13[Anal Add Math CD]

7/24/2019 13[Anal Add Math CD]

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1

Additional Mathematics SPM Chapter 13

Penerbitan Pelangi Sdn. Bhd.

1. (a)y

x01 2 3

3

2

1

(b) y

x01 2 3

3

2

1

2. (a) Gradient =3 13 1

= 1

y-intercept = 0

Equation of the line isy= x.

(b) Gradient = 45

,y-intercept = 3.4

Equation of the line isy= 45x + 3.4

3. (a) y= 1.5

(b) x= 1.7

4. (a) Whenx= 1.5,

y= x

y= 1.5

(b) Wheny= 2.0,

2 = 45x+ 3.4

x= (2 3.4) 54

= 1.75

5. (a) Gradient =6 34 1

= 1

Equation of lineABis

y 3 = 1(x 1)

y=x+ 2

(b) Gradient =5 21 4

= 1

Equation of line PQis

r 5 = 1(t 1)

= t+ 1

r = t+ 6

(c) Gradient = 5

10

= 12

Equation of lineRSis p= 12q+ 5.

6. (a) Wheny= 2, 2 =x+ 2

x= 0

(b) When t= 3.5, r= 3.5 + 6

= 2.5

(c) Whenp= 3, 3 = 12q+ 5

12q= 5 3

q= 2 2 = 4

7. (a) yx= 4x2+ 5x

yx

x

=4x 2

x

+5x

x

y= 4x+ 5

CHAPTER

13 Linear Law


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OR

yx= 4x2 + 5x

yx

x2

=4x 2

x2

+5x

x2

yx

= 4 + 5 1x

= 5 1x+ 4

(b) y=5x

3x

xy= 5 3x2

= 3x2+ 5

OR

y=5x

3x

yx

=5

x2

3

y

x

= 5

1

x2

3

(c)y

x

=1

x2

+7x

xyx=x1

x2

+7x

y = 1x

+ 7

OR

y

x

=1

x2

+7x

x2

yx

= x2

1x2

+7x xy = 1 + 7x

xy = 7x+ 1

(d) y2= 8x2 5x

y2

x

=8x2

x

5x

x

y2

x

= 8x 5

OR

y2= 8x2 5x

y2

x2 =8x2

x2 5x

x2

yx2

= 5 1x+ 8

(e) y=pkx

log10

y= log10

pkx

= log10

p+ log10

kx

= (log10k)x+ log

10p

(f) y= kxn

log10

y= log10

k+ log10

x n

= log10

k+ nlog10

x

= nlog10

x+ log10

k

(g) y=pk(x+ 2)

log10

y= log10

p+ log10

k(x+ 2)

= log10p+ (x+ 2) log10k = log

10p+ xlog

10k+ 2 log

10k

= (log10

k)x+ log10

p+ 2 log10

k

(h) y=k

px 1

log10

y= log10

k log10

px 1

= log10

k (x 1) log10

p

= log10

k xlog10

p+ log10

p

= (log10

p)x+ log10

k+ log10

p

(i) y=k

pnx

2

log10

y= log10

kpnx2 = log

10k log

10pnx

2

= log10

k (log10

p+ log10

nx2)

= log10

k log10

p x2log10

n

= (log10

n)x 2+ log10

k log10

p

(j) y=abx

k

log10

y= log10

abx

k

= log10

abx log10

k

= log10

a+ log10

bx log10

k

= log10

a+ xlog10

b log10

k

= (log10

b)x+ log10

a log10

k

(k) y 4 = abx

log10

(y 4) = log10

(abx)

= log10

a+ xlog10

b

= (log10

b)x+ log10

a

(l) y= 5 + axn

y 5 = axn

log10

(y 5) = log10

a+ nlog10

x

= nlog10

x+ log10

a

(m) y2

= 4x

6x2

2x2y2= 2x24x

6

x2

x2y= 8x 12


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OR

y2

=4x

6

x2

2xy2= 2x4x

6

x2

xy= 8

12

x xy= 12 1x+ 8

(n) xy2= 3x + 5x

xy2x

=3x

x

+5x

x

y2=3

x

+ 5

= 3 1x

+ 5 OR

xy2= 3x + 5x

xy2x

=3x

x

+5x

x

y2x = 3 + 5x = 5x + 3

(o) y=2x

3 + x

y(3 +x) = 2x

3y+ yx= 2x

3y+ yx

y

=2x

y

3 + x=2x

y

xy

=12x+

32

OR

y=2x

3 + x

y(3 +x) = 2x

3y+ xy= 2x

3y+ yx

x

=2x

x

3y

x

+ y= 2

3y

x

= y+ 2

yx

= 13y+ 2

3

8. (a) y= ax2+ bx

yx

=ax2

x

+bx

x

yx

= ax+ b..................................

a= Gradient

=6 2

5 1

=44

= 1

Substituteyx

= 2,x= 1 and a= 1 into ,

2 = 1(1) + b

b= 1

Therefore, a= 1 and b= 1.

(b) y= a+c

x

x(y) =x

a+c

x xy= ax+ c................................

a= Gradient

=5 1

1 4

=4

3

Substitutex= 1,xy= 5 and a= 43

into ,

5 = 43

(1) + c

c= 5 +43

=19

3

Therefore, a= 43

and c=19

3

.

(c) y= abx

log10

y= log10

(abx)

= log10

a+ log10

bx

= (log10

b)x+ log10

a

log10

b= Gradient

=5 2

3 0

= 1 b= 10

log10

a= log10

y-intercept

= 2

a= 102

= 100

Therefore, a= 100 and b= 10.


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(d) y= axn

log10

y= log10

a+ nlog10

x

= nlog10

x+ log10a

n= Gradient

=1 3

1 0

= 2 log

10a= log

10y-intercept

= 3

a= 103

= 1000

Therefore, a= 1000 and n= 2.

(e) y= abx2

log10

y= log10

a+ log10

bx2

= log10

a+ x2log10

b

log10

y= (log10

b)x2+ log10

a............

log10

b= Gradient

= 7 36 1

=45

= 0.8

b= 100.8

= 6.310

Substitute log10

y = 3, x2 = 1 and log10

b =45

into ,

3 =45

(1) + log10

a

log10

a= 3 45

=11

5

a= 158.5

Therefore, a= 158.5 and b= 6.310.

9. (a) (i)x 2 3 4 5 6

y 2.2 2.8 4.0 5.0 5.3

yx 3.1 4.8 8.0 11.2 13.0

y= kx + hx

yx = kx+ h

xy

0

4

6

8

10

12

14

2

2

x

1 2 3 4 5 6

(ii) h=yx -intercept = 2

k= Gradient

=13 (2)

6 0

=52

(b) (i) y= axn

log10

y= log10

axn

= log10

a+ nlog10

x

log10

y= nlog10

x+ log10

a

log10

x 0.3 0.40 0.48 0.54 0.60 0.65

log10

y 0.68 0.94 1.18 1.37 1.57 1.74

0

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

0.2

0.2

log10

x

log10

y

0.2 0.4 0.6 0.8


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(ii) log10

a= log10

y-intercept

= 0.18

a= 0.661

n= Gradient

=1.08 0.4

0.44 0.2

= 2.833

10. (a)

0

2

3

4

6

5

1

x

F

2 4 6 8 10 12

(b) (i) When F= 2.5,x= 5.2

(ii) Whenx= 12.0, F= 5.7

(c) x= 2.5 is wrongly taken when F= 1.0.

The correct value isx= 2.0.

11. (a)v2 19.98 40.07 60.06 79.92 91.20 119.90

s 1.0 2.0 3.0 4.0 5.0 6.0

0

40

60

80

120

100

20

s

v2

2 4 6

(b) v= 9.55 was wrongly taken.

The correct value is v2= 100, that is, v= 10.

(c) (i) When v= 8, v2= 64, s= 3.2

(ii) When s= 3.5, v2= 70

v= 8.37

12. (a) Gradient = 140 224 2

= 59

yx

= 59x2+ c...............................

Substitutex2= 2,yx

= 22 into ,

22 = 59(2) + c

c= 96

Therefore,yx

= 59x2 96

y= 59x3 96x

(b) Whenx= 2.4,y= 59(2.4)3 96(2.4) = 585.2

13. (a) y= axn..........................................

Substitutex= 1,y= 100 into , 100 = a(1)n

a= 100

Substitutex= 3,y= 900 into , 900 = 100(3)n

3n= 9

= 32

n= 2

(b) y= axn

y= 100(x)2

Y= mX+ c

Y=y, X= x2, c= 0, m= 100

Therefore, Y= 100X

When Y= 5,X= a,

5 = 100a

a=5

100

=

1

20

When Y= b,X= 4,

b= 100(4)

= 400

Hence, a=1

20

, b= 400 and c= 0.


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1. py= kx3+ p2x

p,pyp

=kx3

p+p2xp

y=kp x

3

+ px

x,yx

=kp

x2+ p

Compare to Y= mX+ c,

Gradient, m=kp

h 0

0 3=kp

y

x-intercept =p p= h

h

3=k

h

k= 13

h2

2. (a) y= 5x

2

log10

y= log10

5x log10

2

log10

y= (log10

5)x log10

2

(b) Y= mX+ c

c= log10

2

k= log10

2

= 0.301

m= 0 kh 0

log10

5 =0.301

h

h=0.301

log

105

= 0.4306

3. (a) y= 100xp

log10

y= log10

100xp

= log10

100 + log10xp

= 2 +plog10

x

log10y=plog10x+ 2

(b) k= 2 (log10

y-intercept)

p= gradient

=8 2

3 0

=63

p= 2

4. y= 2x2 3

yx2

=2x2

x2

3

x2

yx2

= 3 1x2

+ 2 Y= 3X+ c

Therefore,X= 1x2

and Y= yx2

.

5. (a)yx

1.91 1.84 1.73 1.63 1.58 1.55

1x2

0.21 0.16 0.11 0.06 0.04 0.03

0

1.0

1.5

2.0

0.5

x

x20.05 0.10 0.15 0.20

1

y

(b) (i) y=p

x

+q

p

x

yx

=p

x2

+qp

=p 1x2+qp

Y= mX+ c

p= Gradient

=1.6 1.5

0.05 0

= 2

(ii)qp

=yx

-intercept

q2

= 1.5

q= 3


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6. (a)log

10y 1.08 1.21 1.31 1.43 1.52 1.60

x 1 1.22 1.41 1.58 1.73 1.87

0

1.0

1.5

0.5

x0.5 1.0 1.5 2.0

log10

y

(b) y=pkx

log10

y= log10

pkx

= log10

p+ log10

kx

log10y= (log10k)x + log10p Y= mX+ c

(i) log10

p= log10

y-intercept

= 0.45

p= 2.82

(ii) log10

k= Gradient

=1.5 0.6

1.7 0.25

= 0.6207

k = 4.18

7. (a) log10

y 0.48 0.80 1.12 1.44 1.77 2.09

x 1 0 1 2 3 4 5

0

1.0

1.5

2.0

0.5

x 12 4 6

log10

y

(b) (i) y=pkx 1

log10

y= log10

p+ (x 1) log10

k

log10

y= (log10

k)(x 1) + log10

p

Y= mX+ c

log10

p= (x 1)-intercept

= 0.5

p= 3.16

(ii) log10

k= Gradient

=2.0 1.0

4.8 1.6

= 0.3125

k= 2.05

8. (a)x 1 2 3 4 5 6

log10

y 0.30 0.60 0.90 1.20 1.51 1.81

0

1.0

1.5

2.0

0.5

x

2 4 6

log10

y

(b) (i) y=kx

p

log10

y= log10

kx

p

= log10

kx log10

p

log10

y= (log10

k)x log10p

Y= mX+ c

log10y

-intercept = log10p

0 = log10

p

log10

p= 0

p= 1

(ii) log10

k= Gradient

=1.7 0

5.6 0

= 0.3036

k= 2.01

1. y= kxn 1

log10

y= log10

(kxn 1)

= log10

k+ log10

xn 1

= log10

k+ (n 1) log10x

= (n 1) log10

x+ log10

k

Gradient = n 1

Vertical intercept = log10

k


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2. Gradient =13 3

6 1

=10

5

= 2

Therefore, y = 2x2+ c......................

Substitutex2= 1, y = 3 into ,3 = 2(1) + c

c= 1

Hence, the equation is y = 2x2+ 1,that is,y= (2x2+ 1)2.

3. y=px+ qx32

yx

=px

x

+qx

32

x

=p+ qx

= qx +p...................................... Y= mX+ c

q= Gradient

=8 4

4 6

=4

2

= 2

Substitute x = 4,yx

= 8, q= 2 into ,8 = (2)(4) +p

p= 16

Therefore,p= 16 and q= 2.

4. y= 102x+ 3

log10

y= log10

102x+ 3

= (2x+ 3) log10

10

log10

y= 2x+ 3

Y= mX+ c

log10

y-intercept = 3

Gradient = 2

For pointA,4 3

q 0

= 2

1 = 2q

q=12

For pointB,p 31 0

= 2

p 3 = 2

p= 5

Therefore,p= 5, q=12

.

5. y= ax5

log10

y= log10

a+ 5 log10

x

log10

y= 5 log10

x+ log10

a

Y= mX+ c

log10

a= log10

y-intercept

= 4

a= 104

= 10 000

Gradient = 5

k 4

3 0

= 5

k 4 = 15

k= 19

Therefore, a= 10 000 and k= 19.

6. (a) Gradient =5 2

3 0

= 1

Vertical intercept = 2

The equation of the line is

Y= mX+ c

y= log10

x+ 2

(b) x= a10y 1

log10

x= log10

a+ (y 1) log10

10

log10

x= log10

a+ y 1

y= 1 log10

a+ log10

x

1 log10

a= 2

log10

a= 1

a= 0.1

7. (a) Gradient =10 4

3 1

= 3

Equation of the line is log10

y= 3 log10

(x+ 2) + c

Substitute log10

(x+ 2) = 1 and log10

y= 4 into

the equation,

log10

y= 3 log10

(x+ 2) + c

4 = 3(1) + c

c= 1

Therefore, log10

y= 3 log10

(x+ 2) + 1

y= k(x+ 2)n log

10y= log

10k+ nlog

10(x+ 2)

log10

y= nlog10

(x+ 2) + log10

k

Compare to log10

y= 3 log10

(x+ 2) + 1

Hence, n= 3 and log10

k= 1

k= 10


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(b) Wheny= 0.08, y= k(x+ 2)n

0.08 = 10(x+ 2)3

0.008 = (x+ 2)3

x+ 2 = 3 0.008 = 0.2

x= 0.2 2

= 1.8

8.xy

=3x

+ x

1xxy =

1x

3x

+ x

1y

=3

x2

+ 1

1y

= 3 1x2

+ 1 ..........................

Substitute1

x2

= 3 and1y

= kinto ,

1y

= 3 1x2+ 1 k= 3(3) + 1

= 10

Substitute1

x2

= pand1y

= 16 into ,

16 = 3(p) + 1

3p= 15

p= 5

9. (a) y= rx2+ t....................................

Substitutex= 1,y= 4 into , 4 = r(1)2+ t

r+ t= 4 .......................................

Substitutex= 3,y= 36 into, 36 = r(3)2+ t

9r+ t= 36 ...................................

, 8r= 32 r= 4

Substitute r= 4 into , 4 + t= 4

t= 0

(b) y= rx2+ tbecomesy= 4x2

1x

(y) =1x

(4x2)

yx

= 4x

Therefore, the gradient is 4.

ForA(2,p),

p 02 0

= 4

p= 8

ForB(q, 12),

12 0

q 0

= 4

12 = 4q

q= 3

10. (a) The gradient of the straight line =

11 7

4 2 = 2

Gradient =r 7

3 2

= 2

r 7 = 2

r= 9

(b) log10

y= 2x+ c............................

Substitute log10

y= 7,x= 2 into , 7 = 2(2) + c

c= 3

Therefore, log10

y= 2x+ 3

y= 102x+ 3

11. (a) y= px2+ q...................................

Substitutex= 1,y= 4 into , 4 =p(1)2+ q

p+ q= 4......................................

Substitutex= 3,y= 12 into , 12 =p(3)2+ q

9p+ q= 12 ..................................

, 8p= 8 p= 1

Substitutep= 1 into , 1 + q= 4

q= 3

(b) y=px2+ qbecomes

y=x2+ 3

y

x2

= 1 +3

x2

y

x2

= 3 1x2

+ 1

ForA,

y

x2

=4

12

= 4

1

x2

=1

12

= 1


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ForB,

y

x2

=12

32

=12

9

=43

1

x2

=1

32

=19

0 x21

1

A(1, 4)

x2y

, 1943

B

12. x2y=px.......................................

1x2

(x2y) =1

x2

(px)

y=p 1xIn Diagram (b), gradient =p

=14 7

1 12

= 14Substitutep= 14 into , x2y= 14x

ForA, 12 r= 14(1)

r= 14

ForB, t2(7) = 14t

7t2= 14t

7t= 14

t= 2

13. (a)x 2.5 4.5 6.5 7.0 8.5 9.5

y2 2.5 4.0 5.6 6.0 7.1 7.9

0

4

6

8

2

x

2 4 6 8 10

y2

(b) (i) y2= px+ q

p= Gradient

=6 1

7 0.5

=10

13

q=y2

-intercept = 0.6

(ii) Whenx= 1.5,y2= 1.7

y= 1.3

14. (a)x 1 4 9 16 20 25

yx 4.01 14.02 30.06 52.80 66.0 82.0

0

40

60

80

20

x

10 20 30

yx

(b) (i) y=p

x+ (q+ 1)x

x(y) = x px

+ (q+ 1)x yx =p+ (q+ 1)x yx = (q+ 1)x+ p

p= vertical intercept

= 2

q+ 1 = Gradient

=82 34

25 10

= 3.2 q = 2.2

(ii) Whenx= 10,yx = 34

y10 = 34

y=34

10

= 10.75


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15. (a)log

10x 0 0.30 0.54 0.65 0.699 0.778

log10

y 0.6 1.18 1.69 1.91 2 2.26

0

1.0

1.5

2.0

2.5

0.5

lg x

lg y

0.2 0.4 0.6 0.8

(b) y=pxn2

log10

y= log10

p+n2

log10

x

log10

y =n2

log10

x+ log10

p

log10

p= Intercept of log10

y

= 0.6

p= 3.981

n2

= Gradient

=2.2 1.0

0.8 0.2

= 2

n= 4

(c) Incorrect value ofyis 180.

The correct value is log10

y= 2.2,

that is, y= 158.5

16. (a)x 3.0 3.5 4.0 4.5 5.0 5.6

log10

y 1.4 1.53 1.68 1.83 1.98 log10

p

0

1.0

1.5

2.0

0.5

x

log10y

2 4 6

(b) (i) y= k ax 1

log10

y= log10

k+ (x 1) log10

a

= log10

k+ xlog10

a log10

a

log10

y= (log10

a)x+ log10

k log10

a

log10

a= Gradient

=2.1 0.8

5.4 1 = 0.2955

a= 1.975

log10

k log10

a= Intercept of log10

y

= 0.5

log10

k 0.2955 = 0.5

log10

k= 0.7955

k= 6.24

(ii) Whenx= 5.6,

log10

y= 2.15

y= 141.3

Therefore,p= 141.3

(iii) Whenx= 2.0, log10

y= 1.1

y= 12.6

17. (a)T 0 26.8 32.9 38.0 42.9 46.5

l 0 4.5 5.5 6.3 7.1 7.7

0

10

20

30

40

50

10

l

T

2 4

New

Old

6 8

(b) Gradient =30 05 0

= 6

Therefore, T= 6l T 2= 36l

(c) (i) T= 6l 6

(ii) When T= 6.5, l= 2.1 l = 2.12

= 4.41


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18. (a) y = k(x 1)n

log10

y= log10

k+ nlog10

(x 1)

log10

y = nlog10

(x 1) + log10

k

Y = mX+ c

log10

y 0.7 1.3 1.66 1.79 1.95 2.26

log10

(x 1) 0 0.3 0.48 0.54 0.62 0.78

0

0.5

1.0

1.5

2.0

2.5

lg y

0.2 0.4lg (x 1)

0.6 0.8

(b) (i) Wheny= 100, log10

100 = 2

log10

(x 1) = 0.65

x 1 = 4.467

x= 5.467

(ii) Whenx= 8, log10

7 = 0.85

log10

y= 2.4

y= 251.2

1. y=axn

log2y= log

2(axn)

= log2a+ log

2xn

log2y= nlog

2x+ log

2a

Y= mX+ c

Y= log2y, m= n,X= log

2x, c= log

2a

SubstituteX= 1, Y= 5 into the equation,

5 = m+ c...................

SubstituteX= 3, Y= 11 into the equation,11 = 3m+ c................

, 2m= 6 m= 3

Substitute m= 3 into ,5 = 3 + c

c= 2

n= m

= 3

c= log2a= 2

a= 22

= 4

Therefore, n= 3, a= 4.

2.y

x

2 = ax+

bx

yx2

x = ax+ bxxyx

= ax2+ b

Y = mX+ c

Y=yx , m= a,X= x

2, c= b

Given the gradient =12

a=12

y

x

=1

2

x2+ b

Substitutex2= 4,yx

= 6 into the equation,

6 =12

(4) + b

b = 4

Therefore, a=12

, b= 4

3. (a) y=bx

x+ ab

1y

=x+ abbx

= xbx + abbx

1y

= a 1x+1b

Whenx= 3,y= 1

1x

=13

,1y

=11

= 1

Whenx = 6,y=32

1x

=16

,1y

=1

3

2

=23

,

, 1

0

6 3

1 2

y

1

1

3

1

x


13/14

13



(b) a = Gradient

=

1 23

13

16

= 2

Substitute

1

x =

1

3 and

1

y = 1 into the equation,

1y

= a 1x+1b

1 = 2 13+1b

1b

= 1 23

=13

b = 3

Therefore, a= 2, b= 3.

4. (a)yx

7 8 6 4 2 0 2

x 1 2 4 6 8 10 12

0

4

6

8

10

2

2

x

x

y

2 64 8 10 12

(b) (i) Wrongly recorded y = 7, actual value of

y= 9

(ii) y= ax2+ bx

y

x

= ax2+ bx

x

yx

= ax+ b

b =yx

-intercept

= 10

a= Gradient

=10 0

0 10

= 1

(iii) Wheny= 0,x= 10

Therefore, the horizontal distance of the

point Cis 10 m.

(iv) Whenx= 5,

yx

= 5

y5

= 5

y= 25

Therefore, the height of pointBis 25 m.

5. (a)x 0 10 20 30 40 50

y 20 30 40 50 60 70

0

10

20

30

40

50

60

70

x

y

10 3020 40 50

(b) (i) 4k2x= (y c)2

4k2

x =y c y c= 2kx y= 2kx+ c Y= mX+ C

Y= y, m= 2k,X= x, C= c

c=y-intercept

= 20

2k= Gradient

=70 20

50 0

= 1

k= 12

Therefore, c= 20, k=12

.

(ii) Wheny= 55,

x= 35 x= 352

= 1225

(iii) Whenx= 500,

x= 22.4, y= 42.5


14/14

14



6.f 2 400 196 100 64 49

1m

100 50 25 16.7 12.5

(a), (c)

0

50

100

150

200

250

300

350

400

m

1

f2

20 6040 80

(c)

(a)

100

(b) (i) When m= 0.05 kg,

1

m

=1

0.05

= 20

f 2= 80

f= 9 oscillations per second

(ii) Whenf = 15,

f 2= 152

= 225

1

m

= 56

m=1

56

= 0.01786 kg

= 18 g

(iii) f 2=1

km

= 1k 1m Gradient =

1k

=400 0

100 0

= 4

k=14

7. (a) 2(y+ 1)2= kx+ t

(y+ 1)2=k

2x+t

2

...................

Y= mX+ c

Y= (y+ 1)2, m=k

2

,X= x, c=t

2

Substitutex= 0, (y+ 1)2

= 10 into , 10 =

t2

t= 20

Substitutex= 5, (y+ 1)2= 0, t= 20 into ,

0 =k

2

(5) +202

5k2

= 10

k=205

= 4

(b) (y+ 1)2 =k

2x+t

2

becomes

(y+ 1)2=42x+

202

y2+ 2y+ 1 = 2x+ 10

y2= 2x 2y+ 10 1

= 2(x y) + 9

Y= mX+ c

Y= y2, m= 2,X= x y, c= 9

Gradient = 2

Intercept on the Y-axis = 9

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