y 1.1 Graphs The Rectangular Coordinate System: ( x y · The Rectangular Coordinate System: (x,...
21
Section 1.1 Graphs 55 1.1 Graphs Much of algebra is concerned with solving equations. Many algebraic techniques have been developed to provide insights into various sorts of equations, and those techniques are essential for understanding the basic ideas of calculus and more advanced mathematical topics. In recent decades people have also used calculators and computers for calculating approximate solutions to certain types of equations. Such numerical solution methods are often essential for solving real problems in engineering and science. In this course we investigate problems both algebraically and graphically. It is often useful to consider a situation from several viewpoints: a different direction can provide insights that are not obvious from the original position. The geometrical approach to solving equations has been made more productive by the invention of the graphing calculator. These calculators quickly construct graphs that are fairly accurate and reliable. In pre-calculator years, the graphical approach was much more tedious. In the next two sections (1.2 and 1.3) we will explore various methods to solve equations graphically. (It is assumed that we already know the algebraic methods. Those methods are outlined in sections 2.1 and 2.2 but we will not spend time to review them now.) In this section, we will discuss the rectangular coordinate system, properties of graphs, and how to use the calculator to display a graph. The Rectangular Coordinate System: (x, y)-plane The rectangular coordinate system is a system for labeling points in the plane. The coordinate plane is constructed from two number lines, the x-axis and the y-axis. The x-axis and the y-axis are perpendicular to each other and intersect at the point 0 on both lines. (Perpendicular means that the two lines meet at 90 angles.) The point of intersection is called the origin. See Figure 1 below. y-axis II I origin x-axis III IV
Transcript of y 1.1 Graphs The Rectangular Coordinate System: ( x y · The Rectangular Coordinate System: (x,...
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
