
– 1
1,
solv
e.
9.
Fin
d t
he
per
imet
er o
f th
e tr
iangle
wit
h v
erti
ces
at t
he
poin
ts (
1, 1),
(5, 4)
and
(–1, 2).
10. F
ind t
he
poin
t th
at i
s one
quar
ter
the
dis
tance
fro
m p
oin
t (9
, 12)
to (
–5, 4).
S
ecti
on
1.1
G
rap
hs
6
5
11. A
4 f
oot
by 6
foot
map
is
div
ided
into
squar
es t
hat
are
1-i
nch
by 1
-inch
in
size
. T
he
legen
d o
f th
e m
ap s
tate
s th
at o
ne
inch
on t
he
map
is
equiv
alen
t to
2
mil
es.
a.
If
tow
n A
is
loca
ted a
t poin
t (1
9, 23)
and t
ow
n B
is
loca
ted a
t poin
t (4
5, 60),
what
are
the
coord
inat
es o
f th
e to
wn t
hat
is
hal
f-w
ay
bet
wee
n t
he
two
to
wn
s?
b.
If t
ow
n C
is
loca
ted a
t th
e poin
t (2
7, 49),
fin
d t
he
dis
tance
in m
iles
fr
om
to
wn
C t
o A
. c.
W
hic
h o
f th
e to
wns
A o
r B
is
close
r to
C?
F
or
pro
ble
ms
12
– 1
5, d
raw
th
e g
ra
ph
of
the
equ
ati
on
by
ha
nd
(m
ak
ing
a t
ab
le
of
at
lea
st 6
en
trie
s) a
nd
th
en g
rap
h t
he
equ
ati
on
in
th
e st
an
da
rd v
iew
ing
win
do
w.
12.
2 x
y�
13.
xx
y�
�2
14.
3 x
y�
15.
1
3�
�x
y F
or
pro
ble
ms
16
– 1
7, d
eter
min
e if
th
e g
iven
ord
ered
-pa
irs
are
po
ints
on
th
e
gra
ph
of
the
giv
en e
qu
ati
on
. E
xp
lain
yo
ur
rea
son
ing
.
16.
7
5.2
43
��
�x
ya.
(0
, –7)
c. (
3, 108)
b.
(12, 6935)
d. (–
6, –886)
17.
63
5�
��
xy
a.
�� �
6,
53
c.
�
519
��10
,
b.
(0, 6)
d. (7
.8, 12.5
) F
or
pro
ble
ms
18
– 2
0, u
se t
he
calc
ula
tor
to g
rap
h t
he
foll
ow
ing
eq
ua
tio
ns
an
d
then
use
th
e tr
ace
or
tab
le f
eatu
re o
f th
e ca
lcu
lato
r to
est
ima
te t
he
valu
es f
or
x o
r y.
18.
Fin
d y
when
x =
2, x
= 4
, an
d x
= –
5.
12
��
xy
19.
F
ind x
when
y =
–27, y
= 4
.5, an
d y
= 8
. 6
23
2�
��
�x
xy
20.
F
ind y
when
x =
25 a
nd x
= –
11.
xx
xy
47
52
6�
��
Fo
r p
rob
lem
s 2
1 –
23
, g
rap
h t
he
equ
ati
on
s in
th
e g
iven
win
do
ws.
W
hic
h
win
do
w b
est
dis
pla
ys
a c
om
ple
te g
rap
h o
f th
e eq
ua
tio
n?
J
ust
ify
yo
ur
an
swer
.
21. E
qu
atio
n:
34
78
xx
y�
�a.
W
indow
: S
tandar
d W
indow
b.
Win
do
w:
��
��
5.2
,5.2
,5.
2,
5.2
��
��
xy
c.
Win
do
w:
��
��
5.2
,5.
2,
500
,8
��
��
xy
(N
ote
: �
is
read
as
“is
in”.
)
66
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
22. E
qu
atio
n:
5
100
2�
�x
ya.
W
indow
: S
tandar
d W
indow
b.
Win
do
w:
��
��
5.2
,5.2
,5.
2,
5.2
��
��
xy
c.
Win
do
w:
��
��
5.2
,5.
2,
500
,8
��
��
xy
23. E
qu
atio
n:
2
2)
5.4
()
2.2
(�
��
xx
ya.
W
indow
: S
tandar
d W
indow
b.
Win
do
w:
��
��
10
,10
,200
,0
��
�x
y
c.
Win
do
w:
��
�� 1
,2
,1000
,0
��
��
xy
Fo
r p
rob
lem
s 2
4 –
26
, fi
nd
a v
iew
ing
win
do
w t
ha
t p
rod
uce
s a
co
mp
lete
gra
ph
of
the
equ
ati
on
. T
her
e a
re m
an
y c
orr
ect
an
swer
s. R
emem
ber
th
at
the
gra
ph
nee
ds
to d
isp
lay
all
of
the
inte
rcep
ts (
bo
th x
an
d y
) a
s w
ell
as
all
of
the
pea
k
an
d v
all
eys.
(H
int:
Yo
u m
ay
wa
nt
to u
se t
he
tab
le f
eatu
re t
o h
elp
yo
u.)
24.
xxy
2
�
25.
50
1.2
34
��
��
��
xx
xx
y26. A
car
dia
c te
st m
easu
res
the
con
centr
atio
n y
of
a dye
x se
conds
afte
r a
know
n
amount
is i
nje
cted
into
a v
ein n
ear
the
hea
rt. I
n a
norm
al h
eart
. xx
xx
y179
053
.14
.006
.2
34
��
��
�
27. B
elow
is
a gra
ph i
n t
he
stan
dar
d v
iew
ing w
indow
.
A
fter
adju
stin
g t
he
vie
win
g w
indow
, w
hic
h a
re p
oss
ible
gra
phs
of
the
sam
e eq
uat
ion?
Giv
e your
reas
onin
g a
s to
why i
t is
or
isn’t
the
sam
e gra
ph.
a.
b.
S
ecti
on
1.1
G
rap
hs
6
7
c.
d.
e.
f.
Fo
r p
rob
lem
s 2
4 –
28
, g
rap
h t
he
two
eq
ua
tio
ns
an
d s
tate
wh
eth
er t
he
foll
ow
ing
equ
ati
on
s h
av
e th
e sa
me
gra
ph
.
28.
34
��
xy
and
9999
.2
4�
�x
y
29.
2 xy�
and
xy�
. (
No
te:
Th
e | |
mea
ns
abso
lute
val
ue
fun
ctio
n a
nd
is
found o
n t
he
calc
ula
tor
under
the
[Mat
h]
Num
men
u a
s ab
s.)
30.
��
xy
10
log
� a
nd
xy�
31.
��
2lo
gx
y�
and
�� x
ylo
g2
�
32.
113
���
xxy
and
1
2�
��
xx
y
Fo
r p
rob
lem
29
, if
po
ssib
le,
det
erm
ine
wh
eth
er t
he
two
eq
ua
tio
ns
are
eq
ua
l.
In
ord
er f
or
the
equ
ati
on
s to
be
equ
al,
ea
ch x
-va
lue
mu
st p
rod
uce
th
e sa
me
y-v
alu
e fo
r b
oth
eq
ua
tio
ns.
Ju
stif
y y
ou
r a
nsw
ers.
33.
and
6 )
1(x
y�
�6
54
32
615
20
15
61
xx
xx
xx
y�
��
��
��
34. G
iven
tw
o p
oin
ts, �
�1
1,
yx
and �
�2
2,
yx
, dra
w a
n a
ssoci
ated
rig
ht
tria
ngle
and
use
th
e P
yth
ago
rean
Th
eore
m t
o p
rov
e th
e D
ista
nce
Fo
rmu
la:
��
��2
2
21
xx
d�
��
21
yy�
.
68
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
1.2
Solv
ing E
qu
ati
on
s G
rap
hic
all
y
Part
1:
Th
e Z
ero M
eth
od
Wh
en w
e so
lve
an e
qu
atio
n i
nv
olv
ing
an
un
kn
ow
n q
uan
tity
x, w
e ar
e tr
yin
g
to f
ind
all
of
the
val
ues
fo
r x
that
mak
e th
e st
atem
ent
tru
e.
Tak
e fo
r in
stan
ce t
he
equ
atio
n
. T
his
eq
uat
ion
sta
tes
that
a c
erta
in n
um
ber
x, w
hen
chan
ged
by
“mu
ltip
lyin
g b
y 3
an
d t
hen
su
btr
acti
ng
by
4”
end
s u
p a
t th
e p
oin
t 0
. T
o d
eter
min
e al
l th
e v
alues
x w
ith t
hat
pro
per
ty w
e ca
n u
se a
lgeb
ra:
add 4
to b
oth
sid
es t
o g
et
04
3�
�x
3x
= 4
; d
ivid
e b
oth
sid
es b
y 3
to
fin
d
34�
x. U
sing g
eom
etry
on t
he
num
ber
lin
e
that
solu
tion c
an b
e st
ated
as
foll
ow
s: S
uppose
a c
erta
in n
um
ber
is
moved
3 t
imes
as
far
fro
m t
he
ori
gin
and t
hen
shif
ted l
eft
4 u
nit
s, a
nd e
nds
up a
t 0. T
hen
the
ori
gin
al n
um
ber
was
34
. T
he
geo
met
ric
des
crip
tio
n i
n E
ng
lish
see
ms
more
dif
ficu
lt
and
co
mp
lica
ted
than
tak
ing
th
e al
geb
raic
ste
ps.
T
he
equat
ion 3x
– 4
= 0
ca
n a
lso
be
un
der
sto
od
usi
ng
2-d
imen
sio
nal
p
ictu
res.
In
th
e p
revio
us
sect
ion
, w
e in
tro
duce
d t
he
extr
a v
aria
ble
y a
nd g
raphed
the
equ
atio
n y
= 3
x – 4
. T
o s
olv
e th
e eq
uat
ion 3x
– 4
= 0
w
e ca
n t
hin
k o
f th
is a
s as
kin
g w
hic
h p
oin
ts o
n t
hat
gra
ph h
ave
y =
0. S
ince
th
e se
t of
po
ints
wh
ere
y =
0 i
s th
e x-
axis
, th
e so
luti
on
s to
th
e eq
uat
ion
are
ex
actl
y t
he
x-val
ues
of
the
poin
ts o
n t
he
gra
ph t
hat
lie
on t
he
x-ax
is. T
hes
e ar
e ca
lled
the
x-in
terc
epts
of
the
gra
ph o
f th
e eq
uat
ion
. T
his
idea
moti
vat
es t
he
gra
phic
al m
ethod i
ntr
oduce
d i
n t
his
sect
ion.
43�
�x
y
The
x-in
terc
epts
of
a g
rap
h a
re s
om
etim
es c
alle
d r
oo
ts o
r ze
ros.
(T
hes
e sh
ould
not
be
confu
sed w
ith s
quar
e ro
ots
.) In
oth
er w
ord
s, f
or
the
gra
ph o
f an
eq
uat
ion
a r
oot
or
zero
is
a val
ue
of
x th
at m
akes
y =
0. F
or
exam
ple
, th
e ro
ots
of
the
gra
ph i
n F
igure
1 a
re x
= –
6, x
= –
2, an
d x
= 5
.
y
–6
–2
5
x
F
igure
1
S
ecti
on
1.2
T
he
Zer
o M
eth
od
69
To f
ind t
he
zero
s al
geb
raic
ally
, w
e se
t y
= 0
and s
olv
e fo
r x.
T
o d
o t
his
gra
phic
ally
, w
e use
an a
ccura
te g
raph o
f th
e eq
uat
ion a
nd f
ind w
her
e it
cro
sses
the
x-ax
is.
E
xa
mp
le 1
(F
ind
ing
Ro
ots
)
Fin
d t
he
roots
of
a
lgeb
raic
ally
. 21
10
2�
�x
xS
olu
tio
n:
Sin
ce w
e ar
e as
ked
to f
ind t
he
roots
alg
ebra
ical
ly s
et
equal
to
0.
21
10
2�
�x
x
21
10
02
��
�x
x
This
is
just
solv
ing
a q
uad
rati
c eq
uat
ion.
� � )
7)(3
(0
��
�x
x
We
fact
or
the
rig
ht-
sid
e.
30
��
x or
07
��
x
On
e o
f th
e fa
cto
rs m
ust
eq
ual
zer
o.
� x
= –
3 or
x =
–7
So
lve
for
x.
� H
ere
the
roots
occ
ur
when
x =
–3 a
nd x
= –
7. �
U
nfo
rtunat
ely, al
geb
ra t
echniq
ues
are
not
alw
ays
pow
erfu
l en
ough t
o f
ind
the
roots
. T
his
is
wh
ere
the
gra
phin
g c
alcu
lato
r ca
n h
elp
. W
e ca
n u
se t
he
calc
ula
tor
to g
raph t
he
equat
ion a
nd t
o f
ind a
ppro
xim
ate
val
ues
for
the
roots
. T
his
can
be
done
by u
sing t
he
buil
t-in
root
finder
of
the
calc
ula
tor.
(F
or
som
e of
the
old
er
calc
ula
tor
model
s, t
his
is
not
an o
pti
on. In
stea
d, use
the
zoom
and t
race
funct
ions
to
app
rox
imat
e th
e so
luti
on.)
R
emem
ber
, si
nce
we
are
usi
ng
th
e ca
lcula
tor,
it
is l
ikel
y
that
th
ere
wil
l b
e so
me
loss
of
accu
racy
wh
en f
ind
ing
a r
oot.
E
xa
mp
le 2
(F
ind
ing
Ro
ots
) F
ind t
he
roots
of
the
giv
en q
uan
titi
es g
raphic
ally
.
I.
3
52�
x S
olu
tio
n:
To
do
th
is p
rob
lem
, w
e en
ter
the
po
lyn
om
ial
equ
atio
n
in
to t
he
calc
ula
tor,
and t
hen
pre
ss t
he
[GR
AP
H]
key
to g
et a
pic
ture
of
the
gra
ph i
n
the
stan
dar
d v
iew
ing w
indow
.
53
2�
�x
y
N
ote
: T
his
is
a co
mple
te g
raph a
nd i
t has
tw
o r
oots
.
70
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
Fin
din
g t
he
roo
ts:
To f
ind t
he
roots
we
use
the
buil
t-in
root
finder
. O
n t
he
TI-
83, w
e m
ust
en
ter
the
CA
LC
men
u w
hic
h i
s ac
cess
ed b
y p
ress
ing [
2nd]
and t
hen
[T
race
].
In t
he
CA
LC
men
u, th
e T
I-83 u
ses
the
word
“ze
ro”
inst
ead o
f ro
ot
so w
e se
lect
opti
on 2
.
O
n C
D:
� F
ind
ing
th
e Z
ero
s (R
oo
ts)
of
a
Po
lyn
om
ial
Wit
h t
hat
sel
ecti
on, th
e ca
lcula
tor
asks
whic
h r
oot
we
would
lik
e to
fin
d. I
t does
this
by a
skin
g t
hre
e ques
tions:
the
firs
t tw
o e
stab
lish
an i
nte
rval
and t
he
thir
d a
ppro
xim
ates
the
loca
tion. A
nsw
ers
to t
hes
e ques
tions
can b
e en
tere
d
by u
sing t
he
arro
w k
eys,
or
by t
ypin
g n
um
eric
al v
alues
bef
ore
pre
ssin
g
[EN
TE
R].
In
th
is e
xam
ple
, le
t’s
fin
d t
he
po
siti
ve
roo
t fi
rst.
T
hes
e th
ree
pic
ture
s sh
ow
the
pro
cess
that
the
calc
ula
tor
goes
thro
ugh w
hen
fi
ndin
g r
oots
. T
he
firs
t tw
o s
teps
sele
ct t
he
left
and r
ight
bounds
for
the
inte
rval
. T
his
lim
its
the
val
ues
fo
r x
that
the
calc
ula
tor
must
chec
k w
hen
fi
ndin
g t
he
answ
er. U
sing t
he
arro
w k
eys
we
told
the
calc
ula
tor
that
the
root
is i
n t
he
inte
rval
[0.8
5106383, 1.7
021277].
T
his
see
ms
corr
ect
in t
his
cas
e si
nce
th
e v
alu
es f
or
y ch
ang
e fr
om
neg
ativ
e to
po
siti
ve.
T
he
‘Gu
ess?
’ te
lls
the
calc
ula
tor
wher
e to
sta
rt w
hen
com
puti
ng t
he
root.
S
ecti
on
1.2
T
he
Zer
o M
eth
od
71
Aft
er e
nte
ring t
he
info
rmat
ion a
bove,
the
calc
ula
tor
dis
pla
ys
the
root
or
zero
v
alu
e x
= 1
.2909944. U
sing t
he
sam
e pro
cess
, w
e ca
n f
ind t
hat
the
oth
er
roo
t o
ccurs
wh
en x
= –
1.2
909944. �
Sin
ce t
he
equat
ion i
n t
his
exam
ple
was
a q
uad
rati
c, w
e co
uld
hav
e use
d a
lgeb
raic
met
hods,
fin
din
g t
he
exac
t so
luti
ons
to 0
. I
n t
he
nex
t ex
amp
le t
he
algeb
ra o
pti
on i
s m
uch
har
der
but
the
gra
phic
al m
ethod p
rovid
es f
airl
y a
ccura
te
solu
tio
ns.
53
2�
�x
II.
6
23
24
5�
��
xx
x
So
luti
on
:
Fir
st e
nte
r th
e eq
uat
ion
i
nto
th
e ca
lcula
tor
and
fin
d a
suit
able
vie
win
g w
indow
that
wil
l dis
pla
y t
he
com
ple
te g
raph. T
he
stan
dar
d
win
dow
dis
pla
ys
3 x
-inte
rcep
ts.
62
32
45
��
��
xx
xy
T
his
gra
ph c
onti
nues
ver
y s
teep
ly t
o t
he
left
and r
ight
(bey
ond t
he
vie
win
g
win
dow
). T
his
is
seen
by u
sing t
he
trac
e fe
ature
and c
hec
kin
g t
he
y-val
ues
.
Ther
efore
it
seem
s li
kel
y t
hat
this
is
a co
mple
te g
raph a
nd t
hat
the
gra
ph
does
not
hav
e an
y m
ore
x-i
nte
rcep
ts. O
nce
we
are
confi
den
t th
at w
e hav
e a
com
ple
te g
rap
h, w
e u
se t
he
root
fin
der
on
ce a
gai
n t
o f
ind
th
at t
he
thre
e ro
ots
occ
ur
when
x =
1.2
064313, x
= –
1.7
88648 a
nd x
= –
2.5
50975.
N
ote
that
for
the
last
root,
the
calc
ula
tor
found t
he
val
ue
for
x w
hen
000
1000000000
410
41
2�
��
�y
whic
h i
s a
val
ue
ver
y c
lose
to z
ero. I
t is
not
exac
tly z
ero s
ince
cal
cula
tors
usu
ally
mak
e sm
all
roundin
g e
rrors
.
O
n C
D:
� S
cien
tifi
c N
ota
tio
n a
nd
Y
ou
r C
alcu
latr
or’
s
Use
of
E
72
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
Cal
cula
tors
sel
dom
giv
e tr
uly
exac
t an
swer
s but
they
are
acc
ura
te e
nough
for
man
y p
urp
ose
s.�
T
he
root
or
zero
met
hod t
o s
olv
e eq
uat
ions
beg
ins
in t
he
sam
e w
ay w
e w
ould
use
when
alg
ebra
ical
ly s
olv
ing a
quad
rati
c eq
uat
ion:
we
mu
st m
ake
one
side
of
the
equat
ion 0
. I
n t
he
nex
t ex
ample
s, w
hic
h i
nvolv
e so
lvin
g a
n e
quat
ion w
ith a
si
ngle
var
iable
, w
e m
ight
nee
d t
o s
et o
ne
side
of
the
equat
ion e
qual
to z
ero b
efore
w
hen
can
use
the
gra
ph t
o f
ind t
he
zero
s.
E
xa
mp
le 3
(S
olv
ing
Eq
ua
tio
ns)
S
olv
e th
e eq
uat
ion. R
ound y
our
answ
er t
o t
wo d
ecim
al p
lace
s.
I.
0
34
4�
�x
So
luti
on
:
We
are
asked
to f
ind t
he
val
ues
for
x th
at m
ake
the
exp
ress
ion
equal
to 0
. (
Note
that
ther
e is
no n
eed t
o u
se a
lgeb
ra f
irst
in t
his
sit
uat
ion
since
the
expre
ssio
n i
s al
read
y e
qual
to z
ero.)
T
his
is
the
sam
e as
fin
din
g
the
roots
of
the
equat
ion
. T
o d
o t
his
we
ente
r th
e eq
uat
ion i
nto
the
calc
ula
tor,
gra
ph, an
d a
ppro
xim
ate
the
roots
by u
sing t
he
root
finder
.
34
4�
x
34
4�
�x
y
O
n C
D:
� F
ind
ing
S
olu
tio
ns
of
Eq
uat
ion
s
C
alcu
lato
r w
ork
pro
vid
es r
oots
nea
r to
the
val
ues
when
x =
0.9
3060486 a
nd
x =
–0.9
306049.
R
oundin
g t
o t
wo d
ecim
al p
lace
s w
e hav
e th
e so
luti
ons
x =
0.9
3 a
nd
x =
–0.9
3. (
Solv
e th
e eq
uat
ion a
lgeb
raic
ally
to s
ee t
hat
the
exac
t an
swer
s
are
41
43 �
� ��
x a
nd
41
43 �
� � ��
x. D
o t
hes
e ag
ree
wit
h t
he
calc
ula
tor
answ
ers?
)�
S
ecti
on
1.2
T
he
Zer
o M
eth
od
73
II.
7
73
53�
��
xx
So
luti
on
: In
ord
er t
o u
se t
he
roo
t m
eth
od
fo
r th
is p
rob
lem
, w
e m
ust
fir
st m
ake
on
e si
de
of
the
equat
ion 0
. T
o d
o t
his
, w
e use
som
e al
geb
ra.
12
73
03
��
�x
x
We
sub
trac
t 7x
and
5 f
rom
bo
th s
ides
. �
(Th
is n
ew e
qu
atio
n i
s eq
uiv
alen
t to
th
e fi
rst
so i
t h
as t
he
sam
e so
luti
ons.
)
Fin
din
g s
olu
tions
to t
his
equat
ion i
s th
e sa
me
as f
indin
g w
her
e th
e gra
ph o
f
cro
sses
th
e x-
axis
. E
nte
r th
e eq
uat
ion
in
to t
he
calc
ula
tor,
find a
com
ple
te g
raph, an
d u
se t
he
“zer
o”
feat
ure
to f
ind t
he
x-in
terc
epts
.
12
73
3�
��
xx
y
(N
ote
that
the
gra
ph s
how
n i
s m
issi
ng a
pie
ce b
ut
the
vie
win
g w
indow
is
adeq
uat
e to
fin
d a
root
for
the
equat
ion. C
ould
ther
e be
oth
er r
oots
outs
ide
of
this
win
dow
? E
xper
imen
tati
on w
ith t
he
trac
e fe
ature
or
table
fea
ture
su
gges
t th
at t
he
gra
ph, in
both
dir
ecti
ons,
does
not
ben
d b
ack t
o t
ow
ards
the
x-ax
is. H
ow
ever
, a
true
skep
tic
would
not
be
convin
ced b
y t
his
evid
ence
!)
Th
is t
ells
us
that
th
e ro
ot
occ
urs
nea
r to
x =
2.0
662142.
Aft
er r
oundin
g, th
e so
luti
on t
o
is
app
rox
imat
ely
x =
2.0
7.
If w
e w
ere
to c
hec
k t
his
solu
tion b
y p
luggin
g i
t in
to t
he
ori
gin
al e
quat
ion,
we
would
fin
d t
hat
the
right-
side
and l
eft-
side
are
not
exac
tly e
qual
. T
his
is
due
to t
he
roundin
g o
f th
e an
swer
. I
n t
his
cas
e th
e eq
uat
ion h
ad o
nly
one
solu
tio
n. �
73
57
3�
��
xx
74
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
III.
5
74
3�
�x
So
luti
on
: W
e re
wri
te t
his
equat
ion u
sing a
lgeb
ra.
05
74
3�
��
x
Su
btr
act
5 f
rom
bo
th s
ides
. �
This
pro
ble
m i
s now
red
uce
d t
o a
nal
yzi
ng t
he
gra
ph o
f th
e eq
uat
ion
57
43
��
�x
y. T
o d
o t
his
we
ente
r th
at e
quat
ion i
nto
the
calc
ula
tor,
fin
d
the
com
ple
te g
rap
h, an
d u
se t
he
root
fin
der
.
This
show
s th
at t
he
root
occ
urs
nea
r x
= 1
.6509636.
A
fter
roundin
g, th
e ap
pro
xim
ate
solu
tion t
o t
he
ori
gin
al e
quat
ion i
s x
= 1
.65.
Rem
ark
s:
1.
Does
the
gra
ph a
ctual
ly s
top t
her
e in
the
thir
d Q
uad
rant?
W
her
e is
the
endpoin
t? H
int:
When
is
. 0
74
3�
�x
2.
Can
you f
ind t
he
exac
t so
luti
on a
lgeb
raic
ally
and c
hec
k t
hat
it
mat
ches
th
e es
tim
ated
solu
tio
n x
= 1
.65? �
E
xer
cise
s 1
.2
Fo
r p
rob
lem
s 1
– 1
1, fi
nd
th
e ze
ros,
if
they
ex
ist,
of
the
equ
ati
on
bo
th
gra
ph
ica
lly
an
d a
lgeb
raic
all
y.
1.
4
5�
�x
y2.
3
22
��
�x
xy
3.
3
52�
�x
y
S
ecti
on
1.2
T
he
Zer
o M
eth
od
75
4.
3
22
2�
��
xx
y5.
xx
xy
33
82
3�
��
6.
34
��
xy
7.
10
92�
�x
y
8.
8
22
4�
��
xx
y9.
36
13
24
��
�x
xy
10.
3
24
24
��
�x
xy
11.
���
�60
11
22
��
��
xx
xy
12.
���
�10
72
52
��
��
xx
xy
Wh
en u
sin
g t
he
calc
ula
tor
to f
ind
th
e ze
ros,
Jo
e S
tud
ent
ente
red
th
e fo
llo
win
g
info
rma
tio
n.
Fo
r p
rob
lem
s 1
3 a
nd
14
, ex
pla
in w
ha
t J
oe
did
wro
ng
an
d w
hy
it
did
no
t w
ork
.
13. J
oe
wan
ted t
o f
ind t
he
zero
s of
3
42�
�x
, so
he
ente
red t
he
foll
ow
ing
equat
ion i
n h
is c
alcu
lato
r an
d f
ound t
he
answ
ers
giv
en b
elow
.
76
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
14. T
o f
ind t
he
zero
s of
201
42
3�
�x
x, Jo
e en
tere
d t
he
corr
espondin
g e
quat
ion
in t
he
calc
ula
tor
and l
ooked
at
the
gra
ph.
a) H
e noti
ced t
hat
the
gra
ph a
ppea
rs t
o h
ave
two z
eros.
W
hen
try
ing t
o f
ind
the
zero
on t
he
right,
he
got
this
mes
sage:
b)
Aft
er f
iguri
ng o
ut
what
he
did
wro
ng i
n p
art
a), he
then
tri
ed t
o f
ind t
he
zero
to t
he
left
. W
hen
try
ing t
o f
ind t
his
root,
he
ente
red t
he
foll
ow
ing
info
rmat
ion i
nto
the
calc
ula
tor
and g
ot
an e
rror
mes
sage:
S
ecti
on
1.2
T
he
Zer
o M
eth
od
77
Fo
r p
rob
lem
s 1
5 –
17
, d
eter
min
e g
rap
hic
all
y, b
y u
sin
g t
he
zero
met
ho
d, th
e
nu
mb
er o
f so
luti
on
s to
th
e eq
ua
tio
n. Y
ou
do
no
t n
eed
to
fin
d t
he
roo
ts.
15.
xx
x�
��
45
35
16.
0
10
15
10
57
��
��
xx
x17.
000
,32
16
000
,8
500
32
4�
��
�x
xx
x F
or
pro
ble
ms
18
– 2
1, u
se t
he
zero
fin
der
to
fin
d t
he
ap
pro
xim
ate
so
luti
on
(s)
to
the
equ
ati
on
. (
Ro
un
d y
ou
r so
luti
on
s to
2 d
ecim
al
pla
ce a
ccu
racy
.)
18.
0
34
��
�x
x19.
0
66
5�
��
xx
20. 1
0
06
32
5�
��
�x
xx
21.
42
121
212
xx
x�
�
78
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
1.3
S
olv
ing E
qu
ati
on
s G
rap
hic
all
y
Pa
rt 2
: T
he
Inte
rsec
tio
n M
eth
od
In t
his
sec
tio
n w
e d
iscu
ss a
no
ther
gra
ph
ical
met
ho
d t
o s
olv
e eq
uat
ion
s. T
his
is
th
e in
ters
ecti
on
met
ho
d,
use
d f
or
a sy
stem
of
two
eq
uat
ion
s. B
efo
re w
e d
iscu
ss
this
gra
ph
ical
ap
pro
ach
, w
e re
vie
w t
he
alg
ebra
ic m
eth
od
s u
sed
to
so
lve
a sy
stem
of
equ
atio
ns.
T
wo
co
mm
on
alg
ebra
ic m
eth
od
s to
so
lve
a sy
stem
of
equ
atio
ns
are
elim
inat
ion
an
d s
ub
stit
uti
on
. Y
ou
may
be
fam
ilia
r w
ith
usi
ng
th
ese
met
ho
ds
for
lin
ear
equ
atio
ns.
02
3
72
��
��
yx
yx
2
4
13
22
��
�� x
yxy
x
Lin
ear
Syst
em
Non-l
inea
r S
yst
em
Fig
ure
1
A n
on-l
inea
r sy
stem
is
a sy
stem
in w
hic
h a
t le
ast
one
of
the
equat
ions
is n
ot
a li
nea
r eq
uat
ion. (
See
fig
ure
1.)
W
e w
ant
to b
e ab
le t
o m
ove
bey
ond s
olv
ing l
inea
r sy
stem
s to
those
that
are
non-l
inea
r. S
ince
the
subst
ituti
on m
ethod i
s m
ore
use
ful
than
the
elim
inat
ion m
ethod f
or
solv
ing n
on-l
inea
r sy
stem
s, w
e w
ill
focu
s her
e on
the
subst
ituti
on m
ethod.
Su
bst
itu
tio
n
To u
se t
he
subst
ituti
on m
ethod, so
lve
one
of
the
two e
quat
ions
for
one
of
the
unknow
ns
and s
ubst
itute
this
expre
ssio
n i
nto
the
oth
er e
quat
ion. T
his
pro
duce
s an
eq
uat
ion i
nvolv
ing o
nly
one
unknow
n.
Ex
am
ple
1 (
Su
bst
itu
tio
n)
Solv
e th
e fo
llow
ing s
yst
ems
of
equat
ions
usi
ng t
he
subst
ituti
on m
ethod.
I.
Lin
ear
Syst
em
52
3
12
��
��
yx
yx
So
luti
on
:
Fir
st s
olv
e one
of
the
equat
ions
for
one
of
the
unknow
ns.
T
he
firs
t eq
uat
ion
seem
s si
mple
st t
o h
and
le.
12
��
xy
Add
y a
nd
su
btr
act
1 f
rom
bo
th s
ides
. �
Nex
t, s
ubst
itute
that
form
ula
for
y in
the
oth
er e
quat
ion.
5)1
2(2
3�
��
xx
S
ubst
itute
2x
– 1
in
to t
he
seco
nd e
quat
ion.
� 5
24
3�
��
xx
D
istr
ibute
the
2.
�
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
7
9
52
7�
�x
Com
bin
e li
ke
term
s.
� 7
7�
x
Ad
d 2
to
bo
th s
ides
. �
x =
1
Div
ide
by 7
. � �
We
hav
e fo
und t
he
val
ue
for
x. R
emem
ber
that
ther
e w
ere
two u
nknow
ns.
W
e hav
e fo
und t
he
x-val
ue
bu
t st
ill
nee
d t
o c
om
pu
te y
. T
o d
o t
his
subst
itute
th
e val
ue
that
we
found i
n x
bac
k i
nto
one
of
the
ori
gin
al e
quat
ions.
1)1(
2�
�y
S
ubst
itute
x =
1 i
nto
the
firs
t eq
uat
ion.
1��
�y
S
ub
trac
t 2
fro
m b
oth
sid
es.
� 1
�y
M
ult
iply
both
sid
es b
y –
1.
� T
his
giv
es u
s th
e so
luti
on
(x,
y)
= (
1, 1
). R
emem
ber
to
chec
k t
hat
th
e ord
ered
-pai
r is
a s
olu
tion t
o b
oth
equat
ions.
�
II
.
N
on
-lin
ear
Sy
stem
7
2
42
��
��
�y
xy
x
So
luti
on
:
Ev
en t
ho
ug
h t
his
is
a n
on-l
inea
r sy
stem
, w
e ca
n s
till
use
th
e su
bst
ituti
on
m
ethod. S
olv
e one
of
the
equat
ions
for
one
of
the
unknow
ns
and s
ubst
itute
th
is i
nto
the
oth
er e
quat
ion
. S
ince
the
firs
t eq
uat
ion
is
alre
ady s
olv
ed f
or
y,
we
wil
l ju
st p
lug t
his
val
ue
for
y in
to t
he
oth
er e
quat
ion.
��
74
22
��
��
xx
Subst
itute
y i
nto
th
e se
cond
eq
uat
ion
. �
74
22
��
�x
x
Rea
rran
ge
the
term
s.
� � � 0
32
2�
��
xx
S
ub
trac
t 7
fro
m b
oth
sid
es.
0)
3)(1
(�
��
xx
F
acto
r.
30
��
x or
1
0�
�x
O
ne
of
the
fact
ors
mu
st e
qu
al z
ero
. �
x =
3 or
x =
–1
So
lve
for
x.
� � �
For
each
of
thes
e poss
ible
val
ues
for
x, w
e nee
d t
o f
ind t
he
corr
espondin
g
val
ues
fo
r y.
T
o d
o t
his
, su
bst
itute
eac
h v
alue
of
x bac
k i
nto
one
of
the
ori
gin
al e
quat
ions
and s
olv
e fo
r y.
x =
3:
Subst
itute
x i
nto
the
firs
t eq
uat
ion
. 1
34
32
��
�y
x =
–1
:
Subst
itute
x i
nto
the
firs
t eq
uat
ion.
54
)1(
2�
��
�y
This
giv
es u
s tw
o o
rder
ed-p
air
solu
tions
(3, 13)
and (
–1, 5).�
W
hen
solv
ing a
syst
em o
f eq
uat
ions
we
are
findin
g t
he
ord
ered
-pai
rs (
a, b
) th
at a
re s
olu
tions
to b
oth
equat
ions.
S
ince
a g
raph o
f an
equat
ion r
epre
sents
the
set
of
all
ord
ered
-pai
r so
luti
ons
to t
he
equat
ion, fi
ndin
g t
he
solu
tions
to b
oth
equat
ions
would
mea
n t
hat
we
are
findin
g p
oin
ts t
hat
are
on b
oth
of
the
gra
phs.
T
his
is
the
sam
e as
fin
din
g w
her
e th
e tw
o g
rap
hs
hav
e a
po
int
in c
om
mo
n;
the
pla
ces
wh
ere
they
inte
rsec
t. T
he
gra
ph i
nte
rsec
tio
n f
eatu
re i
n t
he
calc
ula
tor
pro
vid
es u
s w
ith
a
gra
phic
al a
ppro
ach t
o s
olv
ing s
yst
ems
of
equat
ions.
H
ere
are
som
e ex
amp
les
usi
ng
th
is g
rap
hic
al a
pp
roac
h.
80
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
E
xa
mp
le 2
(G
rap
hic
al
Ap
pro
ach
) S
olv
e th
e fo
llow
ing s
yst
ems
of
equat
ions
usi
ng g
raphs.
I.
4
23
12
��
��
�y
xy
x
So
luti
on
: F
irst
input
the
two e
quat
ions
into
the
calc
ula
tor.
T
his
req
uir
es t
hat
we
use
algeb
ra t
o s
olv
e both
of
the
equat
ions
for
y,
12
��
xy
and
234
���
xy
. W
e
know
that
ther
e is
only
one
poin
t of
inte
rsec
tion s
ince
this
is
a li
nea
r sy
stem
(t
wo i
nte
rsec
ting l
ines
).
We
then
nee
d t
o f
ind a
suit
able
vie
win
g w
indow
th
at d
isp
lays
the
po
int
of
inte
rsec
tio
n.
O
n C
D:
� F
ind
ing
S
olu
tio
ns
of
Eq
uat
ion
s
T
he
pic
ture
above
dis
pla
ys
the
two g
raphs
in t
he
stan
dar
d w
indow
. A
fter
ad
just
ing t
he
win
dow
, w
e hav
e a
bet
ter
pic
ture
of
wher
e th
e tw
o l
ines
in
ters
ect.
T
o s
olv
e th
e sy
stem
of
equat
ions,
we
use
the
inte
rsec
tion f
inder
of
the
calc
ula
tor
to l
oca
te t
he
poin
t o
f in
ters
ecti
on
. E
nte
r th
e ca
lc m
enu
an
d s
elec
t o
pti
on
5. T
hen
sel
ect
the
two
gra
ph
s an
d o
ffer
a g
ues
s as
to
wh
ere
the
two
eq
uat
ions
inte
rsec
t.
O
n C
D:
� F
ind
ing
th
e In
ters
ecti
on
of
Tw
o
Po
lyn
om
ials
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
8
1
Aft
er c
om
ple
ting t
his
pro
cess
, th
e ca
lcula
tor
retu
rns
an a
nsw
er o
f (–
6, –11).
T
his
sy
stem
can
als
o b
e so
lved
alg
ebra
ical
ly.�
II
. (
Pre
vio
usl
y d
on
e as
Ex
amp
le 1
, II
) 7
2
42
��
��
�y
xy
x
So
luti
on
:
So
lve
bo
th e
qu
atio
ns
for
y an
d e
nte
r th
e fo
rmu
las
in t
he
calc
ula
tor.
T
he
seco
nd
eq
uat
ion
bec
om
es
72
��
xy
.
S
ince
this
is
a non-l
inea
r sy
stem
, th
ere
may
be
more
than
one
solu
tion. T
he
pic
ture
in
dic
ates
th
at w
e w
ill
hav
e tw
o p
oin
ts o
f in
ters
ecti
on
. S
o w
e ad
just
th
e vie
win
g w
indow
to s
ee b
oth
poin
ts o
f in
ters
ecti
on.
N
ow
use
the
inte
rsec
tion f
inder
tw
ice
to f
ind t
hose
poin
ts o
f in
ters
ecti
on.
Th
e in
ters
ecti
on
fin
der
wil
l lo
cate
only
on
e p
oin
t o
f in
ters
ecti
on
at
a ti
me.
82
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
T
his
giv
es u
s th
at t
he
two
so
luti
ons
to t
he
syst
em o
f eq
uat
ions
are
(–1
, 5)
and
(3
, 1
3).
T
hes
e an
swer
s m
atch
th
e so
luti
on
s fo
un
d a
lgeb
raic
ally
in
E
xam
ple
1,
II.�
As
we
hav
e ju
st s
een
, th
e ca
lcula
tor
can c
om
pu
te a
po
int
of
inte
rsec
tio
n o
f tw
o
gra
ph
s p
rovid
ed w
e fe
ed i
t an
est
imat
ed a
nsw
er. T
o s
olv
e a
sin
gle
eq
uat
ion
we
can
use
the
inte
rsec
tion m
ethod b
y r
ewri
ting t
hat
equat
ion a
s a
syst
em o
f tw
o e
quat
ions.
T
he
foll
ow
ing
ex
amp
les
dem
on
stra
te t
his
tec
hniq
ue.
E
xa
mp
le 3
(In
ters
ecti
on
Met
ho
d)
Solv
e th
e fo
llow
ing e
quat
ions
usi
ng t
he
inte
rsec
tion m
ethod.
I.
4
53
��
�x
xx
So
luti
on
: T
o s
olv
e th
is s
ingle
equat
ion w
e m
ake
it i
nto
a s
yst
em b
y i
ntr
oduci
ng a
new
v
aria
ble
y a
nd s
etti
ng e
ach s
ide
of
the
equat
ion e
qual
to y
:
xx
y5
3�
� a
nd
4�
�x
y.
Th
e x-
coord
inat
es o
f th
e ord
ered
-pai
r so
luti
ons
to t
his
syst
em c
orr
espond t
o
solu
tio
ns
of
the
ori
gin
al e
qu
atio
n. T
o s
olv
e a
syst
em o
n t
he
calc
ula
tor
we
gra
ph
th
e tw
o e
qu
atio
ns
and
fin
d t
he
inte
rsec
tio
ns.
W
e en
ter
thes
e eq
uat
ion
s
into
the
calc
ula
tor
as
Y a
nd
XX
53
1�
�4
2�
�X
Y.
L
ookin
g a
t th
e st
andar
d w
indow
, w
e se
e th
ree
poin
ts o
f in
ters
ecti
on.
Ex
per
ien
ce w
ith
su
ch g
rap
hs
tell
s u
s th
at t
her
e ar
e n
o m
ore
in
ters
ecti
on
poin
ts. I
t is
not
nec
essa
ry t
o a
dju
st t
he
vie
win
g w
indow
, but
in o
rder
to g
et
a m
ore
usa
ble
pic
ture
we
wil
l ad
just
th
e y-
axis
val
ues
dow
n b
y 5
unit
s.
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
8
3
U
nli
ke
solv
ing a
syst
em o
f eq
uat
ions,
we
nee
d t
o g
ive
only
the
val
ues
for
x as
the
solu
tions
inst
ead o
f an
ord
ered
-pai
r. B
ecau
se o
f th
is, w
e fi
nd t
hat
the
solu
tio
ns
are
app
rox
imat
ely
x =
–2.7
3, x
= 0
.73 a
nd x
= 2
. N
ote
that
we
could
hav
e so
lved
this
pro
ble
m u
sing t
he
root
met
hod b
y r
ewri
ting t
he
ori
gin
al e
quat
ion
as
. (
This
equat
ion c
an a
lso b
e so
lved
algeb
raic
ally
. C
an y
ou f
ind t
he
exac
t so
luti
ons
usi
ng a
lgeb
ra?)
04
63
��
�x
x
II.
t
97
52
3�
��
t S
olu
tio
n:
Lik
e pro
ble
m I
, to
solv
e th
is w
e in
troduce
anoth
er v
aria
ble
y a
nd
so
lve
the
syst
em. H
ow
ever
, w
hen
ente
ring t
he
val
ues
in
to t
he
calc
ula
tor,
we
wil
l le
t t
be
repre
sente
d b
y x
. A
lso n
ote
that
this
pro
ble
m i
s not
easy
to s
olv
e usi
ng
the
algeb
ra t
ools
that
we
hav
e av
aila
ble
. (C
an y
ou f
ind t
he
exac
t so
luti
ons?
)
A
gai
n, lo
okin
g a
t th
e st
andar
d w
indow
, w
e se
e th
ree
poin
ts o
f in
ters
ecti
on.
Ho
wev
er, th
is t
ime
to g
et a
“n
icer
” pic
ture
, w
e w
ill
adju
st t
he
y-ax
is v
alu
es
up b
y 5
unit
s.
84
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
Rem
emb
er t
hat
we
let
x re
pre
sen
t t a
nd w
e ar
e tr
yin
g t
o f
ind t
he
val
ues
of
t th
at s
atis
fy
. T
hes
e v
alu
es a
re a
ppro
xim
atel
y t
= –
4,
97
52
3�
��
tt
t = –2.5
616 a
nd t
= 1
.56156. W
e do n
ot
nee
d t
o g
ive
an o
rder
ed-p
air
solu
tio
n s
ince
ou
r o
rig
inal
pro
ble
m w
as a
sin
gle
eq
uat
ion
.�
III.
6
27
22
��
��
xx
x
So
luti
on
:
This
pro
ble
m w
ill
also
involv
e se
ttin
g u
p a
nd s
olv
ing a
syst
em o
f eq
uat
ions.
T
his
pro
ble
m h
as n
o s
imple
solu
tion b
y u
sing a
lgeb
ra a
lone.
H
ere
we
can
see
tw
o p
oin
ts o
f in
ters
ecti
on
on
the
scre
en a
nd t
he
gra
ph
s d
on
’t s
eem
to
in
ters
ect
else
wh
ere.
W
e u
se t
he
inte
rsec
tio
n f
ind
er t
o
com
pu
te t
he
po
ints
of
inte
rsec
tio
n.
T
his
sh
ow
s th
at t
he
two
so
luti
on
s to
ou
r o
rig
inal
eq
uat
ion
are
ap
pro
xim
atel
y
x =
-1.9
5898 a
nd x
= 4
.30939. �
S
ince
we
dis
cuss
ed a
lgeb
raic
met
ho
ds
and
gra
ph
ical
met
ho
ds
to s
olv
e a
syst
em o
f eq
uat
ion
s, n
ow
is
a g
oo
d t
ime
to m
enti
on
th
at i
t is
po
ssib
le t
o u
se a
m
ixtu
re o
f al
geb
ra a
nd
gra
ph
ing
to
so
lve
a sy
stem
of
equ
atio
ns.
E
xam
ple
4 u
ses
an
algeb
raic
/gra
phin
g h
ybri
d a
ppro
ach t
o s
olv
e th
e pro
ble
m o
rigin
ally
done
in e
xam
ple
1, II
and r
e-done
in e
xam
ple
2, II
.
E
xa
mp
le 4
(H
yb
rid
Met
ho
d)
Solv
e th
e fo
llow
ing s
yst
em o
f eq
uat
ions.
72
42
��
��
�y
xy
x
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
8
5
So
luti
on
: T
o u
se t
he
root
met
hod, fi
rst
we
must
use
subst
ituti
on a
nd a
lgeb
ra t
o r
ewri
te
the
equ
atio
n
��
74
22
��
��
xx
Subst
itute
y i
nto
th
e se
cond
eq
uat
ion
. �
74
22
��
�x
x
Rea
rran
ge
the
term
s.
� � 0
32
2�
��
xx
S
ub
trac
t 7
fro
m b
oth
sid
es.
En
ter
this
equ
atio
n i
nto
th
e ca
lcula
tor
and
vie
w t
he
gra
ph
.
No
te t
hat
the
gra
ph
is
a q
uad
rati
c an
d a
ll o
f th
e ro
ots
are
vis
ible
. U
se t
he
roo
t fi
nd
er t
o f
ind
th
e v
alu
es.
R
emem
ber
that
this
is
giv
ing u
s only
the
x-co
ord
inat
es o
f our
ord
ered
-pai
r so
luti
on
so
we
nee
d t
o s
ub
stit
ute
eac
h v
alu
e in
to o
ne
of
the
ori
gin
al
equat
ions
to f
ind y
.
x =
–1
:
54
)1(
2�
��
�y
� � x
= 3
:
13
4)
3(2
��
�y
This
giv
es u
s th
e ord
ered
-pai
rs o
f (–
1, 5)
and (
3, 13).
N
ote
that
the
hybri
d
appro
ach i
nvolv
es u
sing l
ess
algeb
ra t
han
the
algeb
raic
appro
ach (
Exam
ple
1
, II
) an
d d
oes
n’t
req
uir
e u
s to
ad
just
th
e v
iew
ing
win
do
w t
o f
ind
th
e se
cond
p
oin
t o
f in
ters
ecti
on
(E
xam
ple
2,
II).
�
A
s w
e hav
e ju
st s
een i
n t
he
firs
t th
ree
sect
ions
of
the
book, gra
phin
g
calc
ula
tors
all
ow
us
to s
olv
e m
ore
ty
pes
of
equ
atio
ns
then
we
cou
ld d
o j
ust
by
usi
ng
al
geb
ra. H
ow
ever
, to
quo
te a
rec
ent
mo
vie
, “w
ith
gre
at p
ow
er c
om
es g
reat
re
sponsi
bil
ity.”
N
ow
that
you c
an u
se t
he
calc
ula
tor
as a
tool
to s
olv
e eq
uat
ions,
y
ou
nee
d t
o b
e aw
are
that
it
is n
ot
alw
ays
the
bes
t o
pti
on
. T
he
calc
ula
tor
som
etim
es
cau
ses
yo
u t
o s
pen
d m
ore
tim
e th
an n
eces
sary
on a
pro
ble
m, o
r it
mig
ht
no
t g
ive
reli
able
an
swer
s. T
his
is
wh
ere
yo
u w
ill
nee
d t
o e
xer
cise
res
po
nsi
bil
ity
. H
ere
are
86
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
som
e ex
amp
les
sho
win
g t
hat
cau
tio
n i
s so
met
imes
nec
essa
ry w
hen
wo
rkin
g w
ith
a
calc
ula
tor.
E
xa
mp
le 5
(P
ote
nti
al
Pit
fall
s)
I. S
olv
e usi
ng t
he
inte
rsec
tion m
ethod.
yx
yx
x�
�� 2
3
2
S
olu
tio
n:
As
we
wil
l se
e, t
his
“in
no
cen
t” l
oo
kin
g n
on
-lin
ear
syst
em i
s p
arti
cula
rly
tr
ouble
som
e. A
s usu
al, w
e in
put
the
two e
quat
ions
into
the
calc
ula
tor
and
vie
w t
he
gra
ph i
n t
he
stan
dar
d w
indow
.
A
s w
e ca
n s
ee, th
e st
andar
d w
indow
does
not
clea
rly s
how
what
is
hap
pen
ing f
or
the
posi
tive
val
ues
of
x. S
o w
e ad
just
the
win
dow
and v
iew
th
e g
rap
h a
gai
n.
F
rom
th
is g
rap
h, it
ap
pea
rs t
hat
th
ere
are
two
pla
ces
wh
ere
the
gra
ph
s m
eet.
S
o w
e u
se t
he
inte
rsec
tion
fin
der
to
fin
d t
hem
. (T
his
is
wh
ere
it g
ets
inte
rest
ing
.)
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
8
7
This
is
not
a m
ispri
nt.
W
hen
you t
ry t
o f
ind t
he
seco
nd i
nte
rsec
tion p
oin
t,
whic
h o
ccurs
when
x =
1, th
e ca
lcula
tor
retu
rns
to y
ou t
he
poin
t (0
, 0)
as t
he
resu
lt. W
hy
is
this
? W
ell,
at
this
po
int
the
two
cu
rves
mee
t b
ut
do
no
t cr
oss
. T
hey
are
tan
gen
t to
eac
h o
ther
at
this
inte
rsec
tio
n p
oin
t. T
he
po
int
(1, 2
) is
a s
olu
tio
n t
o t
he
syst
em o
f eq
uat
ion
s, b
ut
the
calc
ula
tor
can
no
t fi
nd
it
. T
his
is
one
of
those
tim
es w
hen
you n
eed t
o c
hoose
your
met
hod o
f so
lvin
g
the
pro
ble
m w
isel
y. T
his
pro
ble
m i
s n
ot
too
dif
ficu
lt t
o s
olv
e u
sin
g a
lgeb
ra
(or
even
th
e h
yb
rid
met
ho
d).
�
II. S
olv
e th
e g
iven
pro
ble
m u
sin
g t
he
roo
t m
eth
od
.
03
34
��
��
xx
x
S
olu
tio
n:
In t
his
pro
ble
m, w
e en
ter
the
equ
atio
n i
nto
th
e ca
lcu
lato
r an
d l
oo
k a
t th
e gra
ph. A
ll t
he
feat
ure
s of
the
gra
ph c
an b
e se
en i
n t
he
stan
dar
d v
iew
ing
win
dow
so a
ll w
e tr
y t
o u
se t
he
root
finder
to d
eter
min
e th
e ze
ros.
W
hen
tr
yin
g t
o d
o t
his
, w
e g
et a
n e
rror.
T
he
calc
ula
tor
has
a p
rob
lem
her
e si
nce
it
is l
ook
ing
fo
r th
e p
lace
in
th
e in
terv
al w
her
e th
e y-
val
ues
chan
ge
sign. T
his
does
n’t
hap
pen
wit
h t
his
gra
ph s
ince
the
val
ues
for
y ar
e al
way
s nonneg
ativ
e (g
reat
er t
han
or
equal
to
zero
).
Th
ere
is a
way
to
so
lve
this
pro
ble
m u
sin
g t
he
roo
t m
eth
od
, th
ou
gh
. W
e re
call
the
foll
ow
ing
fac
t:
0�
a o
nly
when
a =
0.
So, w
e so
lve
by g
raphin
g
and
fin
din
g t
he
roo
ts.
30
34
��
��
xx
x3
34
��
��
xx
xy
88
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
T
his
sh
ow
s th
at t
he
two
so
luti
on
s to
th
e o
rig
inal
eq
uat
ion
are
ap
pro
xim
atel
y
x =
–1.4
75096 a
nd x
= 1
.2372475. �
II
I. S
olv
e th
e fo
llo
win
g e
qu
atio
n.
23
10
xx
��
��
�
So
luti
on
:
To
do
th
is p
rob
lem
we
nee
d t
o e
nte
r th
e eq
uat
ion i
nto
the
calc
ula
tor
and
vie
w t
he
gra
ph.
W
e fi
nd t
he
two v
isib
le r
oots
by u
sing t
he
zero
or
root
finder
.
T
his
giv
es u
s tw
o s
olu
tio
ns
at a
pp
rox
imat
ely
x =
–1.9
82 a
nd
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
8
9
x =
1.8
56
69
. T
his
gra
ph
lo
ok
s v
ery
sim
ilar
to
a g
rap
h o
f a
qu
adra
tic
equ
atio
n e
xce
pt
for
the
fact
that
it
app
ears
to
sto
p. D
oes
th
is g
rap
h e
xh
ibit
th
e sa
me
beh
avio
r as
th
e g
rap
h f
rom
Ex
amp
le 3
, II
I in
sec
tio
n 1
.2?
In t
his
cas
e th
e g
rap
h t
hat
we
see
do
esn
’t t
ell
the
wh
ole
sto
ry. T
he
alg
ebra
ic
form
ula
hel
ps
expla
in t
he
dif
ficu
lty. F
rom
2
31
xx
y�
��
��
we
see
that
a v
alu
e o
f y
is d
efin
ed w
hen
ever
0
���
x, th
at i
s, w
hen
ever
��
�x
. W
e
get
an
id
ea o
f th
e b
ehav
ior
by
lo
ok
ing
at
the
tab
le s
tart
ing
at
x =
–3.1
5 a
nd
goin
g b
y s
teps
of
0.0
1.
T
he
sign c
han
ge
of
y-v
alu
es b
etw
een x
= –
3.1
2 a
nd x
= –
3.1
1 i
ndic
ates
that
th
ere
is a
zer
o i
n t
hat
in
terv
al. T
he
[CA
LC
] fe
atu
re o
f th
e ca
lcu
lato
r co
mp
ute
s th
at z
ero
more
acc
ura
tely
.
(C
an y
ou
fin
d a
win
do
w m
akin
g t
he
calc
ula
tor
dis
pla
y t
his
gra
ph
more
ac
cura
tely
?) T
his
ex
amp
le s
ho
ws
that
cal
cula
tor
gra
ph
s ca
n s
om
etim
es b
e m
isle
adin
g a
nd
in
com
ple
te.�
E
xer
cise
s 1
.3
Fo
r p
rob
lem
s 1
– 6
, so
lve
the
syst
ems
of
equ
ati
on
s a
lgeb
raic
all
y b
y s
ub
stit
uti
on
an
d a
lso
by
gra
ph
ing
. C
hec
k y
ou
r a
nsw
ers
to b
e su
re t
ha
t th
ey a
re t
he
sam
e
for
bo
th m
eth
od
s.
Fo
r ea
ch p
rob
lem
dec
ide
wh
ich
met
hod
pro
vid
es a
mo
re
effi
cien
t m
eth
od
of
solu
tio
n.
Ex
pla
in t
he
rea
son
s fo
r y
ou
r ch
oic
es.
90
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
1.
4
2
13
��
��
�y
xy
x
2.
61
13
��
��
yx
xy
3.
50
��
��
sr
sr
4.
2
3
02
��
��
��
yx
yx
5.
x
xy
xx
y4
53
2
2
��
�
��
�
6.
x
yx
y2
72
��
�
7.
15
01
75
15
45
03
25
85
2
2
��
��
��
�
xx
yx
xy
8.
xx
yx
y5.
23
53
4
22
��
��
F
or
pro
ble
ms
9 –
10
, fi
nd
ap
pro
xim
ate
so
luti
on
s to
th
e eq
ua
tio
ns
usi
ng
th
e
inte
rsec
tio
n m
eth
od
. (
Ro
un
d y
ou
r a
nsw
ers
to 2
dec
ima
l p
lace
s.)
9.
4
53
23
��
��
xx
x10.
xx
x24
35
23
��
� F
or
pro
ble
ms
11
– 1
8, so
lve.
Y
ou
ma
y u
se a
ny
met
ho
d t
ha
t w
ork
s. Y
ou
ma
y
wa
nt
to t
hin
k a
bo
ut
the
pro
ble
m b
efo
re y
ou
ch
oo
se y
ou
r st
rate
gy
. W
hen
ap
pro
pri
ate
, ro
un
d y
ou
r a
nsw
ers
to 2
dec
ima
l p
lace
s.
11.
0
23
43
��
�x
x12.
xx
��
34
13.
22
36
13
xx
��
14.
2
34
12
xx
xx
��
��
15.
03
34
��
��
xx
x
S
ecti
on
1.3
T
he
Inte
rsec
tio
n M
eth
od
9
1
16.
y
xy
xx
�
�� 2
3
4
4
17.
42
121
212
xx
x�
�
18.
15
2�
�xx
Fo
r p
rob
lem
s 1
9 a
nd
20
, fi
nd
an
ex
act
(N
o d
ecim
als
all
ow
ed!)
so
luti
on
fo
r th
e
equ
ati
on
in
th
e giv
en i
nte
rval.
Y
ou
can
use
a c
alc
ula
tor
to f
ind
an
ap
pro
xim
ate
so
luti
on
. T
his
dec
ima
l v
alu
e sh
ou
ld t
hen
be
con
ver
ted
to
th
e
exa
ct v
alu
e, e
.g.
3732050808
.�
1.
19.
�� 1,
0,
02
32
23
��
��
�x
xx
x3
20.
�� 1,
0,
02
25
12
23
4�
��
��
�x
xx
xx
12
Ap
pli
cati
on
s
21. A
cco
rdin
g t
o d
ata
fro
m t
he
Nat
ional
Cen
ter
for
Ed
uca
tio
nal
Sta
tist
ics
and
th
e C
oll
ege
Bo
ard
, th
e av
erag
e co
st y
of
tuit
ion a
nd f
ees
(in t
housa
nds
of
do
llar
s) a
t pu
bli
c fo
ur-
yea
r in
stit
uti
on
s in
yea
r x
is a
pp
rox
imat
ed b
y t
he
equ
atio
n
195
.2
114
.00039
.00044
.2
3�
��
�x
xx
y,
wh
ere
x =
0 c
orr
esp
on
ds
to 1
99
0. I
f th
is m
od
el c
on
tin
ues
to
be
accu
rate
, in
w
hat
yea
r w
ill
tuit
ion
and
fee
s re
ach
$4
,00
0?
22. A
ccord
ing t
o d
ata
from
the
US
Dep
artm
ent
of
Hea
lth a
nd H
um
an S
ervic
es,
the
cum
ula
tiv
e n
um
ber
y o
f A
IDS
cas
es (
in t
housa
nds)
dia
gnose
d i
n t
he
Unit
ed S
tate
s duri
ng 1
982 –
1993 i
s ap
pro
xim
ated
by
75
.17
81
.14
223
.3
2�
��
xx
y,
wh
ere
x =
0 c
orr
esp
on
ds
to 1
98
0. I
n w
hat
yea
r d
id t
he
cum
ula
tiv
e n
um
ber
of
case
s re
ach 2
50,0
00?
92
Ch
apte
r 1
G
rap
hs
an
d C
alc
ula
tors
C
hap
ter
1 R
evie
w
Mid
po
int
an
d D
ista
nce
F
or
pro
ble
ms
1 –
4,
fin
d t
he
dis
tan
ce b
etw
een
th
e tw
o p
oin
ts,
an
d f
ind
th
e
mid
po
int
of
the
lin
e se
gm
ent
join
ing
th
em.
1.
(9,
7)
and
(7
, 9
) 2
. (–
23
.5,
45
.1)
and
(–
1.7
5,
79
) 3
. (7
, 0
) an
d (
–5
0,
0)
4.
(0,
4.7
5)
and
(0
, –
92
) F
or
pro
ble
ms
5 &
6,
fin
d t
he
mid
po
int
of
the
two
x-i
nte
rcep
ts o
f th
e g
rap
h o
f
the
equ
ati
on
.
5.
6
07
2�
��
xx
y6
.
71
22
��
�x
xy
Co
mp
lete
Gra
ph
s F
or
pro
ble
ms
7 &
8,
dra
w a
co
mp
lete
gra
ph
of
the
equ
ati
on
. F
ind
an
d l
ab
el a
ll
of
the
zero
s.
7.
9
51
67
45
64
52
3�
��
�x
xx
y8
. x
xx
xy
23
61
25
73
4�
��
� S
yst
ems
of
Eq
ua
tio
ns
Fo
r p
rob
lem
s 9
& 1
0, so
lve
the
syst
em o
f eq
ua
tio
ns.
9.
yx
yx
x�
��
��
18
5
5 2
3
10
.
yx
xy
xx
��
�
��
��
45
10
25
3
2
2
Fo
r p
rob
lem
s 1
1 –
15
, so
lve
if p
oss
ible
.
11
. 1
34
��
�x
x
12
. x
x�
��
1
13
.
15
17
10
25
44
25
��
��
xx
xx
14. 1
0
35
105
35
85
23
4�
��
��
xx
x
15. T
he
pro
fit
earn
ed i
n a
month
by t
he
law
fir
m D
ewey
, C
hea
tum
, an
d H
ow
e is
model
ed b
y t
he
equat
ion
, w
her
e x
is t
he
num
ber
of
case
s han
dle
d. K
now
ing t
hat
the
law
fir
m c
an h
andle
at
most
40 c
ases
a
month
, how
man
y c
ases
do t
hey
nee
d t
o h
andle
to e
arn $
41,3
56.2
2 a
month
?
xx
y1500
756
.9
2�
��
A.4
An
swer
s to
Od
d-N
um
ber
ed E
xer
cise
s
Ch
ap
ter
1
Sec
tio
n 1
.1, p
ag
e 6
4
Po
ssib
le g
rap
hs:
b, c,
d
*
Oth
er i
nte
rpre
tati
on
s p
oss
ible
bas
ed o
n g
oo
d
reas
on
ing
.
1.
d =
15
.55
6,
�
21,
21 ��
29. Y
es
3. d
= 6
, �
� 11
,7
31. N
o
5. d
= 8
1.2
16
, �
� 19
,20
33. P
oss
ibly
Tru
e. C
an’t
pro
ve
by
loo
kin
g a
t th
e g
rap
h.
7.
�� �
��
0,
2,
21
21
xx
xx
�d
S
ecti
on
1.2
, p
ag
e 7
4
1.
54�
�x
9. 1
3.5
61
11. a.
(32, 41.5
)
b.
54
.40
6 m
iles
3. N
o z
eros
c. B
5. x
= 3
, 0, or
–11
13.
7. x
= 3
or
–3
9. x
= –
3, 3, –2, or
2
15.
11. x
= –
15, 4, or
12
13
. Jo
e fo
rgo
t p
aren
thes
es i
n
the
nu
mer
ato
r.
17. a.
Yes
b. N
o
c.
Yes
15. 3
d. N
o
17. 2
19. y
= –
27
, x
= 3
y =
4.5
, x �
0.4
5
19. x �
-1.7
521
y =
8, N
o v
alue
of
x ex
ists
21. x
= 0
or
appro
x.
2.2
074
21. b
23. b
Sec
tio
n 1
.3, p
ag
e 8
9
25. M
ust
show
the
foll
ow
ing
po
ints
:
1.
�
513
5��,
6
(-3
.51
8, 0
), (
11
.30
8, 0
)
(0, 50),
(8.1
5, 224.7
2)
3.
25,
25�
�s
�r
2
7. N
ot
po
ssib
le g
rap
hs:
a, e,
f
A
nsw
ers
to O
dd-N
um
ber
ed E
xer
cise
s A
.5
5.
��
,5
����
3,1
,415
2�
7. (2
, 140),
(3, 240)
9. x
= 1
.40
11.
21�
�x
13.x
= 3
.92 o
r –3.9
2
15. x
= 1
.24 o
r –1.4
8
17. x
= 0
or
2.2
1
19.
32�
x
21. 2001
Ch
ap
ter
1 R
evie
w,
pa
ge
92
1.
8
3.
57
5.
(12, 0)
& (
–5, 0):
(3.5
, 0)
7. R
oots
at
–13, –12, an
d
561
9.
(1.3
4, 9.0
7)
11.
32�
13. x �
4.5
4
15. 36 c
ases
top related