## 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

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.

If ABD = 60o, BOC = 80o and O is the centre of the circle, find with reasons BEC

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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 0.3 0.24 0.2 0.16 0.137

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 1.3 2 2.8 4.4 5.7 I 1.5 1.8 2.1 2.5 2.9 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)

r = 3 units

2.  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)

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)

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.

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 0.6 0.65 0.7 Log R 1.56 1.69 1.81 1.91 2

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 1T 2.04 3.33 4.17 5 6.25 7.3

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)

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

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

y -3 -4 -8

c  (i) solution 0.8

-1.5

And -3.2

(c) 1, -2, -3

15.

(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 0.3

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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