standard graph

8/14/2019 Standard Graph

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In this chapter, studcnts wili:r (a) usc a graphic calcuhtor to dra\r and compare the graphs ofa variety oftunolions:(b) unclcrstand th relationshjp betu'eeo a graph and ils associated algebraic equation, and inparlicLrlar show tnmiliaity with the grrphs oflhe slandard equalions sucb as cllipse andhyperbola:

3 (c) undersland the characlcrislics ol gmphs wilh thc hclp of a graphic crlculalor, locaic thc lllmingpoints. alrd delcmrinc lhc asyrnptotes (horizontal, \'erlical and obliquo). axcs ofs}rlnnclry. andtestriclious on lhe possible valuos ofr and-f;E (d) sketch the graph of a ralionai function whcn thc dcnoninalor is a lincar cxprcssion ftrd thenullleralor is cilhcr a linear or quadratic cxpressrcn, and delermine the cqualions ofasynplotcs,ir,rrr .e\': ,r', . ir'r tr e r tc.. .r.rl r''n, e poir ' .:&x+l) arz +b\-tcY= *".1 '=-,k*"E (l.) select lhe rpproprjatc "windorr- ofr graphic calcularor that would clisplay thc crilioal lcalures ofthe lu clions whon skctching graphs.

I23456'/.

-l-hincs to include $hcn skctchinq a eraph-t andy axcs"r and I intcrccpts (if^ny)as)'rnptotes (if any) and label the equation ofthe aslanptotesstatioiary points (ifany) and iabel the coordinalesequatiol oftlle graphthc radi s and the coo.dinates ofthc ccntrc (applicable to circles)the semi-ma.jor and semi-minor axes alld the coordinates ofdrc centre(applicable to eilipses)

The excllLsioi ofany ofthe above nertioned features mav lead to loss ofcredit duringexaminations. (x ir) I (v t,\ ,A. Graphs of the form a' ' b (Note : both r: and.), are ofdcgrce 2)

,se1: Whcn h-0. k=0 .l * .rt- =,(a) The Ellipse

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'I he equation represents an cllipse with celtre (0, 0) and its axes of symmetryarc the.r-axis and thc-},-axis. When r: 0: -t : ! l), when J = 0, :! : t d.If a > b, then a length of semi-najor axis; b: length ofsemirninor axis.If d < b, lhen D : lcngth of semi-najor axis; ,r : length of semi-minor axis.

,4..4' is called the major axis.BR'is called lhe rninor axis

.1,f is callcd thc rninor axisBB' is r:alled thc ma.jor axis.

trxample 1Skctch the eliipse I *L = r ..84Solution:The equation ofthe ellipse cdn be written as

PJ2,i Z

ase2: Wher h+0, k+0{r ,)' , (r -efThe equntion f, + 5j =t represents an ellipse with centre (/r, /r) an


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(b)

(it-l) (] 2llil)' u 4 ='flence, the coordinates ofthe centre ofeliipse isThc CircleWhcn rr . 15, thc cquation

!+l+0,0)'='is corrmonly expressed as (..- t'| , (1 - *)' =.'It rcpresents a circle with centre ( lr, i ) and radius a.Hence- rvhen n=t=0. !- * l-=1 is acircle,Ll'ccntre at (0, 0) and radius a uDits.(The ecluation can also be written as r'1 + -y) = 11r 1

Example 3Skctch orl separate diagranls, the graphs of(D l, *J =r (ii)Solution

{i' I *I I \\hich c:n,lsn bc crprcs.cJ99u. "2 + -r,2 = 9, is the eqlration of a circlewith centre and radius

(ii) (;+l)2 +(y-2)2 = 5 is theequation ofa circle with centre and radius

(:r+ l)2 +(1,-U)2 =S

l2.

How do we check ifthe circle passes through the oriein?By substitutirg the coordinates (0, 0) into the equationofthe circle, orBy using p)4hagoras theorem to check that the distancebetween the origin an


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Gcnerf,l Form ol thc Equation of an Ellipse/Circle'l hc gencral cquation ofan ellipsc or circlc is givcn bywhcrc o, I, c, d and e are constants and lr + 0, l) + 0.

a.r t +h1 t +,n+J1+e=0

The cquation ofan ellipse and the cquatiol ola cirole are similar- Both equationsi) do not have any 'r.I,' tcrms, andii) the coeificients ol-r2 and y2 (i.e. a ancl l) are both ofthe same sign.

Diffcrcncc between the equation olan ellipsc and thc cquation of a circle:Circle Coeliicient of-r2 : Coeflicient ofy'?Ellipse Cocfficicnt of;2 + Cocfficicnt ofy?

Example 4Sketch the lbllowing graphs on separatc diagrairs.(i) -r'z+ y'? +8-r- 2y+13 = 0(ii) 9r' +,11' + 36-r - 0(iii) 4-r: +16r+9-r,: + 15 = 0Solution:(t r'+ y'+81 2),+13=0

(r+4)r +(), l)'] 4=0(r.+4)? +(J l)r =2'?Centre ofcirole:Radius:

(ii) 9:r'z+ 41,'z+36,r = 0

9(x+2)')+4y')-36=0

(x+ 2.lr l,lt ' l='Centre ofellipse:

( 4, 1)

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(iii) 4rc? + 16r+ 9y? + l5 = 04 (x'? + 4-r) +9y2 +15 = 01 (r +2)1 - 16 +9yt +15=O4 (t +2)' +9y' =1

Cenhe ofellipse:Note on use of GC:GC APPS, CONICS supports the graphing ofcircles and ellipse using the sta dardforms ofequation for both, and the general.foml ofequationfor circle- LIo*-eNer,sludents are expected to be able to carry out 'completing llle square' \Nheneyerexact values of lhe charactelistics ofthe conics are requtred or when unltnowns are

(' ,, \v k)4. Grapbs of the torm a - -: 'The [IvperbolaA hyperbola comprises two disconnected curves called itsarms oi branches. The special feature ofa h),?erbola is its twoasymploles. These two as)4rptotes make equal angleswith the coordinate axes and closs each other at lhe centerrl1'the hperbola (but not necessarily at the origin) and havelyadrents ot i .The diagram on the right shows a sketch ofahvoerbola ofeouation l- l- = tThe centre ofa h)?erbola is given bydetermined.

(i.e. h: k: O).(, , t ) and the jr-intercepts, r = tp are to be

SDecial case:ln the case when a : 6, the hlperbola becom es a rectangular hyper6ol4 becausethe two as)4nptotes are perpendicular to each other.

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Example 7Sketch the graph of l-!=1. State, if a,,y, the equatiom of the as)mptotes, the4la\ial inlsrcepts 3nd the stalronary pornls.Solution:22L=L t - ,2 =1"2 -331 4

. Points ofintersection with axes:As r ) oo, 1x2 4) >' ]


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(ii)hh..t:,,+ .


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Example 9'1,?lSketch the gaph of y = --=\l

Solution3-2r ^ I' t ) x-?3 Horizontal asymptote :? Vertical asFnptote:r=o.y= 1')

3Y =O,x=tc. Graphs of Rational Functions ofthe form

The graph of equation , = *'i?;" should be expressed in proper fiaction,vieldinslhe lorm Y=h I lt_L.' dx+eHence, ) = i:r+/ is the oblique aslmptote and r=-i is the veitical aslmptote.Example 10Sketch the gaph of y -' '- ' . Stale. ifany.x2(i) the points ofinte$ection with the axes,(ii) the equations ofthe aslmptotes, &(iii) the exact coordinates of the stationary points.

S.slgg.s4.:

)- \)= r + I + -+ ( This can be obtained using long division or the 'juggling' method.)

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(D Using GC, the points ofintcrscction with axes: (0, -l12)(i, As r -> ur,y +Equations ofasymptotes: (Oblique asymptote)(Venical asynptote)( iii)

o_Y r' 4x+1dr (_. :)'For stationary points,

=l=2 16 or 2+JJ= The stationary points are 1Z + JT, : + 2.f)and (2 J3,3 2Jt)

Do You realise?i) r: 1, 1"gory= I,l>o;I).lir v = Ii L {in ErcmDle 8)-rl

1= -3, 1"g ot1 = ! ,t.o)xx3 -2x' \2

ii)iii) (in Examplc 9)

.,-2,t--.,These rectangular l)?rbolas are actually cases of y = la----l]---l which crnbewritten as llal Lj- -s(')+dx,. dxtewhere fO:) can be a constant, a linear expression or a quadratic expression, leading tog(jr) being a zero, a non-zero constant and a linear expression respectively.

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D. Paramctric EquationsA curvc can be clefined pararnct.ically by the equations -r = l(r) rud 1, = g(l),where f and g are funclions of/. a new vatiablc.

'l he equations ofthc graphs sketched belore llis section are called ca esian*equations (i-c. expressed in tenns ol'-r and ], only).In some cases, relalionships belwecn r and }. ar'e so corDplicatcd that it is easirto express r. and ), cach in tcrms of a third variable. callod a parineter. Sornctimcs,simplcr Cartcsian equations may allso be expressed pamnelricaliy il1 order 1o modelhow r and I behave rvith respect to 1lle tlrird parameter.

+ lhc Ltrtc.'kt tootlinate systent ( i e. llle t r plok, ius uty.at.l b\ nNth.tnutuntn, Rc i Dts.trtat(tj96 l6.tA) ||ho\rds dlso u hishlr inl antul phibsartuL sdcnli.\!tnltvLteL!lxampleSupposc that the fliliht path of two airplanes ciln be traced on a 2 D planc.Supposc that one airplane moves along the line 1

: 2-r i 3whilc thc other airplane moves along the line -t = 3,\ - 2

(ri http.//hro line.,etlsrcenhoursai t AIPoktPatun/PARAMIQ I ITM)Bven though the lines intersect, the equations themselvcs do not tcll us wltether therewill bc a rnid air collision. To bc ablc to mathematically model this scenario. we canuse paramctric cquations- We introduce thc variable I for time and wite r and -t, ds afunction oft.Consider thc flight path oithc airplanc givcn by y: Lt + \.It can be described usingany ofthe follo*,ing scts ofcquations:

A. jr:1, y:2t+1,orB. x:2t, y:4t+lThese two sets ofequations describe the same line, J,,:2-y r 1 but thc sccond set ofequations indicates that the speed is twice that ofthc speed captured by the lirst set.

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Erample llA curvc has paramctric equations r=l+12- t=21where I is a non-negativc parameler.(i) Sketch the curve.(ii) What are the coordinatcs ofthe poiit at /: .l? Indicate this point on thecurvc in (i).(iiD Find lhe equalion ofthc cun'e in cartesian lbmr and skctch this crlrvc-Solution:(i) S/ep ,i isu,itch calculator mode to Paratnelric modc.

, The calculator is now tn pdtdnrctric equalions Dbdt..t/ep 2r Entcr the parametric equations.

Slep J: Graph the lirnction and set lhe wirdo'"v.

question carelully aDd adjust the values ofTntn and T-," ltccord

Pl+t1 Pldtt Fl')tl\,1irll1+1.I,J 1T EzTzrC

[Note: it is important to sct T,ri. : 0 in lhe window because it is givcn that t is anon-negativc parameter. We need to .cad the range ofvaiues oft glven in thc

tJIHn0tJTnin=BTr4ax= 1ETstep=. E 1Hn i n=BHnax= 1B.tVr'r i n=B

bcI EngB 12345678PoI seqDDIlflGll Sir.tula+br. |.P"tltHor-iz E-T

trt I H0[trltTsLep= , 01Hr"ri n=BHr'rex= 1BVr,r i n=BYr'rax= 1Ecq-t l


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: du,'dx: de,'dt: dx,'dt(ii)

(iiD Plotl Plot? PlotSrYrE.I(4{X-1))

=r*L4L=*-l4v'=q(x t)

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AppendixThe Straieht LincThe equation of a straight line may be expressed as. ax+ by+c:O (general l-om), oro y = zx i c, where m: gadient, c:y-intercept (Fadient-intercept form).Example(j) r=o (li) r=2 ( iii) 21 +]l 4:0

Parallel or Pcrpcndicular lines:'Iwo lines having gradients rnr and m, respectiveiy arc(i) parallei if m, = rr,(ii) perpendicular if m,m2: IExample (i) The straight lines 2t +3-i, 7:0and y=-lt*t are parallel.3(ii) The straight lines 3x + 41, 2 :0 and 4rf 3l = 5 are perpendicular

to each other.The Parabolaln general, a parabola. with an axis ofs)4nmehy parallel to the.l,-axis, can be expressed as

y=a(,+u)'+c, e+0.. with an axis ofs),mmetry parallel to ther-il\is, can be expressed as(t+")'=n ; 6, "*gSome examples of y = a(x+b)'+c, a+0.i y=+(x-2)'+1 b) v = (x + t)'7 +4

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q-p9S!4IlS!-L: When,: c:0, rvc havc(i)

the equation ofa parabola u,ith axis ofsymmetry -y = 0 ancl venex (0,0) .Example:a) (wh rd -(r.))

("Upward" Parabola)Parabola)

("Right Hand" Parabola)C. (l)Graphsof y=tr1'. zeZ+- r>l

.v'J. 0) b)

(ii)the equation ofa parabola with aris (]1 s),rnmetry _r = 0

Example:and vertex (0,0) .

("Leff Hand" Parabola)

Case 1: fliseven (v: lcr',E.kx'. Case 2: n is odd (v : kx'- /cr' - ...)fr o)

("Downward"

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Example ISkctch on the san]e diagram(i) the graphs of .y = i'? and I = -r4Solution(i) For the graphs of y = r'? and y = -1',

Note:(iD

(ii) lhe graphs ol 1= -tr

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Example 2Sketch on the same diagamI(i) the graphs of t: ).2 and 1j| =ir4 (ii) the graphs of I ,t,

Solution:(i)

Differentiation1. The Quotient Rule:Let?l and v be two functions ofr.

du dvollll dr dra" I ",t

"o 1[3]=dx\1+1,/ (;-r)i1z"i-z,i(l+r)(r')' 2r(2r) - z(l-r)("*r)'2. A stationary point is a point on the curve *he.e $= 0dir

Nature of roots to an equation

Quadratic Equation ax' + bx+ c =O, Lt .-DtnD 4ACI l he rnots ire 2a2. The naturc ofthe roots depends on the drscriminant D=br 4ac(a) IfD > 0, roots are Real and Distinct(b) IfD : 0, roots arc Real and Equal(c) IfD > 0, roots ar Rdal(d) IfD < 0, roots are Complex, imaginary ot uhreal

6'*tf

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[Jsc of GC for Gr:rphing Techniqucs: Standard graDhsA: Inlportant Notes

l) How to switch to "dottcd" modcFor rnosl graphs on the G.C. due to screen resoluliorl linlitations, il qray or may not bebettcr to work ir "dotted" mode. You can change each graph individually as shown inlater cxanrplcs or do it automatically fol allgraphs like so:scroll down tu DO I using thc arrow keys,, .:2nd>. . for " - 9" and NoTYou will noticc that for begiining examples the steps to using thc calculator is verytlelailed- As you rnovc along. gaining cxpcricnce in using the calculator, many keyingproccsses will be regarded as assuned knowledge.B: The Parabola

How to plot a graph using the function

I -, - t>. lsce (iii)]. < GRAPH> fsee (iv)lHorv ro use the ZOOM firnction

(iDy2 = 2x =.> -r, : t.'Di (Drarv I = .,D.v and y: "Ex ):Use the a.row keys to move to the llrst variable Yl.rnd . r' . ', X.l,r),n, ,) . [scc(i)]Use the arrow keys to move to the seconcl variable Y2, Use the arrow keys to move to Y VARS lsee (ii)l

VJEJ(2X)

(iiD

1) ZoomBox:o , ,. Use the arow keys to move the blinking cursor to any comer ofbox lsee (i)l. Usc thc arrow keys to move the culsor in ordcr to create a box which covers thc areayou want to zoom in [see (ii)]

(t

. Isee (iii)]

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p*, l.ull,inu y:tL? +bx+c, a>0,saya:1.b:A,c: 3;2) Zoolnln:. t]se the nrrow kcys to nrcve 1o the first variable Y I. - .:X, t,0,n>, , < >, .-l>, '-GRAPHt [sce (i)]. Usc the anow keys to scroll to the point where you want the cenhe ofzoom t') he. ..ZOONl:'. . fsee (ii)l

3) Zoomout:. tjse the arrow keys to soroll 1o thc point \there. , .:J>.


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Cl: Graphs of l,: kr-"'For,t>0. 'r even, say /. : L5, r'?=8To draw graph:. Use the arrow kcys to movc to thc first variable Yl. , , . .:X.'f ,e ,n>, , .:X,T,o,n>',


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Clhoosc your 2nd cune by ntoving the cursor-rvilh thc arrorv kevs again Isee (iii)]ENTFR LNIER -(i)

IIo\r'to show the {able and use the lunctiot

Note:In order to find lhe coordinales oflhe intelsectjon point that yotl want, you can eithcl movethe cu$or near to the inte$ection point al'tc. choosing lhe 2nd curve or guess a valuc ofxwhich is closc to the actuai value.

lsee (iv)l(iii)

Ifthe talrlc ofvalues don't give you thc points that you want,valuc:. . . Use thc arrow keys to move to "ATbl". flhange the Tblstart valuc to 0.6. Changc the increment value to 0.1 lscc (i)]. flse --2nd>. ,


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Sirnilar steps k) thc carlicI ones but please ersurc that you haveincludcd the brackets whcre appropriate i.e. ,r ^(l/2) instcad ofr "1/2.Use the ZOOM 1'unotion ifneccssarv.To comparc the y coordinatos ofeach cur.,,e at the same,r coordinate:. ':2nd>-

lf the tablc ofvalues doesn't givo you the poinls thal you want, you can adjust theincrerr)ent value using the firnction:. [see (i)]o , , , , lsee (ii)]. , , fsee (iii)]

Zoorn hl

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Nole: To change the Zoom Factor

Notice that near the -r axis, the cuNe scdns disjoinled; thjs is duc to the caiculiltorresolution.Also, aftcr you ZOOM h i.e. , , some ofyoul ci.cles appears too big, tlis jsdue to the ZOOM factor. You can teset the ZOOM 1'actor.. . Use lhe arrow keys lo move the cursor to MIMORY lsel. (i)]

. Change XFact value to 3. Clhangc YFact vaiue to 3 lsee (ii)]. .

(t

BeloreZoomFactoradjusted ZoomFactoradjusted

Ifyou have thc ncrver calculdtor modcls you can achieve a nicer circle (or ellipse) bypressing thc button and scroll down to look for an application called "CONICS".Follow thc instructiclns careftrlly from the application on how to draw your circle (orellipse). Note that without having this application the calculalor still allows you to draw acircle (or ellipsc) as showl abovc.

Eranrple 5(ii)Draw thc circie (x+ 1)'+ (1- 2)? = 5 - iv = 21J5 - (t + l)'?. r"yin r=rE G*tf rn.vi.. To draw y=z a,G-1t*rf ,teyiD 2-Yl lbr Y2.

fo drara I -7.V5 ry I)'.keyrnJ Yllor Y-].Follow the similar carlier steps to draw Y2 and Y3.

How to select the graph to be drawnTo prevent Yl from being drawn, use the arow keys to move to the":" sign and press to remove the highlight on the ":"sign, so when you press , Yl will not be drawn

ZOOM Box ZOOM Squar

=f::

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Example 4: Skctch thc cllipse Q A*01=t-, =z,K"y i" ,'' = fi ;i, t)- ror vr.1o,hrr, I ) Jin.t1,-l) leirn2 Yt tbrY2loJrr\\ v : r Jr" .1 ' l1: lc1 rn 2 Yl turY3.

ZOOMSlandard ZOOMIractor 2. Follow the similar earlier stcps to draw Y2 and Y3.Note:To prevent Yl fronr bcing drawn- use the at.row keys to highlight thl3 "=" sign ancl press


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How to check lbr oblique as)'lnptotes6To conlinn that the line / = 1{t,r are thc oblique asnnptotcs:. a"r," r=ruff; forY3 byputting ,


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Example: S\ct,hrlrcp,rpl.ni ' 2\ J a,.,..,'.'ur..l(i) the points ofintersection with the axes.(iD thc cquatioDs ofthe as)mptoles, &(iii) thc stationary points.r?+2r+3 6--Y+3+ 11 rl

To draw thc erart!'r+)'+ier in r .-. -' lor Y I Jnd TOOM :c\ordrngly.rlYou should realize lhat there is a veftical asymptoie at ]l: = I . You can observ(j the valuesnumcrically by looking at the GC tablc ofvalues as discussed carlicr on.

'l o confirn that thc linc -y = -r + 3 _are the oblig!!9jq4!p!q!9I. Key in I =.t+3 1br Y2.. Try zooninS fbr large values ()1r- Also use the'fllAClB lunction to detennine which

linc rcfcrs to what graph. Observe hor,' irs .\ + o . r -'(r+3) ilnd \ -+co!r + ()c+3)..

. You can itlso ohservc the trend by looking at thc valucs numcrically using the tablevalues. As lhei values increase. the Yl values tend to be very close to Y2.

FIow to find stationary points using the functionFirst you need to look at which pai ofthe graph which has a possibility ofhavingstationary points (maximum or minimum points). Use the ZOOM f'unction ifneccssary.

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Sdy we use

standard graph

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