tutorial 6 graphing techniques solutions

8/12/2019 Tutorial 6 Graphing Techniques Solutions

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2014 SAJC JC2 H2 Mathematics

Tutorial

Tutorial 5 : Graphing Techniques Page 1 of 22

Tutorial 6: Graphing Techniques

DIY Questions

1. Identify and sketch the curves represented by the following equations.

(a) 067 22 yxx

(b) 2

4 xy

(c) 222 13)5()12( yx (d) 01362 2 xyy

(e)2

1

x

xy

(a) 067 22 yxx

22

2

22

22

22

222

2

2

5

2

7

4

25

2

7

04

25

2

7

064

49

2

7

062

7

2

7

2

72

yx

yx

yx

yx

yxx

This is the equation of a circle with radius 2

5

, centre at

0,2

7

.

(b) 222222 244 yxxyxy

This is the equation of a semi-circle with

radius 2 , centre at 0,0 . It is a semi-circle

and not a circle because 24 xy and

not 24 xy

It cuts thex-axis at 2 2x and x


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(c) 222 13512 yx

1

5

13

12

13

113

5

13

12

2

2

2

2

2

22

2

22

yx

yx

When 0x ,5

13y , When 0y ,

12

13x

The equation is an ellipse with centre 0,0 .

The length of minor axis is6

13and the

length of major axis is526 .

(d) This equation has 2y term but no 2x , it

will be a parabola.

2

2

2 6 3 1 0

2 6 1 3

y y x

y y x

2

2 2

2

2 2

2

2

2 13

3 2

2 3 3 3 12

3 2 2 2 2

2 3 3 1

3 2 2 2

2 3 7

3 2 4

2 3 7

3 2 6

x y y

y y

y

y

y

When 0x ,

2

2

2 6 1 0

6 6 4 2 1

2 2

6 28

4

3 7 3 7or

2 2

y y

y

y

y y

When3

1,0 xy .

This is a quadratic curve with2

3y as the

line of symmetry, minimum x value at

2

3,

6

7.


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(e)

2

11

2

1

xy

x

xy

1,As yx

1y is the horizontal asymptote.2x is the vertical asymptote.

When 0x ,2

1y .

The graph is a rectangular hyperbola.

It cuts the y-axis at2

1y

When 1,0 xy

It cuts thex-axis at 1x


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Tutorial


2. Given that C is the curve that represents the equation4

142

x

xxy .

(i) Show thatycannot take values between 2 and 6.

Hence find the turning points of the curve.

(ii) Sketch the graph of C, showing clearly its behaviour as it approaches its asymptotes.

[Ans : (i) Turning points = 2,3 and 6,5 ]

(i)

4

1

4

142

xx

x

xxy

0)41()4(

144

1)4)(

2

2

yxyx

xyxxy

xxy(

For equation to have real solutions, the discriminant must be 0

2

2

2

4 4 1 1 4 0

16 8 4 16 0

8 12 0

2 6 0

y y

y y y

y y

y y

2 6y or y

Therefore y cannot take values between 2 and 6.

The turning points will be at 2y and 6y .

When 2y

14424

14 22

xxx

x

xx

3

03

096

2

2

x

x

xx

When 6y

14464

146

22

xxx

x

xx

5

05

02510

2

2

x

x

xx

The maximum point is 2,3 and the minimum point is 6,5 .


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ii) 4x is a vertical asymptote.As x , xy .

The line xy is an oblique

asymptote.

When 0y ,

2

2

4 1 0

4 4 4 1 1

2

4 16 4

2

4 12

2

4 2 3

22 3

x x

x

It cuts the x-axis at 0,32 and 0,32


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Tutorial


3. Sketch the following parametric equations, indicating the axes intercepts on the graphs.

a) , s2cos inx y for 0 .

b)2 , 2x t y t for t

c)2 3cos , sinx t y t for 0

2t

a) b) c)

Note:

Must remember to change

scale to square.

When 0x ,0cos

2

2

sin 12

y

Therefore theyintercept is 0,1 .

When 0y ,

0

0

in

,

s

When 0, 2cos 0 2x

When , 2cos 2x

Thereforexintercepts are

2, 0 and 2,0 .

Note:

The domain is t , you cando this by setting Tmin=-10

and Tmax=10.

In general, you should start

with small range of tand

increase if the graph is cut off

on the screen.

x,yintercepts 0,0

When 0x ,2os 0

2

c t

t

3

2sin 1y

Therefore theyintercept is

0,1 .

When 0y ,3

0

in 0s t

3cos 0 1x

Therefore thexintercept is

1,0 .


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Practice Questions

4. The curve C has equationax

aaxxy

2

32 22

, where a is a negative constant.

(i) Obtain the equations of the asymptotes of C.

(ii) Find ddyx

and deduce that C has no stationary point.

(iii) Draw a sketch of C and mark in the coordinates of the points of intersections of C with

the coordinate axes.

[Ans : (i) Equations of the asymptotes of C : xy and ax 2 (ii)

2

2

d 31

d 2

y a

x x a

]

(i) The vertical asymptote is ax 2 .

ax

aaxx

y 2

32 22

By long division,ax

axy

2

3 2

As x , xy . The line xy is an oblique asymptote.

(ii)

22

2

31

ax

a

dx

dy

02 2 ax , x , ax 2 and03 2 a .

Hence

02

32

2

ax

a, x , ax 2

1dx

dy, x , ax 2

Therefore there is no turning point.

(iii)When 0x , a

a

ay

2

3

2

3 2

.

When 0y ,

2 2

2 2

2 30

2

0 2 3

0 3

3

x ax a

x a

x ax a

x a x a

x a or x a

The curve passes through the points

0,a , 0,3a and

a

23,0


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(ii) When the two curves intersect,

22

2

2

2

2

2

2 22

1 21

6 3 2

22 6

2

2 26

2

2 2 6 2 (shown)

x x

x

xx

x

xx

x

x x x

(iii) Using the GC

Repeat the same process for the other intersection.

Thex-coordinates are 0.515, 2.45x .


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Tutorial


6. [MJC/07/P1/Q9]

A sketch of the curve2

( )x ay

x b

,

where aand bare constants, is shown in the diagram. The curve has stationary points at

(1, 0) andP.

(i) State the values of aand b.

(ii) Find the equation of the oblique asymptoteL.

(iii) Determine the coordinates of the turning pointP.

(iv) Copy the above sketch, and on the same diagram, draw a sketch of

the curve 2 2 16x y , showing clearly the axial intercepts.

(v) Hence show that the equation4 22 6 68 63 0x x x

has exactly two real roots.

[Ans: (i) b= 2, a= 1 (ii) 4y x (iii) ( 5, 12) ]

(i) 2( )x ay

x b

b= 2, a= 1

(ii) 2 2( 1) 2 1 94

2 2 2

x x xy x

x x x

Equation of oblique asymptoteLis 4y x

(iii) Using GC, coordinates ofP= ( 5, 12)

y

L

2 O 1

P

x


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Tutorial


(iv)

(v) 222

2 2 4 2

2 2 4 3 2 2

4 3 2 4 3 2 2

4 2

( 1)16

2

( 2) ( 1) 16( 2) 0

( 4 4) ( 4 6 4 1) 16( 4 4) 0

4 4 4 6 4 1 16 64 64 0

2 6 68 63 0

xx

x

x x x x

x x x x x x x x x

x x x x x x x x x

x x x

Since the graphs of2( 1)

2

xy

x

intersects the graph of 2 2 16x y at exactly two

points, there are exactly two real roots for the given equation.

y

2 O 1 4

4

P

x

2( 1)

2

xy

x

2 2

16x y

4

4


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Tutorial


7.

A sketch of the curvedx

cbxaxy

2

, where cba ,, and d are constants, is shown, not to

scale in the diagram. The equations of the asymptotes, also shown in the diagram, are 2x and 32 xy .

(i) Write down the value of d .(ii) Find the value of a and show that 7b .(iii) Given that the curve has a stationary point where 1x , find the value of c and the

x-coordinates of the other stationary point.

(iv) Copy the above sketch and, by drawing a sketch of another suitable curve in the

same diagram, find the number of real roots for the equation

02872 234 xxxx .[Ans : (i) 2d (ii) 2a (iii) 8c , Other stationary point is at 3x ]

(i) 2x is the vertical asymptote implies the denominator is 2x . Therefore, 2d .

(ii) 2

2

ax bx cy

x

By long division

dx

abcabaxy

222

As x , abaxy 2 .

2 2 3ax b a x

Comparing coefficients:

2a

7

322

32

b

b

ab

(iii)

2

632

2

227232

x

cx

x

cxy

2

6d2

d 2

cy

x x

When 1x ,d

0d

y

x

Hence

2 2

8 6d 22 2

d 2 2

y

x x x

Whend

0d

y

x

2

2

20 2

2

2 1

2 1

1, 3

x

x

x

x

The other stationary point is at 3x .


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Tutorial


8

21

6

21

620

2

2

c

c

c

(iv)

2

232

2

872 2

xx

x

xxy

4 3 2

2 2

2 2

2 2

2

2

2 7 8 2 0

2 7 8 2 0

2 7 8 22 7 8

12

2 7 8 1

2

x x x x

x x x x

x x x xx x x

x

x x

x x

By sketching the curve2

1

xy on the same diagram as

dx

cbxaxy

2

, we can find

the number of real roots by counting the number of intersection points.

From the graph, we see two intersection points. Hence there are 2 solutions.


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Tutorial


8. [SAJC/07/P1/Q7]

The curve C has equation y =1

22

x

xa wherex 1 and a is a non-zero constant .

(i) Show that if C has no stationary points, then 2 < a< 0 . [3]

(ii) It is given that the line y=x1 is an asymptote of C. Find the value of a. [2]

(iii) Sketch C, showing clearly the asymptotes and coordinates of any intersections with the

coordinate axes. [3]

[Ans: ii) 1a ]

7(i)

(ii)

2

2

2

2

1

2 2

( 1)

axy

x

dy ax ax

dx x

Since C has no stationary points, there are no real roots for 0222 axax

02

0)2(4

0)2(4)2( 2

a

aa

aa

1

2

1

22

x

aaax

x

axy

Asymptote isy = ax + a.

a=

1

(iii)Sketch of

1

22

x

xy :

2 2

y= x1

x= 1

2


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Tutorial


iii) 4x t 4t x 2( 4)y a x

For the curves Cand Qto

cut at 2 distinct points, maximum

point of Qmust be at

the point (4, 8).

8a

10. Convert the following parametric equations to Cartesian equations:

a)2 , 2x t y t for t

b) , scos inyx for

c) 1x tt

, 1y tt

(2010/P1/Q11iii)

[Ans : (i) 2 4y x (ii) 2 2 1x y (iii) 2 2 4x y ]

a)2

2

y t

yt

Substitute into x

2

2

2

4

yx

y x

b)

2 2

cos

cos

x

x

2 2

sin

sin

y

y

Remember the trigonometry identity2 2

cos 1sin .

2 2 1x y

This is a circle with centre 0,0 and radius 1.

c)

2

2

2 2

2

1

1

12

x tt

x t t

x tt

2

2

2 2

2

1

1

12

y tt

y t t

y tt

2 2 2 2

2 2

2 2

2

2

2

2

1 12 2

4

12 2

x y t tt t

x y

x y

This is a hyperbola with asymptotes y x and y x andxintercepts 2,0 and 2,0

2

2-2

(4, 8)

C

P


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Tutorial


11. A curve is defined parametrically by the equations,

t

tx

1 ;

t

ty

1

2

t, 1t

(i) Find the Cartesian equation of the curve, expressing your answer in the form

fy x .

(ii) Sketch the curve. Label your graph clearly, indicating any asymptote(s) and stationary

point(s).

(iii) By sketching another suitable graph on the same diagram as in (ii), determine the

number of real roots of the equation 3 2f( ) 6x x x .

[Ans : (i) Cartesian Equation :x

xy

1

2

(iii) Number of real roots = 2]

(i)

2

2 2 2

2

2

2

,1 1

1

1 1

1 1111

1

t tx y

t t

x xt t

xt

x

x

x x x xy

x xxx

xyx

(ii) To find the equation of the asymptotes of the curve

2

1

11

1

, 1

xy

x

y xx

x y x

2

2

d 11

d (1 )

d 10 1 0 or 2

d (1 )

When 0, 0 min . When 2, 4 max .

y

x x

yx x

x x

x y x y

Equation of the asymptotes: 1 xy and 1x

Cartesian Equation of the curve


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(iii) 3 2 3 2f( ) 6 f( ) 6x x x x x x .

We need to find the number of intersections between the curve representing

623 xxy and f( )y x

From the sketch, there are 2 intersection points between the graphs, hence, 2 real roots

to the equation.


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Supplementary Exercises

1. The curve Chas equation

2

422

x

xxy

i) Find the equations of the asymptotes of C.ii) Draw a sketch of C, which should include the asymptotes, and state thecoordinates of the points of intersection of Cwith the axes.

iii) On the same diagram draw a sketch of the curve

2

1

xy

iv) Hence show that the equation

0242 234 xxxx

has exactly 2 real roots.

[Ans : (i) 2x , 4xy ]

i)2

422

xxxy

By long division,2

44

xxy .

The vertical asymptote is 2x .

The oblique asymptote is 4xy .

When 2,0 yx .

When 0y ,

51

2

202

2

41442

042

02

42

2

2

x

x

x

xx

x

xx

224

1

2

44

xdx

dy

xxy

When 0dx

dy,

0or422

42

2

410

2

2

xx

x

x

When 2,0 yx


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Tutorial


When

1024

4424,4

2

yx

2

2

22

234

1

2

42

242

0242

xx

xx

xxxx

xxxx

From the diagram above, there are two intersections.Therefore 0242

234 xxxx has two real solutions.


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Tutorial


2. [CJC/06/Promo/Q7]

Given the graph of y=x2+ a

x+ b, where a> 0 and b> 0,

(i) State the coordinates of the intersection(s) of the graph with the axes. [1]

(ii) Find the equations of the asymptote(s). [2]

(iii) Draw a sketch of the curve, labelling the equations of its asymptotes and coordinates of

any intersection with the axes. [2]

[Ans : i) 0,a

b

, ii) ,x b y x b ]

i)0,

a

b

ii) 2b ay x b

x b

,x b y x b iii)

x=b

y=xbb

a


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Tutorial

3. [CJC/06/Promo/Q8]Find the cartesian equations and coordinates of the intersections of the following curves with the

xandyaxes (if any):

(i) x= t2,y= t4+ 1 [2]

(ii) x=2 sec ,y= tan [2]

On separate diagrams, sketch the curves in (i) and (ii), indicating clearly the equation(s) of any

asymptotes. [4]

[Ans : i)y= x2+ 1 ; (0, 1), ii) 2

21 , 2,0 , 2,0

4

xy ]

i) y= x2+ 1 ; (0, 1)

As x , 2y x x

Note:2

x x !

Since2x t , it means 0x , hence x x .

Therefore the oblique asymptote is y x

ii) x4

= sec2,y

2= tan

2

tan2+ 1 = sec

2

y2+ 1 =x

2

4

Intercepts : (2, 0) , (2, 0)

1

y=x

2

y=2

xy=

2

x

tutorial 6 graphing techniques solutions

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