Standard Graph

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 in parlicLrlar show tnmiliaity with the grrphs oflhe slandard equalions sucb as cllipse and hyperbola:

3 (c) undersland the characlcrislics ol gmphs wilh thc hclp of a graphic crlculalor, locaic thc lllming points. alrd delcmrinc lhc asyrnptotes (horizontal, \'erlical and obliquo). axcs ofs}rlnnclry. and testriclious on lhe possible valuos ofr and-f; E (d) sketch the graph of a ralionai function whcn thc dcnoninalor is a lincar cxprcssion ftrd the nullleralor 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 ' .:

Y= E (l.) select lhe rpproprjatc

the

-l-hincs

lu

&x+l)

arz

*".1

+b\-tc

'=-,k*"

"windorr- ofr graphic calcularor that would clisplay thc crilioal lcalures of

clions whon skctching graphs.

to include $hcn skctchinq a eraph

I

-t andy axcs

2

"r and

3

as)'rnptotes

intcrccpts (if^ny) (if any) and label the equation ofthe aslanptotes statioiary points (ifany) and iabel the coordinales equatiol oftlle graph thc 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)

4 5

6 '/.

I

The excllLsioi ofany ofthe above nertioned features mav lead to loss ofcredit during examinations.

(x ir) I (v t,\ A. Graphs of the form a' ' b dcgrce 2)

(a)

The Ellipse

,se1:

Whcn

h-0. k=0

.l

*

c9-1

.rt-

=,

,

(Note : both r: and.), are

of

'I he equation represents an cllipse

with celtre (0, 0) and its axes of symmetry

arc the.r-axis and thc-},-axis. When

r:

If If

a d < b, lhen D : a > b, then

: ! l), when J = 0, :! : t d. axis; b: length ofsemirninor axis. axis; ,r : length of semi-minor axis.

0: -t

length of semi-najor

lcngth of semi-najor

major axis. BR'is called lhe rninor axis

,4..4' is called the

trxample

.1,f BB'

is callcd thc rninor axis is r:alled thc ma.jor axis.

1

Skctch the eliipse .84I

*L

=r

.

Solution: The equation ofthe ellipse cdn be written as

PJ2,i

Z

ase2: Wher h+0, k+0

{r ,)' +, (r -ef =t represents an ellipse with centre (/r, /r) an
The equntion

Example

2

Slerch rhe ellinse

'

!i-l) )

(

v li' t6

-l

Solution: The equation ofthe ellipse can be written as

c9-2

(it-l)

u(] 42ll ='

il)'

flence, the coordinates ofthe centre ofeliipse is (b)

Thc Circle Whcn rr

.

15,

thc cquation

!+l+0,0)'=' is corrmonly expressed as

(..-

t'|

, (1 -

It rcpresents

a circle

Hence- rvhen

n=t=0. !- * l-=1

with centre ( lr,

i

*)'

=.'

) and radius a. is acircle,

Ll' (0, ccntre at 0) and radius a uDits.

(The ecluation can also be written as r'1 + -y) =

11r

1

Example 3 Skctch orl separate diagranls, the graphs

of

(D l, *J =r

(ii)

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

Solution

{i'

I *I 99 u.

"2

with

I

\\hich c:n,lsn bc crprcs.cJ

+ -r,2 = 9, is the eqlration of a circle

centre

(ii) (;+l)2 +(y-2)2 a

and radius

= 5 is theequation

circle with centre

of

and radius

How do we check ifthe circle passes through the oriein? l By substitutirg the coordinates (0, 0) into the equation ofthe circle, or 2. By using p)4hagoras theorem to check that the distance between the origin an
c9-3

Gcnerf,l Form ol thc Equation of an Ellipse/Circle 'l hc gencral cquation ofan ellipsc or circlc is givcn by whcrc o,

I,

a.r

t +h1 t

+,n+J1+e=0

c, d and e are constants and lr + 0, l) + 0.

The cquation ofan ellipse and the cquatiol ola cirole are similar- Both equations

i)

do not have any 'r.I,' tcrms, and

ii)

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: Coeliicient of-r2

Circle

Ellipse

:

Coeflicient ofy'?

Cocfficicnt of;2 + Cocfficicnt ofy?

Example 4 Sketch 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 = 0 Solution: (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,l

t ' l='

Centre ofellipse:

c9-4

4,

1)

4rc? +

(iii)

4

16r+ 9y? + l5 =

(x'? + 4-r)

0

+9y2 +15 = 0

- 16 +9yt +15=O 4 (t +2)' +9y' =1 1 (r +2)1

Cenhe ofellipse:

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

(' ,, \v 4. Grapbs of the torm a -:

k) '

The [Ivperbola A hyperbola comprises two disconnected curves called its arms oi branches. The special feature ofa h),?erbola is its two asymploles. These two as)4rptotes make equal angles with the coordinate axes and closs each other at lhe center rl1'the hperbola (but not necessarily at the origin) and have lyadrents ot

i

.

The diagram on the right shows a sketch hvoerbola ofeouation

l- l-

=

t

The centre ofa h)?erbola is given by

ofa

(i.e. h:

(, , t

k:

O).

) and the jr-intercepts, r =

tp

are to be

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

c9-5

Example 7

l-!=1. 4l

Sketch the graph of

a\ial inlsrcepts

State,

if

a,,y, the equatiom of the as)mptotes, the

3nd the stalronary pornls.

Solution:

22

L=L t 31 v-x,! .$' .

,2 =1"2 -3 4

a)

Points ofintersection with axes:

As r

) oo,

1x2

' Soas

4)

>

]<" -ot

r..,,o,

'

/r . .i:(r'-4)

t

l/4

y-+ ''' .?

t

.. Fqualion\ olss1mproles: I

r .6 -jr

t; r - | vr )

Stationary points:

Graphs of Rational Functions ofthe The Rectaneular Ilyperbola

(i)

y:

Eg.

Ah a+___:L cx+

>O d ,c

When r -+

co,

when r' -+

-co, J" -+ 4

Aq.mptotes:

Jr'

y -+ a*

:

d, :v

=

r,=dx+b form ' cx+d

4

c9-6

hh.. 't:,,+ cx+d . c


(ii)

Wlen -r > co, y -> a* When j )_co,Jl).l Asymptotcs: -y =

17,

jr:

!

Note:

i)

As)Tnptotes (ifthey exist) ofa 1'unction are lines or curves that the gaph ofthe flnction tends to at extreme values ofr ory. A RECTANGULAR HYPERBOLA is a h)?erbola whose as),rnptotes arc at right angles.

ii)

Example 8 Sketch the srooh

of -v: {i-1.

rl

Solution

n+l

=

Horizontal as].rnptote

When ;r

) n-1

:

- I = 0, (why do we consider this?)

we obtain

-

Vertical aslmptote:

\914:

1* is anumber slightlybigg€r than 1, e.g. 1.001 and l- is a number slightly smaller than l, i.e. 0.999.

c9-7

Example 9 '1,?l Sketch the gaph of y =

--= \l

Solution

3-2r ^

'

t )

I

x-?

3

Horizontal asymptote

?

Vertical asFnptote:

:

r=o.y= ') 1 3

Y

c.

=O,x=t

Graphs of Rational Functions ofthe form The graph of equation

, = *'i?;"

vieldinslhe lorm 'Y=h I Hence,

) = i:r+/

should be expressed in proper fiaction,

lt_L. dx+e

is the oblique aslmptote and

r=-i

is the

veitical aslmptote.

Example 10 Sketch the gaph of y

(i) (ii) (iii)

the points

-' x2'- ' . Stale. ifany.

ofinte$ection with the axes,

the equations ofthe aslmptotes,

&

the exact coordinates of the stationary points.

S.slgg.s4.:

)-

\)

= r + I + -+

( This can be obtained using long division or the

c9-8

'juggling' method.)

(D

Using GC, the points ofintcrscction with axes: (0, -l12)

(i,

As r -> ur,y +

Equations

(

iii)

ofasymptotes:

o_Y

dr

r'

(Oblique asymptote) (Venical asynptote)

4x+1

(_. :)'

For stationary points,

=l=2 =

16 or 2+JJ

The stationary points are and

1Z

+ JT,

: + 2.f)

(2 J3,3 2Jt)

Do You realise?

i) r: 1, I).

1"gory=

lir -rl v = Ii L

{in ErcmDle 8)

These rectangular

I,l>o;

l)?€rbolas

ii) 1= iii)

are actually cases

of

' y=

! -3, 1"g ot1 = ,t.o) xx 3

-2x

\2

.,-2,t--.,

la----l]---l

(in Examplc 9)

which crnbe

written as

j-

llal dx,.

L

-s(')+ dxte

where fO:) can be a constant, a linear expression or a quadratic expression, leading to g(jr) being a zero, a non-zero constant and a linear expression respectively.

c9-9

D.

Paramctric Equations A curvc can be clefined pararnct.ically by the equations where f and g are funclions of/. a new vatiablc. 'l he equations

-r =

graphs sketched belore llis section are called equations (i-c. expressed in tenns ol'-r and ], only).

ofthc

l(r) rud

1,

= g(l),

ca esian*

In some cases, relalionships belwecn r and }. ar'e so corDplicatcd that it is easi€r to 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 model how r and I behave rvith respect to 1lle tlrird parameter. +

lhc Ltrtc.'kt tootlinate systent ( i l6.tA) ||ho\rds dlso u hishlr

(tj96

t r plok, ius uty.at.l b\ nNth.tnutuntn, Rc antul phibsartuL sdcnli.\!tnltvLteL

e. llle

inl

i Dts.trtat

!lxample Supposc that the fliliht path of two airplanes ciln be traced on a 2-r Supposc that one airplane moves along the line 3 whilc thc other airplane moves along the -t 3,\ - 2

1: line =

(ri http.//hro line.,etlsrcenhoursai

t

2 D planc.

i

AIPoktPatun/PARAMIQ I ITM)

Bven though the lines intersect, the equations themselvcs do not tcll us wltether there will bc a rnid air collision. To bc ablc to mathematically model this scenario. we can use paramctric cquations- We introduce thc variable I for time and wite r and -t, ds a function oft. Consider thc flight path oithc airplanc givcn by any ofthe follo*,ing scts ofcquations:

y:

Lt +

\.It

can be described using

jr:1,

y:2t+1,or B. x:2t, y:4t+l A.

r

These two sets ofequations describe the same line, J,,:2-y 1 but thc sccond set of equations indicates that the speed is twice that ofthc speed captured by the lirst set.

c9-10

Erample A curvc

ll has paramctric equations

r=l+12-

t=21

where I is a non-negativc parameler.

(i) (ii) (iiD

Sketch the curve. What are the coordinatcs ofthe poiit at /: .l? Indicate this point on the curvc in (i). Find lhe equalion ofthc cun'e in cartesian lbmr and skctch this crlrvc-

bcI

Solution:

(i)

B

S/ep ,i isu,itch calculator mode to Paratnelric modc.

Eng

12345678

PoI

,
DDI

lflGll Sir.tul a+br. |.P"tlt



Hor-iz E-T

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

Pl+t1 Pldtt Fl')tl

\,1irll1+1. I,J

1T

EzT

zrC

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

: [Note: it is important to sct T,ri. 0 in lhe window because it is givcn that t is a non-negativc parameter. We need to .cad the range ofvaiues oft glven in thc question carelully aDd adjust the values ofTntn and T-," ltccord

tJIHn0tJ

Tnin=B Tr4ax= 1E Tstep=. E 1 Hn i n=B

Hnax= 1B .tVr'r i n=B trt

I H0[trl

tTsLep= , 01 Hr"ri

n=B

Hr'rex= 1B Vr,r

cq-t l

seq

i n=B

Yr'rax= 1E

(ii) :

: :

du,'dx

de,'dt dx,'dt

(iiD

Plotl Plot? PlotS

rYrE.I(4{X-1))

=r*L4 L=*-l 4

v'=q(x

t)

c9-12

Appendix The Straieht Linc The equation of a straight line may be expressed as . ax+ by+c:O (general l-om), or o y = zx i c, where m: gadient, c:y-intercept (Fadient-intercept form).

Example

(li) r=2

(j) r=o

(

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: I

Example (i)

(ii)

7:0and y=-lt*t are parallel. 3 :0 + lines 3x 41, 2 and 4rf 3l = 5 are perpendicular

The straight lines 2t +3-i, The straight

to each other.

The Parabola ln 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, "*g Some examples

of y = a(x+b)'+c,

a+0.

i y=+(x-2)'+1

b)

c9-13

v

=

(x + t)'7

+4

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)

i,

:3rr,

(whcrc d > o)

(wh rd

h)

("Upward" Parabola)

-(r.)

("Downward"

Parabola)

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

_r

= 0 and vertex (0,0)

.

Example:

r)

.v' = 2x, (where a > 0)

J.

b)

<0)

("Leff Hand" Parabola)

("Right Hand" Parabola)

C. (l)Graphsof y=tr1'. zeZ+- r>l Case

1:

fliseven

(v: lcr',E.kx'.

Case

2:

n is

odd (v

:

kx'- /cr' - ...)

fr<0 y+

c9-11

Example I Skctch on the san]e diagram (i) the graphs of .y = i'? and

I

=

(ii)

-r4

lhe graphs

ol 1= -tr

Solution

(i)

For the graphs

of

y = r'? and

y = -1', I a d lor-1< r < 0,

<-rt fbr0.r< ro = rr lbrr: 1, 0, I rt - t' L,rr 'l unJ t xo

Note:

(iD

Thc two curvcs intorscct at (0,0).

Fol the graphs

of .y : ;r n5l

(

I

I,l)

and

( I , l).

.i = ,ri,

.rt<.rt for0-rt lbr l<-r<0 andr>l Notc: The two culves intersect at (0,0),

c.(2) Glaohs of

Case

1:

v-[1'.

r?

(l,l)

"eV,'

and

(-1,-l).

n>l

Case

is even

2: r is odd ( .1,,=

(k-0)

c.(3)

I craDhs of Y =kt" Case

1:

z is even (1=

(r'0)

n€Z+- n>l /oi

I

..)

c9-15

Case

2:

I

a is odd

(r'0)

(v = t,' ,l' = h5...)

Example 2 Sketch on the same diagam

(i)

the graphs

of t:

I

).2 and j|

1

=ir4 (ii)

the graphs

of

ollll

du dr

I

,t,

Solution:

(i)

Differentiation

1.

The Quotient Rule:

Let?l and v be two functions

dx\1+1,/ "o 1[3]=

2.

ofr.

a" I

dv

dr

",t

(;-r)i1z"i-z,i(l+r)

(r')'

2r(2r)

6'*tf

A stationary point is a point on the curve *he.e

$= dir

-

0

Nature of roots to an equation Quadratic Equation ax' + bx+ c =O I

l he rnots

ire

, Lt -DtnD

. 4AC

2a

2.

ofthe roots depends on the drscriminant D=br 4ac IfD > 0, roots are Real and Distinct IfD : 0, roots arc Real and Equal IfD > 0, roots ar€ Rdal IfD < 0, roots are Complex, imaginary ot uhreal

The naturc

(a) (b) (c) (d)

c9-16

z(l-r) ("*r)'

[Jsc of

GC for Gr:rphing Techniqucs: Standard graDhs

A: Inlportant Notes

l)

How to switch to "dottcd" modc For rnosl graphs on the G.C. due to screen resoluliorl linlitations, il qray or may not be bettcr to work ir "dotted" mode. You can change each graph individually as shown in later cxanrplcs or do it automatically fol all graphs like so:

scroll down tu DO I using thc arrow keys, , .:2nd>. .
2)

Ali()r, all variables arc

Before, only Y I is {loried

dotted.

Tlre dilGrence between <(-)> md < > <( )>,':9> for " - 9" and NoT

You will noticc that for begiining examples the steps to using thc calculator is very tlelailed- As you rnovc along. gaining cxpcricnce in using the calculator, many keying proccsses will be regarded as assuned knowledge.

B: The Parabola How to plot a graph using the function

.y2

= 2x =.>

-r,

: t.'Di

(Drarv I = .,D.v and

y:

"Ex

Use the a.row keys to move to the llrst variable

.rnd

. r' . ', X.l,r),n, ,) .

):

Yl

(t

(iD

VJEJ(2X)

[scc(i)]

Use the arrow keys to move to the seconcl variable Y2

<( )>, Use the arrow keys to move to

I -, - t>.

.

<

GRAPH>

Y VARS

lsce (iii)] fsee (iv)l

lsee

(ii)l

(iiD

Horv ro use the ZOOM firnction

1) ZoomBox:

o

. . .

, <1>, 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 area

you want to zoom in [see (ii)] Isee (iii)]

c9-1'/

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 <4>- .:X, t,0,n>, <-t:>, < >, .-l>, '-GRAPHt [sce (i)] Usc the anow keys to scroll to the point where you want the cenhe ofzoom t') he ..ZOONl:'. <2>. fsee (ii)l

3)

Zoomout: tjse the arrow keys to soroll 1o thc point \there you want the centrc ofzoom to be. , .:J>.
. .

IZIris Zoomstdndard: This is used when you nccd the graph to be dralvn using the del'ault rvindow lsee (iv)l r -:ZOOM>, <6:.,

4)

LIow to find the i!-intercepts using the function

<2nd>.:TRACE><2> lsce (i)l Us(] the arrow keys to move the culsor to tho lcft hand side ofthe intercepl which you are interestcd in linding and press <.ENTEII> [see (ii)] Use thc anow keys to move the cursor to the right hand side ofthe intcrcept wlich you are intercsted in linding and press [scc (iii)] Isee (iv)]

.

(i)

(ii)

ort)

Go through the same procedure in order to find th(- other it-interccpt.

How to find ther-interccpt usirg thc function

o

<2nd><1><0>

c9

18

Cl: Graphs For,t>0.

'r

of

l,:

kr-"'

even, say /.

:

L5,

r'?=8

To draw graph:

. .

Use the arrow kcys to movc to thc first variable Yl <1>, <.>, <5>. .:X.'f ,e ,n>, <^-', <8>,

'fo zoom for a bcttff vil]w: . .ZOOM-:

.

LJse

.

the a.row keys

b

select the appropriale zoom (Zoomln,

Zoonout or ZoornBox)

<(;ItAPH>

For other /c and n values, say To draw othc. graphs:

. .

A

-

1,

r-

3 (rdd)

Use the anow keys to move to the first variable Y I

, <( )>.

Example I Thc graphs of Drarv .1

.

:

-r,

:

-r:r

.: I >, .:X,T,o,n>',

and .I

:

<">. <3>,

15.

xr:

. <-X.l,e.n-:. <^>.. <3:,.

Ho\\'to chanse the fbnnat ofa line e.g. to chaDge fo dotted line.

Dlaw )'

:

in a dolled line: Use the arow keys to movc to variablc Y2 llso the anow keys to move to the nost l.lls of that row , , , , ,

.

. .

rr5

I

To entery - rs and plot both graphs: . , <^>. <5>. To zoom for a better view:

.



. .

Use the arrow kcys to solcct the app.opriate zoom

functiol



How to find intorscction points

'fo tlnd

.

. e

an intersection point between the graphs:

<2nd>, , <5> [sce (i)] Use tlle arrow keys to choose your fiIst curve by moving the

c9- l9

cursor

[see

(ii)]

Clhoosc your 2nd

ENTFR

cune by ntoving the cursor-rvilh thc arrorv kevs again Isee (iii)]

LNIER

-

lsee

(iv)l (i')

(iii)

(i)

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

IIo\r'to show

the {able and use the

lunctiot

To comparc lhe-}' coordinates of cach curve at thc samc.t coordinate: . ..2nd>,
the increnent

Ifthe talrlc ofvalues don't give you thc points that you want, you can adiust valuc: . <2nd>. . 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>. to rctum to the tablc ofvalues.

y: tr ": For /.>0. r cven. sav i:2.

C2: Graphs of

Draw

. o

J,

n:2.

r.e.

:ffifi

F:

y=3

=3: <-t2>, < ) >

, <2>,<->, < ( >, ,

Use the ZOOM function ifnecessary

To observe relationship with 1 =

. . . .

(iv)

j;2

:

Use the anow keys to move to Y2

, <:>, Use tho arow keys to move

to

Y

1E2/
VARS

<1>, <1>,

Examplc 2

I

Graphsofy=r:

L

alal

1

L

. Draw r,=x2

and

c9-20

!V

Sirnilar steps k) thc carlicI ones but please ersurc that you have includcd 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 the

increrr)ent value using the firnction: . <2 d:'. . [Jsc the anow keys to move to " A'l'bl" . Change thc incrernenf valuc to {).2 . <0:-, <.>, <2>,.-2nd--,
D: Circles & Ellipses Example 5(i) D.aw thc circle -t'] + ),r =

9

+ | = 1.,6

,r'2

Drarv l'i hirlf oI circle: r <1. >, <2nd>, <9>, <->- ,

,<)>., , <5>,

,

.

lsec (j)l lsee (ii)l

Zoom standard Zoom square

<2> Usc the atrow keys to moYc youa cursor up betwccn 0 and the m,L\. pt lsee (iii)]

Draw 2''d half of circle): (')

.

Use the alTow kcys to move to the second variable Y2, <(-)>,



.

o .

Use the arrow keys to move to YVARS <1>, <1>, < CRAPH> [see (i)] , <6>, , <5>, lsee (ii)] , <2>, fsee (iii)]



c9-21

Zoorn hl

Nole:

To change the Zoom Factor

Notice that near the -r axis, the cuNe scdns disjoinled; thjs is duc to the caiculiltor resolution. Also, aftcr you ZOOM h i.e. , <2>, some ofyoul ci.cles appears too big, tlis js due to the ZOOM factor. You can teset the ZOOM 1'actor.

.

(t



.

Use lhe arrow keys lo move the cursor to

. . .

Change XFact value to 3 Clhangc YFact vaiue to 3 lsee

MIMORY lsel. (i)]

(ii)]

<2nd>.

Belore ZoomFactor adjusted

ZoomFactor adjusted

have thc ncrver calculdtor modcls you can achieve a nicer circle (or ellipse) by pressing 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 (or ellipse). Note that without having this application the calculalor still allows you to draw a circle (or ellipsc) as showl abovc.

Ifyou

Eranrple 5(ii) Draw thc circie (x+ 1)'+

.

r"yin r=rE

.

To draw

(1-

G*tf

2)? = 5

-

iv =

21J5 - (t

+

l)'?

rn.vi.

y=z a,G-1t*rf

,teyiD 2-Yl lbr Y2. fo drara I -7.V5 ry I)'.keyrnJ Yllor Y-].

ZOOM

Follow the similar carlier steps to draw Y2 and Y3.

Box

ZOOM Squar

How to select the graph to be drawn To 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

":"

c9-22

=f

: :

Example 4: Skctch thc cllipse

K"y

i"

,'' =

1o,hrr,

fi ;i,

t)-

Q

A*01=t-,

ror

vr.

I ) Jin.t1,-l) leirn2

=z,

l6

-s(,v

82

Yt

tbr

lc1 rn 2 Yl

tur

Vr a2+v l

Y2

loJrr\\ v : r Jr"

.1

' l1:

l)'?

ZOOM

ZOOM

Slandard

Iractor 2

Y3. 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

El HyDerbolas F\ample 7: slcrulr llu rranh

'1

(D (ii) (iii)

nf '

',

I

\lar(. il rn).

thc pojnts ofintcrseclion with the axes, lhe equations olthc aslmptotcs, the stationary points.

To draw the cuNe:

\1..' \ t.,' l\

L

.

Kev

&

in r : - /i..r2 3 for Yl

-,/

'r/'l

i.

. Draw.l,=-./lr2 -l ,keyro Yl forY2. v4

Follow the similar earlier steps to draw

Y2.

How to find the interscction points with the r-axis

. . r o .

Key

irl : 0 into Y3.

Change the lincs in Y3 to thick. Do a <2nd>, and choose INTERSECTION. Choose the hyperbola as first curve and the liney = 0 as your second line. Prcss when prompted for gxessing point, you will obtain th€ inter\ection with the axis.

r

Similarly for the other parts ofthe graph. c9,23

How to check lbr oblique as)'lnptotes To conlinn that the line

. a"r," r=ruff;

6

/ = 1{t,r

are thc oblique asnnptotcs:

forY3 byputting <{>,<( ):', , ,

in front ofthe function. Try zooming lor large values of,r-.

.

Note that after zooming, the lines look like 2 slraight lines. Use the and move yout cursors on the lines. When you do this thc cquatio Thc screen as shown.

ZOOM Box Try zoomiDg to other areas to confirm the obliquc as)orptotes. E\ample 9r Sketch lhe graph of

:I,

. I 2r r r 2 r-2 I

To draw the graph:

.

in Kev-i2

,,=B

torYl.

r

How to check for vertical and horizontal asymptotes You should realize by now that the vertical asynplote for the rectangllar hlperbola is at -r

You can estimate the value by looking at the table ofvalues on yol]I GC and observe how the values tend to be as the.n value approaches the discontinuity point from the left and liom the right. ln thct, from your table of values, an error is indicated at n : 2.

=

2, you can aiso observe Aom the same table of For your horizontal aslqnptote at ]) values. Scroll down for very large values oft and observe to what value the-t value approaches.

You rotice that on older calculato.s, the curve seems to be joining with a ve ical line (which is not the asymptote). This is due to the programming limitatioff of the calculator. The newel TIu4+ (upgraded ROM) will not have the same problern.

c9-24

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!

Ker

-' 'r+)'+i in r .-. rl

lor Y I Jnd TOOM :c\ordrngly.

You should realize lhat there is a veftical asymptoie at ]l: = I . You can observ(j the values numcrically by looking at the GC tablc ofvalues as discussed carlicr on.

.

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

'l o

linc rcfcrs to what graph. Observe hor,' irs .\

r

.

+

+

o

. r -'(r+3) ilnd \ -+co!

()c+3)..

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

FIow to find stationary points using the function

First you need to look at which pai ofthe graph which has a possibility ofhaving stationary points (maximum or minimum points). Use the ZOOM f'unction ifneccssary.

c9-25

Sdy we use
.7

Follow these steps to find thc maximum point: . <2nd>, , <4> . tJse the arow kcys to move the culsor to choosc tho Lcft Bound lsee (i)l . . tlse the arrow kcys to move the cursor to choose the Right Bound lsee (ii)l . , . l he coordinates ofthe naximum point will bc refleoted on the screen. fsee (iii)] (i)

(iiD

(D)

Say we change the window 10 enlarge the area on the

riglt

and remove the oblique

asymptote.

Follow thesc stcps to find the minimum point: . <2nd>, , <3> o Use the arrow kcys to move the cuNor to choose the Lelt Bould lsee (iv)l . . Use the anow keys to movc the cu$or to choose the Right Bound lsee (v)l . , . The coordilates of thl- minimum point will be reflected on the scrccn- lscc (vi)l (i")

(")

c9-26

('i)