Graphs and Equations

8/8/2019 Graphs and Equations

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Graphsandequations

A summary of everything that we now know which will help us to sketch curves of the form

y = ax 2 + bx + c

1. If a is positive, the curve is U-shaped.

If a is negative, the curve is an upside-down U.

2. The value of c tells us the y -intercept. The curve crosses the y -axis at (0,c ).

3. We can factorize (or use the formula) to find whether and where the curve cuts the x -axis.

If b2 – 4ac is negative, the curve does not cut the x -axis at all.

4. We can complete the square to find where the least value of the curve is (or the greatest value, if it is an inverted U-shape). We shall see in Section 8.E.(b) that this can also be found by usingcalculus.

If the curve does cut the x -axis, substituting the midway value of x between the cuts into theequation for y gives the least value of y (or the greatest value of y if the curve has an inverted U-shape).

Stretching and shifting – new functions from old

(1) Adding a fixed amount to a function

What happens if we go from f ( x ) to f ( x ) + a, where a is some given constant number? Here

are two examples, both taking a = 3.

(a) f ( x ) = 2 x + 1 (b) f ( x ) = x 2 so f ( x ) + 3 = 2 x + 4. so f ( x ) + 3 = x 2 + 3.

I show sketches of the two pairs of graphs below in Figure (a) and (b).

We see that the effect of adding 3 to f ( x ), so that y = f ( x ) + 3, is to shift the graph up by

3 units.

(2) Adding a fixed amount to each x value

What will happen if we add a fixed amount to each x value instead, so that we go from f ( x ) to f ( x + a) ineach case? Again, we look at two examples, taking a = 3.



(a) f ( x ) = 2 x + 1 (b) f ( x ) = x 2 so f ( x + 3) = 2( x + 3) + 1 = 2 x + 7. so f ( x + 3) = ( x + 3)2.

Notice that, to find f ( x + 3) from f ( x ), we just replace x by ( x + 3). This time, the effect is to slide thewhole graph 3 units to the left.

(3) Multiplying the original function by a fixed amount

What will happen if we go from f ( x ) to a f ( x ) where a is some given constant number?

Working with the same two examples as before, and with a = 3 again, we get

(a) f ( x ) = 2 x + 1 so 3f ( x ) = 6 x + 3

(b) f ( x ) = x 2 so 3f ( x ) = 3 x

2

Sketches of the two pairs of graphs are shown below in Figure (a) and (b).

This time, the whole graph has been pulled away from the x -axis by a factor of 3, so that every point is

now three times further away than it was originally. Therefore the only points on the graph, which willremain unchanged, are those on the x -axis itself.

(4)Multiplying xbyafixedamount

What will happen if we go from f ( x ) to f (ax )?

Taking our same two examples, with a = 3, we have

(a) f ( x ) = 2 x + 1 so f (3 x ) = 2(3 x ) + 1 = 6 x + 1

(b) f ( x ) = x 2 so f (3 x ) = (3 x )

2 = 9 x 2.

Notice that we simply replace x by 3 x to find f (3 x ) from f ( x ).

I show sketches of the two pairs of graphs below in Figure (a) and (b). This time the stretching effect ismore complicated because it only affects the part of the function involving x . Any purely number parts



remain unchanged. The points which are unaffected by the stretching are those where the graphs cutthe y -axis, so x = 0.

Notice too that the strength of the effect now depends upon the power of x . Having (3 x )2 in example 4(b)

gives a more extreme effect than the 3 x 2 in 3(b), since the 3 is also being squared here.

Asummaryofsomeeffectsoftransformingfunctions

(1)Transforming f( x )to f( x )+ashiftsthewholeof f( x )upwardsbyadistancea.Wehave

(2)Transforming f( x )into f( x+a)shiftsthewholeof f( x )backadistancea,becausethecurveisgettingto

eachofitsvaluesfaster,byanamounta.Wehavefig.(b)

(3)Transforming f( x )intoaf( x )stretchesouteachvalueof f( x )byafactora.Wehavefig.(c)

(4)Transforming f( x )into f(ax )hasamorecomplicatedeffect,sincehowmuchaaffectseachpartof f( x )

dependsonwhatishappeningto xitselfin f( x ).Forexample,if f( x )= x 2+ x+1,then f(ax )=a2 x 2+ax+1.

Eachtermhasbeenaffecteddifferently.Thereforeitisnotpossibletoshowthiscaseononesketch;the

changeinshapewilldependentirelyuponthefunctionconcerned.





Evenfunctions

Afunctionisevenifitissymmetricalaboutthe y ‐axis.Forthesefunctions, f( x )= f(– x )foranyvalueof x .

Oddfunctions

Afunctionisoddifrotationthroughahalf‐turnleavesitunchanged.Thisisthesameassayingthatthe

functionreversesitssignifitisreflectedinthe y ‐axis,so f( x )=– f(– x ).

Seeifyoucandecidewhichof(a),(b),(c)and(d)haveinversefunctions.(a)and(c)willeachhavean

inversefunctionbecauseeachvalueof yisgivenbyonlyonepossiblevalueof x ,but(b)and(d)willonly

haveinverserelations.

With(b)forexample,if y=4then xcouldbe+2or–2.

If y= x 2then x= y 1/2.Theinverserelationis x: x 1/2,and x 1/2canbeeither+or–.

ThesketchinFigure(a)showsthegraphsof y= x 2

anditsinverserelation y= x

1/2

.However,ifwesaythat xcannotbenegative,sothatwerestrictthedomainof y= x 2tovaluesof xwhicharegreaterthanorequalto0

(whichwewriteas x≥0),thenweshallhaveaperfectlygoodinversefunctionwhichis y=√ x .Thisis

showninFigure(b).Thesymbol√istakentomeanthepositivesquarerootonly.



Findingandworkingwiththeequationswhichgivecircles

Howcanwefindtheequationofthecurvewhichgivesaparticularcircleintermsof xand y ?

Wewillstartbyconsideringthesimplestcasewhichisacircleofradiusrsymmetricallyplacedsothatits

centreisattheorigin.

AnypointPonit,withcoordinates( x , y ),mustbeadistancerfromtheorigin,so x 2+ y 2=r 2byPythagoras’

Theorem.Theequationofthecirclewithcentre(a,b)andradiusrisgivenby( x–a)2+( y–b)2=r 2.



Theequationofthecirclewithcentre(a,b)andradiusrcanalsobewrittenas

x 2–2ax+ y 2–2by+c=0wherec=a2+b2–r 2.

Foranequationlikethistogiveacircleitmustfitthefollowingconditions.

(1)Theremustbeequalcoefficientsof x 2and y 2.Thecoefficientisthenumberwhichtellsushowmany

we’vegot.Thecoefficientof3 x 2is3.Thecoefficientof y 2is1.Iftherearenotermsin x ,say,thenthe

coefficientof xiszero.

(2)Theremustonlybe,atthemost,termsin x 2, y 2, x , yandanumber.(Wemustn’thaveanytermswith xy ,

forinstance.)

(3)Thevalueofr 2mustbepositivesothatwehaveaphysicallypossiblelengthfortheradius.

Example(1)Findwhether,andifsowhere,thelines(a) y=2 x–4(b)3 y= x+11and(c) y=3 x+6cutthe

circlewhoseequationis x 2–4 x+ y 2–2 y–5=0.Drawasketchshowingthethreelinesandthecircle.

Answer:(a)Iftheline y=2 x–4cutsthecircle,thevaluesof xand yatthepointswhereitcutsmustfitboth

theequationsofthecircleandoftheline.

Thismeansthatwecanput y=2 x–4intotheequationofthecircletofindthepossiblevaluesof x .

Thisgivesus

x 2–4 x+(2 x–4)2–2(2 x–4)–5=0

x 2–4 x+4 x 2–16 x+16–4 x+8–5=0

5 x 2–24 x+19=0

(5 x–19)( x–1)=0 x=1or x=19/5.

Substitutingthesevaluesof xbackintheline y=2 x–4givesusthecorrespondingtwovaluesfor yof–2and

18/5.Sotheline y=2 x–4cutsthecircleatthetwopointswithcoordinates(1,–2)and(19/5,18/5).

(b)Tofindiftheline3 y= x+11cutsthecircle,wecanrewriteitsequationas

x=3 y–11andsubstitutethisfor xintheequationofthecircle.Thisgivesus

(3 y–11)2–4(3 y–11)+ y 2–2 y–5=0

9 y 2–66 y+121–12 y+44+ y 2–2 y–5=0

10 y 2–80 y+160=0

y 2–8 y+16=0

( y–4)2=0.

Thetwopossiblecuttingpointshavecometogetherheretogivethesinglepointforwhich y=4and

x=12–11=1.Thismeansthattheline3 y= x+11justtouchesthecircle–itisatangent toit.

Thepointofcontacthasthecoordinates(1,4).

(c)Thistime,weput y=3 x+6intheequationofthecircle.Thisgivesus

x 2–4 x+(3 x+6)2–2(3 x+6)–5=0

x 2

–4 x+9 x 2

+36 x+36–6 x–12–5=010 x 2+26 x+19=0.

Usingthequadraticformulaonthisequation,witha=10,b=26andc=19givesb2–4ac=–84,sowecan’t



findanyvaluefor xwhichwillsatisfythisequation.

Thismustmeanthatthelinemissesthecirclecompletely.

Forthesketch,weneedthecentreandtheradiusofthecircle.

Wehave

x 2–4 x+ y 2–2 y–5=0

( x–2)2–4+( y–1)2–1–5=0

so( x–2)2+( y–1)2=10.

Thecentreofthecircleisatthepoint(2,1)anditsradiusis

Straightlinesandcircles

• Substitutingtheequationofthelineintotheequationofthecirclewillgiveyouaquadraticequation

in xor y .

Therearethenthreepossibilities.

• Theequationhastworoots.Thismeansthatthelinecutsthecircleintwopoints.

• Theequationhasonerepeatedroot.Thismeansthatthelineisatangenttothecircle–itjust

touchesit.

• ‘b2–4ac’isnegative,andtheequationhasnorealroots.Thismeansthatthelinemissesthecircle

altogether.

FindingtheequationsoftangentstocirclesAnytangenttoacirclemustbeperpendiculartotheradiusgoingtothepointofcontact.Thegradientofthe

tangentwillthentellustheslopeorgradientofthecircleatthispointofcontact.

Example(1)Findtheequationsofthefourtangentstothecircle x 2–6 x+ y 2–4 y–12=0withpointsof

contact(a)(7,5),(b)(–1,–1),(c)(8,2)and(d)(3,7).Drawasketchshowingthecircleandthesefour

tangents.

Answer: Westartbyfindingthecentreandradiusofthecircle.Wehave

x 2–6 x+ y

2–4 y–12=0=( x–3)

2–9+( y–2)

2–4–12.

Sotheequationofthecircleisalsogivenby( x–3)2+( y–2)2=25.Itscentreisatthepoint(3,2)and

itsradiusis5units.IhavedrawnasketchofthiscircleinFigureshowingthefirsttangentthatwe

shallfind.Ithinkthatitwillhelpyouintheworking,whichfollowsifyousketchinhowyouthinkthe

10



otherthreetangentswillgo.

(a)Thefirsttangenttouchesthecircleatthepoint(7,5).Theradiustothepointofcontactjoins(3,2)to

(7,5),soitsgradientis

x2- x1

y2- y1

= 7- 3

5- 2

= 4

3

Demo

Graphs and Equations

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