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Graphical Methods Questions

 

1.  The equation of a circle is given as

2x2 + 2y2 – 8x + 5y + 10 = 0. Find the radius of the circle and the coordinates of its centre. (3 mks)

2.  The equation of a circle is given by x2 + 4x + y2 – 5 = 0. Find the centre of the circle and its

radius.

3.  The equation of a circle is x2 + y2 + 6x – 10y – 2 = 0. Determine the co-ordinates of the

centre of the circle and state its radius  

 

4.  In the diagram below ABE is a tangent to a circle at B and DCE is a straight line.

Image From EcoleBooks.comIf ABD = 60o, BOC = 80o and O is the centre of the circle, find with reasons BEC  

 

 

 

 

 

ecolebooks.com

 

 

 

 

 

 

 

 

5.  Obtain the centre and the radius of the circle represented by the equation:

  x2 + y2 – 10y + 16 = 0  

 

6.  Complete the table below, for the function y = x3 + 6x2 + 8x

x

-5

-4

-3

-2

-1

0

1

x3

-125

 

-27

-8

 

0

1

6x2

 

96

54

 

6

0

6

8x

-40

 

-24

  

0

8

y

  

3

0

 

0

15

 (a) Draw a graph of the function y = x3 + 6x2 + 8x for – 5 x 1 and use the graph to estimate

the roots of the equation x3 + 6x2 + 8x = 0

 (b) Find which values of x satisfy the inequality x3 + 6x2 + 8x -1 > 0

 

7.  Sketch the curve of the function y = x3– 3x + 2 showing clearly minimum and maximum points

and the y – intercept.  

 

8.  Show that 4y2 + 4x2 = 12x – 12y + 7 is the equation of a circle, hence find the co-ordinates

of the centre and the radius

 

9. Two variables R and P are connected by a function R = KPn where K and n are constants.

The table below shows data involving the two variables

P

3

3.5

4

4.5

5

R

36

49

64

81

100

 

(a) Express R = KPn
in a linear form  

(b) Draw a line graph to represent the information above  

(c) Find the values of constants K and n  

(d) Write down the law connecting R and P  

(e) Find the value of P when R = 900

 

10. A circle of radius 3cm has the centre at (-2, 3) . Find the equation of the circle in the

form of x2 + y2 + Px + qy + c = 0  

 

11.  In an experiment, the values of two quantities V and T were observed and the results recorded as

shown below.

 

V

0

2

4

6

8

10

T

0.49

0.30

0.24

0.20

0.16

0.137

Image From EcoleBooks.com

It is known that T and V are related by a law of the form a

b + V

where a and b are constants.

a) Draw the graph of I against V  

  T

 b) Use your graph to find;

  i) The values of a and b.

ii) V when T = 0.38

iii) T when V = 4.5  

 

12.  Find the equation of the tangent to the curve y = 2x3 + x2 + 3x – 1 at the point (1, -5)

expressing you answer in the form y = mx + c

 

13.  Given that :-  243 = (81)-1 x ( 1/27) x determine the value of x

14.  Show that 3x2 + 3y
2
+ 6x – 12y – 12 = 0 is an equation of a circle hence state the radius and

 centre of the circle  

 

15.  (a) Fill in the table below for the function y = -6 + x + 4x2 + x3 for -4
x
2

 

x

-4

-3

-2

-1

0

1

2

-6

-6

-6

-6

-6

-6

-6

-6

x

-4

-3

-2

-1

0

1

2

4x2

  

16

  

4

 

x3

       

y

       

  (b) Using the grid provided draw the graph for y = -6 + x + 4x2 + x3 for -4 x  2    

 (c)  (i) Use the graph to solve the equations:-

 (i) x3 + 4x2 + x – 4 = 0

 (ii) -6 + x + 4x2 + x3 = 0

 (iii) -2 + 4x2 + x3 = 0

 

16.  The table below shows the results obtained from an experiment to determine the relationship

 between the length of a given side of a plane figure and its perimeter

 

Length of side  (cm)

1

2

3

4

5

Perimeter P(cm)

6.28

12.57

18.86

21.14

31.43

 

(a) On the grid provided, draw a graph of perimeter P, against   

 (b) Using your graph determine;

  (i) the perimeter of a similar figure of side 2.5cm

  (ii) the length of a similar figure whose perimeter is 9.43cm  

  (iii) the law connecting perimeter p and the length

 (c) If the law is of the form P = 2k + c where k and c are constants, find the value of k  

 

17.  In an experiment with tungsten filament lamp, the reading below of voltage (V) current (I),

 power (P) and resistance (R)were obtained. It was established that P was related to R by

a law P = a Rn – 0.6. Where a and n are constants.

V

1.30

2.00

2.80

4.40

5.70

I

1.50

1.80

2.10

2.50

2.90

P

0.73

2.05

3.28

7.44

10.62

R

0.89

1.13

1.33

1.78

1.99

Plot a suitable line graph and hence use it to determine the value of a and n

18.  Find the gradient of a line joining the centre of a circle whose equation is x2 + y2 – 6x = 3 – 4y

and a point P(6,7) outside the circle..

 

19.  a) Complete the table below for the function y = -x3 + 2x2 – 4x + 2.  

 

x

-3

-2

-1

0

1

2

3

4

-x3

27

8

 

0

 

-8

  

2x2

18

8

2

0

    

-4x

 

8

 

0

   

-16

2

2

2

2

2

2

2

2

2

y

 

26

 

2

 

-6

 

-46

 b) On the grid provided below draw the graph of -x3 + 2x2 – 4x + 2 for – 3 ≤ x ≤ 4.  

 c) Use the graph to solve the equation -x3 + 2x2 – 4x + 2
= 0.  

 d) By drawing a suitable line on the graph solve the equation. –x3 + 2x2 – 5x + 3 = 0.  

 

20.  Determine the turning point of the curve y = 4x3 – 12x + 1. State whether the turning

point is a maximum or a minimum point.  

 


21.  (a) Complete the table below for the equation of the curve given by y = 2x3 – 3x2 + 1  

X

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

2x3

-16

 

-2

 

0

 

2

 

16

  

-3x2

-12

  

0.75

0

-0.75

    

-27

1

1

   

1

      

y

-27

-12.5

  

1

     

13.5

 (b) Use the table to draw the graph of the function y = 2x3 – 3x2 + 1  

c) Use your graph to find the values of x for :-

 (i) y > 0

 (ii) The roots of the equation 2x3 – 3x2 + 1 = 0

 (iii) 2x3 – 3x2 = 9  

 

22.  Find the radius and the centre of a circle whose equation is :

  2x2 + 2y2 – 6x + 10y + 9 = 0

 

 

Graphical methods Answers

 

1.  x2 + 4x + y2 = 5

x2 + 4x + ( ½ x 4)2 + y2 = 5 + (½ x 4)2

 

(x + 2)2 + (y + 0)2 = 5 + 4

 

(x + 2)2 + (y + 0)2 = 9

  Centre (-2,0)

 

Radius 9

  r = 3 units

 

Image From EcoleBooks.com2.  x2 + 6x + (3)2 + y2 – 10y + (5) = 2 + 9 + 25

 

(x + 3)2 + (y – 5)2 = 36

  (x – 3)2 + (y – +5)2 = 62


∴ centre (-3, 5)

Radius 6 units

 

3.   CBE = 400 ( alt.segiment theoren)

  BCE = 1200 (Suppl. To BCD = 600alt. seg.)

  (40 + 120 + E) = 1800 (Angle sum of  )


 BEC = 200

 

4.  X2 +Y2 – 10Y + 25 = 25 – 16

(X -0)2 + (Y – 5)2 = 9

(X – 0)2 + (Y – 5)2 = 32

Centre (0, 5)

Radius = 3

 

5.

x

-5

-4

-3

-2

-1

0

1

x3

-125

-64

-27

-8

-1

0

1

6x2

150

96

54

24

6

0

6

8x

-40

-32

-24

-16

-8

0

8

y

-15

0

3

0

-3

0

15

x3 + 6x2 + 8x >1

Between

(i) x = -3.85 0.1 and x -2.15 0.1

(ii) x > 0.5 0.1

 

6.    y = x3 – 3x + 2

x = 0, y = 2

(0, 2) ⇒ y – intercept.

 

 dy = 3x2 – 3 = 0

 dx x2 = 1

x = ∓ 1

 

 x = 1 y = 0

  Point (1, 0) min point

  x = -1, y= 4

 

  Point (-1, 4) max point.

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

7.  4x2 – 12x + 4y2 + 12y = 7

x2 – 3x + y2 + 3y = 7/4

x2 – 3x + (3/2)2 + y2 + 3y + (3/2)2 = 7/4 + 9/4 + 9/4 = 25/4

(x – 3/2)2 + (y + 3/2)2 = 25/4

Centre (1,5, -1.5) Radius 2.5units

 

8.  Log R =nlog p + log K

 

Log P

0.48

0.54

0.60

0.65

0.70

Log R

1.56

1.69

1.81

1.91

2.00

Gradient = 2 – 0.6

0.7

= 1.4 = 2

0.7

Log R intercepts = 0.6 = logk

K= 4

The law connecting R and P is R=4P2

900 = 4P2

P2 = 900

4

225 = P2

9.  (x +2)2 (y-3)2 = 32

X2 + 4x + 4 + y2 – 6y + 9 = 32

X2 + y2 + 4x – 6y + 4 = 0  

10.  

V

0

2

4

6

8

10

1

T

2.04

3.33

4.17

5

6.25

7.30

 T = a

b + V


I = b + V

 T a


I = 1V + b

  T a a

 y = mx + C

 

 b) (i) 1 = Grad ⇒ ∆y = 7.3 – 5 = 2.3 = 0.575

a ∆x 10 – 6 4

 

a = 1.739

 

 b = y – Intercept ⇒2.04

 a

b = 2.04 b = 2.04 x 1.739

1.739 = 3.547556

b ≃ 3.548

 

(ii) T = 0.38

I = 2.63 shown on graph

T

  V = 1

  -1

  (iii)  I = 4.45

 T

 T = (4.45)

  = 0.2247

≃0.22

 

 

11.  y = 2x3 + x2 + 3x -1

dy = 6x2 + 2x + 3

dx

gradeindent at (1, -5)

= 6 + 2 + 3= 11

y-(5) = 11

x -1

y + 5 =11x -11

  y = 11x -16

 

12.  35 = 3-4 x 3-x

35 = 3-4-x

-4 –x = 5

-x = 9

x =-9

13.   x2 + 2x + 1 + y2 – 4y + 4 = 4 + 1 + 1

(x+1)2 + (y-2)2 = 9

Centre (-1, 2)

Radius 3units

 

14.  c)

X

-4

-3

-2

-1

0

1

2

-6

-6

-6

-6

-6

-6

-6

-6

X

-4

-3

-2

-1

0

1

2

4x2

64

36

16

4

0

4

16

X3

-64

-27

-8

-1

0

1

8

Y=-6+x+4x2+x2

-10

0

0

-4

-6

0

20

Image From EcoleBooks.com

  y = x3 + 4x2 + x -6

  0 = x3 + 4x2 + x -4

  y = -2

 

(iii)  y = x3 + 4x2 + x – 6

  0 = x3 + 4x2 + 0 – 2

y = x – 4

 

x 1 0 -2

Image From EcoleBooks.comy -3 -4 -8

 

 

 

 

 

 

 

 

 

 

 

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

c  (i) solution 0.8

-1.5

And -3.2

(c) 1, -2, -3

 

 

 

Image From EcoleBooks.comImage From EcoleBooks.com15.  

Image From EcoleBooks.com

 

 

Image From EcoleBooks.com

 

 

 

 

 

Image From EcoleBooks.com

Image From EcoleBooks.com

 

Image From EcoleBooks.com

 

Image From EcoleBooks.com

 

Image From EcoleBooks.com

 

 

Image From EcoleBooks.com

 

 

 

Image From EcoleBooks.com

 

 

 

 

Image From EcoleBooks.com

 

Image From EcoleBooks.com

 

(i) P = 15.75cm

(ii) l=1.5cm

(iii) m = 35- 25 = 10 = 6.667

  5.5 – 4.0 1.5

 

(c) choose P(5,31.4)

p – 31.4 = 10

  l -5 1.5

p-31.4 = 100

 l-5 1.5

15p – 471 = 100k – 500

15p = 100l – 29

15 15

2k = 100

  15

k= 100 = 3.33

  2 x 15

c =1.93

P + 0.6 = arh

Log (P + 0.6) = log a + n log R

= n log R + log 9

P + 0.6

1.33

2.65

3.85

8.04

11.22

Log (P + 0.6)

-0.13

0.42

0.59

0.91

1.05

Log R

-0.05

0.05

0.12

0.25

0.30

 

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Log 0.3 = ¼ = 0.25

Log a = 0.3

 

 

 

17.   x2 + y2 – 6x = 3 – 4y

  x2 – 6x + (-6/2)2 + y2 + 4y + (4/2)2 = 3 + (-6/2)2 + (4/2)2

 

 (x – 3)2 (y + 2)2 = 3 + 9 = 4

  (x – 3)2 (y + 2)2 = 16

 C (3, -2)

 

 Gradient ∆y = 7 – -2 = 3

∆x 6 – 3

 

 

 

 

 

x

-3

-2

-1

0

1

2

3

4

-x3

27

8

1

0

-1

-8

-27

-64

2x2

18

8

2

0

2

8

18

32

-4x

12

8

4

0

-4

-8

-12

-16

2

2

2

2

2

2

2

2

2

y

59

26

9

2

-1

-6

-19

-46

 

 b) Check on the graph paper.

 c) x = 0.5 + 0.1

d) –x3 + 2x2 – 5x + 3 = 0

  Line to allow: y = x – 1

 

x 0 1

y -1 0

 

  x = 0.65

 

19.  Dy/dx = 12x2 – 12

12x2 – 12 = 0

  12(x2 – 1) =0

  x = 1

 x = -1

 

  At x = 1 At x = -1

0

1

2

-2

-1

0

GRD = 12

0

36

36

0

-12

– 0 + + 0 –

 

(1,7) (-1, 9)

Minimum maximum

20.  (a) table

(b) plotting

  scale

  smooth curve

(c) (i) -0.5 < x < 1 and x>1

(iii) x = 2.5  0.1

 

21.  2x2 + 2y2 – 6x + 10y + 9 = 0

x2 + y2 – 3x + 5y + 9/2 = 0

x2 + y2 – 3x + 5y = -9/2

x2 – 3x + 9 + y2 + 5y + 25 = 8.5 – 4.5

4 4

(x – 3)2 + (y + 5)2 = 4

  2 2

Radius = 2 units

Centre = (1.5, -2.5)

 

 

 

 

 


 




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