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1.2 Distance e BetweenTwoPoints; Circles
7
17. Marketresearchtellsyouthatifyousetthepriceofanitemat\$1.50,youwillbeabletosell
5000items;andforevery10centsyoulowerthepricebelow\$1.50youwillbeabletosell
another1000items.Letxbethenumberofitemsyoucansell,andletP bethepriceofan
item.(a)ExpressP linearlyintermsofx,inotherwords,expressPintheformP =mx+b.
(b)ExpressxlinearlyintermsofP.
18. Aninstructor r gives a100-pointﬁnalexam,anddecides that ascore 90or above willbea
oftheformy=mx+bwhichappliestoscoresxbetween40and90.
Giventwopoints(x
1
,y
1
)and(x
2
,y
2
),recallthattheirhorizontaldistancefromoneanother
is∆x=x
2
−x
1
andtheirverticaldistancefromoneanotheris∆y=y
2
−y
1
.(Actually,the
word“distance”normallydenotes“positive distance”. . ∆xand∆yare e signed d distances,
butthisisclearfromcontext.) Theactual(positive)distancefromonepointtotheother
isthe length h of f the e hypotenuse e of a a right t triangle e with legs s ∆x and d ∆y, asshown in
ﬁgure1.4. ThePythagoreantheoremthensaysthatthedistancebetweenthetwopoints
isthesquarerootofthesumofthesquaresofthehorizontalandverticalsides:
distance=
p
(∆x)2+(∆y)2=
p
(x
2
−x
1
)2+(y
2
−y
1
)2.
Forexample,thedistancebetweenpointsA(2,1)andB(3,3)is
p
(3−2)2+(3−1)=
5.
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(x
1
,y
1
)
(x
2
,y
2
)
∆x
∆y
Figure1.4
Distancebetweentwopoints.
Asaspecialcaseofthedistanceformula,supposewewanttoknowthedistanceofa
point(x,y)totheorigin.Accordingtothedistanceformula,thisis
p
(x−0)2+(y−0)=
p
x2+y2.
p
x2+y=r,or,ifwe
squarebothsides: x
2
+y
2
=r
2
theorigin. Thespecialcaser=1iscalledtheunitcircle;itsequationisx2+y=1.
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8
Chapter 1 1 AnalyticGeometry
pointCifandonlyif
p
(x−h)2+(y−k)=r,i.e.,ifandonlyif
(x−h)
2
+(y−k)
2
=r
2
.
2
+(y−5)
2
=
25.Ifweexpand(y−5)
2
=y
2
−10y+25andcancelthe25onbothsides,wecanrewrite
thisas: x2+y2−10y=0.
Exercises
a) (0,0)
d) (0,3)
b) (5,6)
e) (0,−3)
c) (−5,−6)
f) (3,0)
2. For r eachpairof points A(x
1
,y
1
) andB(x
2
,y
2
) ﬁnd(i)∆xand∆yingoingfromAtoB,
(ii)theslopeofthelinejoining AandB,(iii) the equationofthelinejoiningAandB in
theformy=mx+b,(iv)thedistancefromAtoB,and(v)anequationofthecirclewith
centeratAthatgoesthroughB.
a) A(2,0),B(4,3)
d) A(−2,3),B(4,3)
b) A(1,−1),B(0,2)
e) A(−3,−2),B(0,0)
c) A(0,0),B(−2,−2)
f) A(0.01,−0.01),B(−0.01,0.05)
3. Graphthecirclex
2
+y
2
+10y=0.
4. Graphthecirclex
2
−10x+y
2
=24.
5. Graphthecirclex
2
−6x+y
2
−8y=0.
6. Findthe e standard equationof the circle passing through (−2,1) and tangent to the e line
3x−2y=6atthepoint(4,3). Sketch. (Hint: Thelinethroughthecenterofthecircleand
thepointoftangencyisperpendiculartothetangentline.)
A function y y =f(x) ) isa rule for determining y y when n you’re given avalue of x. . For
example,theruley=f(x)=2x+1isafunction. Anyliney=mx+biscalledalinear
function. The e graphofafunctionlookslikeacurveabove(or below) the x-axis,where
foranyvalueofxtheruley=f(x)tellsyouhowfartogoabove(orbelow)thex-axisto
reachthecurve.
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1.3 Functions
9
Functionscanbedeﬁnedinvariousways: byanalgebraicformulaorseveralalgebraic
formulas,byagraph,or byanexperimentallydeterminedtableofvalues. . (Inthe e latter
case, the e table givesa bunch of pointsin the plane, , which we e then interpolate e with h a
smoothcurve.)
Givenavalueofx,afunctionmust giveyouatmostonevalueof y. . Thus,vertical
linesarenotfunctions.Forexample,thelinex=1hasinﬁnitelymanyvaluesofyifx=1.
Itisalsotruethatifxisanynumbernot1thereisnoywhichcorrespondstox,butthat
isnotaproblem—onlymultipleyvaluesisaproblem.
x
2
. Youcandrawthegraphofthisfunctionbytakingvariousvaluesofx(say,atregular
intervals) and plotting the points s (x,f(x)) ) = = (x,x
2
). Then n connect t the e points s with h a
smoothcurve.(Seeﬁgure1.5.)
The two examplesy=f(x) =2x+1 and y=f(x) =x
2
arebothfunctionswhich
canbeevaluatedat anyvalue ofxfromnegative inﬁnitytopositiveinﬁnity. . For r many
functions, however, it t onlymakessense to take e x in n some e interval or outside of f some
“forbidden”region.Theintervalofx-valuesatwhichwe’reallowedtoevaluatethefunction
iscalledthedomainofthefunction.
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y=f(x)=x
2
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y=f(x)=
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.
y=f(x)=1/x
Figure 1.5
Somegraphs.
Forexample,thesquare-rootfunctiony=f(x)=
xistherulewhichsays,givenan
x-value,takethenonnegativenumberwhosesquareisx. Thisruleonlymakessenseifx
ispositiveor zero. . Wesaythat t thedomainofthisfunction isx≥0,or more formally
{x∈R|x≥0}. Alternately,wecanuseintervalnotation,andwritethatthedomainis
[0,∞). (Inintervalnotation,squarebracketsmeanthat t theendpointisincluded,anda
parenthesismeansthattheendpointisnotincluded.) Thefactthatthedomainofy=
x
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10
Chapter1 AnalyticGeometry
isx≥0 meansthat inthe graphofthisfunction((seeﬁgure 1.5) wehavepoints(x,y)
onlyabovex-valuesontherightsideofthex-axis.
Anotherexampleofafunctionwhosedomainisnottheentirex-axisis: y=f(x)=
1/x,thereciprocalfunction. We e cannot substitute x=0in thisformula. . The e function
makessense,however,foranynonzerox,sowetakethedomaintobe: x=0. Thegraph
ofthisfunctiondoesnothaveanypoint(x,y)withx=0. Asxgetscloseto0fromeither
side,thegraphgoesoﬀtowardinﬁnity.Wecalltheverticallinex=0anasymptote.
Tosummarize,tworeasonswhycertainx-valuesareexcludedfromthedomain of a
functionarethat (i)wecannot dividebyzero,and (ii)we cannot takethe squareroot
ofa negative number. . Wewillencounter r some other waysinwhichfunctionsmight be
undeﬁnedlater.
Another reason whythedomainofafunctionmightberestrictedisthatinagiven
situationthex-valuesoutsideofsomerangemighthavenopracticalmeaning.Forexample,
if y y is the e area of a a square of f side e x, then n we can write y y = = f(x) = x
2
. In n a purely
mathematicalcontextthedomainofthefunctiony=x
2
isallx.Butinthestory-problem
contextofﬁndingareasofsquares,werestrictthedomaintox>0,becauseasquarewith
negativeorzerosidemakesnosense.
Inaprobleminpuremathematics,weusuallytakethedomaintobeallvaluesof x
atwhichtheformulascanbeevaluated. Butinastoryproblemtheremightbefurther
restrictionsonthedomainbecauseonlycertainvaluesofxareofinterestormakepractical
sense.
In astoryproblem,often lettersdiﬀerentfrom m xandy y are used. . Forexample,the
3
πr
3
.
Also,lettersdiﬀerentfromfmaybeused.Forexample,ifyisthevelocityofsomethingat
(andtplayingtheroleofx).
Theletterplayingtheroleofxiscalledtheindependent variable,andtheletter
playingtheroleofyiscalledthe dependent variable(becauseitsvalue“dependson”
thevalueoftheindependentvariable). Instoryproblems,whenonehastotranslatefrom
Ifonlywordsandnolettersaregiven,thenyouhavetodecidewhichletterstouse.Some
EXAMPLE1.3
bycuttingoutasquareofsidexfromeachofthefourcorners,andthenfoldingthesides
upandsealing themwithducttape. . Finda a formulafor thevolume V V of f theboxasa
functionofx,andﬁndthedomainofthisfunction.
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1.3 Functions
11
Heretheboxwegetwillhaveheightxandrectangularbaseofdimensionsa−2xby
b−2x.Thus,
V =f(x)=x(a−2x)(b−2x).
Hereaandbareconstants,andV isthevariablethatdependsonx,i.e.,V isplayingthe
roleofy.
Thisformulamakesmathematicalsenseforanyx,butinthestoryproblemthedomain
ismuchless. Intheﬁrstplace,x x mustbepositive. . Inthesecondplace,itmustbeless
thanhalfthelengthofeitherofthesidesofthecardboard.Thus,thedomainis
0<x<
1
2
(minimumofaandb).
Inintervalnotationwewrite: thedomainistheinterval(0,min(a,b)/2).
EXAMPLE 1.4
Circle of radius s r r centered d at t the origin
The equation for
thiscircle isusuallygiveninthe formx
2
+y
2
=r
2
. To o writethe equationinthe form
y =f(x) wesolvefor y,obtainingy =±
r2−x2. But t thisisnotafunction, , because
whenwesubstitute avaluein(−r,r)for xthere aretwocorresponding valuesofy. . To
get a function, , we e must choose one of the two signsin front ofthe square root. . If f we
choose thepositivesign,for example,wegetthe uppersemicircley =f(x)=
r2−x2
(seeﬁgure1.6).Thedomainofthisfunctionistheinterval[−r,r],i.e.,xmustbebetween
−randr(includingtheendpoints).Ifxisoutsideofthatinterval,thenr
2
−x
2
isnegative,
andwecannottakethesquareroot.Intermsofthegraph,thisjustmeansthatthereare
nopointsonthecurvewhosex-coordinateisgreaterthanrorlessthan−r.
−r
r
.
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Figure1.6
Uppersemicircley=
r−x2
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12
Chapter1 AnalyticGeometry
EXAMPLE 1.5
Findthedomainof
y=f(x)=
1
4x−x2
.
2
negative(because
wecannottakethesquarerootofanegative)andalsothex-valuesthatmake4x−x
2
zero
(becauseif4x−x
2
=0,thenwhenwetakethesquarerootweget0,andwecannotdivide
by0). Inotherwords,thedomainconsistsofallxforwhich4x−xisstrictlypositive.
Wegivetwodiﬀerentmethodstoﬁndoutwhen4x−x
2
>0.
Firstmethod. Factor r 4x−x
2
asx(4−x). The e product oftwonumbersispositive
when either both are positive or botharenegative,i.e., if either x >0 and 4−x>0,
orelsex<0and4−x<0. Thelatteralternative e isimpossible,since if xisnegative,
then4−xisgreaterthan4,andsocannotbenegative. Asfortheﬁrstalternative,the
and4>x(thisissometimescombinedintheform4>x>0,or,equivalently,0<x<4).
Inintervalnotation,thissaysthatthedomainistheinterval(0,4).
Secondmethod. Write4x−xas−(x2−4x),andthencompletethesquare,obtaining
(x−2)2−4
·
=4−(x−2)2. For r thisto be positive we need(x−2)<4,which
everythinggives0<x<4.Bothofthesemethodsareequallycorrect;youmayuseeither
inaproblemofthistype.
seen (in the incometaxproblem, for example). . For r example, , suppose e that y y =v(t) ) is
the velocityfunctionfor acarwhichstartsout fromrest (zero velocity) at time t=0;
constantspeed20m/secfor 15seconds;andﬁnallyappliesthebrakestodecreasespeed
thethreetimeintervals:ﬁrsty=2x,theny=20,theny=−4x+120. Thegraphofthis
functionisshowninﬁgure1.7.
Not all functionsare given n by formulasat t all. . A A function n can be given by an ex-
perimentallydeterminedtable of values, , or r bya description other thana formula. . For
example,thepopulationyoftheU.S.isafunctionofthetimet: wecanwritey=f(t).
forvarioust—butyoucan’tﬁndanalgebraicformulaforit.
1.3 Functions
13
10
25
30
0
10
20
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t
v
Figure 1.7
Avelocityfunction.
Exercises
Findthedomainofeachofthefollowingfunctions:
1. y=f(x)=
2x−3
2. y=f(x)=1/(x+1)
3. y=f(x)=1/(x
2
−1)
4. y=f(x)=
p
−1/x
5. y=f(x)=
3
x
6. y=f(x)=
4
x
7. y=f(x)=
p
r2−(x−h),whererandharepositiveconstants.
8. y=f(x)=
p
1−(1/x)
9. y=f(x)=1/
p
1−(3x)
10. y=f(x)=
x+1/(x−1)
11. y=f(x)=1/(
x−1)
12. Findthedomainofh(x)=
(x
2
−9)/(x−3) x=3
6
ifx=3.
13. Supposef(x)=3x−9andg(x)=
x. Whatisthedomainofthecomposition(g◦f)(x)?
(Recall that composition is deﬁned d as s (g◦f)(x) ) = = g(f(x)).) ) What t is the e domain of
(f◦g)(x)?
14. Afarmerwantstobuildafencealongariver.Hehas500feetoffencingandwantstoenclose
arectangularpenonthreesides(withtheriverprovidingthefourthside).Ifxisthelength
ofthesideperpendiculartotheriver,determinetheareaofthepenasafunctionofx. What
isthedomainofthisfunction?
15. Acan n inthe shape of a cylinder r is to be made with h atotal of 100square centimeters s of
materialintheside,top,andbottom;themanufacturerwantsthecantoholdthemaximum
ofthefunction.
14
Chapter1 AnalyticGeometry
16. Acanin n the shape of acylinder r is tobemade toholda volume of one liter (1000 cubic
centimeters). Themanufacturerwantstousetheleastpossiblematerialforthecan. Write
ofthecan;ﬁndthedomainofthefunction.
Manyfunctionsinapplicationsarebuiltupfromsimplefunctionsbyinsertingconstants
in various s places. . It t isimportant to understandthe eﬀect t such h constants s have e on the
appearanceofthegraph.
Horizontal shifts. . If f you replacex x by y x−C everywhereitoccursin theformula for
f(x),thenthegraphshiftsoverCtotheright. (IfCisnegative,thenthismeansthatthe
graphshiftsover|C|totheleft.) Forexample,thegraphofy=(x−2)
2
isthex
2
-parabola
shiftedovertohaveitsvertexatthepoint2onthe x-axis. . Thegraphofy=(x+1)
2
is
thesameparabolashiftedovertotheleftsoastohaveitsvertexat−1onthex-axis.
Vertical shifts. . If f youreplacey byy−D, thenthegraph moves up D units. . (IfD D is
negative,thenthismeansthatthegraphmovesdown|D|units.) Iftheformulaiswritten
intheformy=f(x)andifyisreplacedbyy−Dtogety−D=f(x),wecanequivalently
moveD totheother side oftheequationandwrite y=f(x)+D. . Thus,thisprincipal
canbestated: togetthegraph h ofy=f(x)+D,takethegraphofy=f(x)and moveit
D unitsup. . Forexample,thefunctiony=x
2
−4x=(x−2)
2
−4canbeobtainedfrom
y=(x−2)(seethelastparagraph)bymovingthegraph4unitsdown.Theresultisthe
x
2
-parabolashifted2unitstotherightand4unitsdownsoastohaveitsvertexatthe
point(2,−4).
Warning. Donotconfusef(x)+Dandf(x+D).Forexample,iff(x)isthefunctionx
2
,
thenf(x)+2isthefunctionx2+2,whilef(x+2)isthefunction(x+2)2=x2+4x+4.
EXAMPLE 1.6
Circles
Animportantexampleoftheabovetwoprinciplesstarts
withthecirclex
2
+y
2
=r
2
saw,thisisnotasinglefunctiony=f(x),but rathertwofunctionsy=±
r2−xput
together;inanycase,thetwoshiftingprinciplesapplytoequationslikethisonethatare
notintheformy=f(x).) Ifwereplacexbyx−Candreplaceybyy−D—gettingthe
equation(x−C)2+(y−D)=r2—theeﬀectonthecircleistomoveit Ctotheright
ushowtowritetheequationofanycircle,notnecessarilycenteredattheorigin.
Wewilllater wanttousetwomore principlesconcerningtheeﬀectsofconstantson
theappearanceofthegraphofafunction.
1.4 ShiftsandDilations
15
Horizontal dilation. . Ifxisreplacedbyx/AinaformulaandA>1,thentheeﬀecton
thegraphisto expanditbyafactorofAinthex-direction(awayfromthey-axis). . IfA
isbetween0and1thentheeﬀectonthegraphistocontractbyafactorof1/A(towards
they-axis).Weusetheword“dilate”inbothcases.
Forexample,replacingxbyx/0.5=2xhastheeﬀectofcontractingtowardthey-axis
bya factor of2. . If f A isnegative,we dilate bya factor of |A|andthen ﬂip about the
y-axis. Thus,replacing g xby−x hasthe eﬀect oftakingthemirror imageofthe graph
withrespecttothey-axis. Forexample,thefunctiony=
−x,whichhasdomainx≤0,
isobtainedbytaking thegraphof
xandﬂippingitaroundthey-axisinto thesecond
Vertical dilation. . If f y isreplaced byy/B B ina a formula and B >0,then theeﬀecton
thegraph isto dilateitbya factorof B B in n thevertical direction. . Note e thatifwe have
afunctiony=f(x),replacingy byy/Bisequivalenttomultiplyingthefunctiononthe
rightbyB: y=Bf(x). . Theeﬀectonthegraphistoexpandthepictureawayfromthe
x-axisbya factorof B ifB >1,tocontract it towardthe x-axisbya factorof 1/B if
EXAMPLE1.7
Ellipses
Abasicexampleofthetwoexpansionprinciplesisgiven
byanellipseofsemimajoraxisaand semiminoraxis b. . Wegetsuchanellipseby
starting with theunit circle—thecircle of radius1 centeredat the origin,the equation
ofwhichisx
2
+y
2
=1—anddilatingbyafactorof ahorizontallyandbyafactor ofb
vertically. Togettheequationoftheresultingellipse,whichcrossesthex-axisat±aand
crossesthey-axisat ±b,wereplacexbyx/aandy byy/bintheequationforthe unit
circle.Thisgives
x
a
·
2
+
y
b
·
2
=1.
Finally, ifyou want t toanalyze afunction thatinvolvesbothshiftsanddilations,it
isusuallysimplest to workwiththe dilationsﬁrst,andthen theshifts. . Forinstance,if
youwanttodilateafunctionbyafactorofAinthex-directionandthenshiftCtothe
right,youdothisbyreplacingxﬁrstbyx/Aandthenby(x−C)intheformula. Asan
example,supposethat,afterdilatingourunitcirclebyainthex-directionandbybinthe
16
Chapter1 AnalyticGeometry
ellipsewouldhaveequation
µ
x−h
a
2
+
µ
y−k
b
2
=1.
Notewellthatthisisdiﬀerentthanﬁrstdoingshiftsbyhandkandthendilationsbya
andb:
x
a
−h
·
2
+
y
b
−k
·
2
=1.
1
2
3
−1
1
2
3
4
−2
−1
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Figure1.8
Ellipses:
x−1
2
2
+
y−1
3
2
=1ontheleft,
x
2
−1
2
+
y
3
−1
2
=1onthe
right.
Exercises
Startingwith the graphof y=
x,thegraphof y= 1/x,andthe graph ofy =
1−x(the
upperunitsemicircle),sketchthegraphofeachofthefollowingfunctions:
1. f(x)=
x−2
2. f(x)=−1−1/(x+2)
3. f(x)=4+
x+2
4. y=f(x)=x/(1−x)
5. y=f(x)=−
−x
6. f(x)=2+
p
1−(x−1)2
7. f(x)=−4+
p
−(x−2)