Graphical Methods Questions and Answers - Form 3 Topical Mathematics

Questions

1. The equation of a circle is given by x2 + 4x + y2 – 5 = 0. Find the centre of the circle and its radius.
2. 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.
3. In the diagram below ABE is a tangent to a circle at B and DCE is a straight line.
If ABD = 60
o, BOC = 80o and O is the centre of the circle, find with reasons ∠BEC
4. Obtain the centre and the radius of the circle represented by the equation:
x2 + y2 – 10y + 16 = 0.
5. 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
1. 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.
2. Find which values of x satisfy the inequality x3 + 6x2 + 8x -1 > 0
6. Sketch the curve of the function y = x3– 3x + 2 showing clearly minimum and maximum points and the y – intercept.
7. Show that 4y2 + 4x2 = 12x – 12y + 7 is the equation of a circle, hence find the co-ordinates of the centre and the radius.
8. 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
1. Express R = KPn in a linear form
2. Draw a line graph to represent the information above
3. Find the values of constants K and n
4. Write down the law connecting R and P
5. Find the value of P when R = 900
9. 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
10. 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
T =    a
b + V
where a and b are constants.
1. Draw the graph of I/T against V
2. Use your graph to find;
1. The values of a and b.
2. V when T = 0.38
3. T when V = 4.5
11. 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.
12. Given that :- 243 = (81)-1 x ( 1/27)x determine the value of x
13. Show that 3x2 + 3y2 + 6x – 12y - 12 = 0 is an equation of a circle hence state the radius and centre of the circle.
1. 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 x3 y
2. Using the grid provided draw the graph for y = -6 + x + 4x2 + x3 for -4≤x ≤2
3. Use the graph to solve the equations:-
1. x3 + 4x2 + x – 4 = 0
2. -6 + x + 4x2 + x3 = 0
3. -2 + 4x2 + x3 = 0
14. 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 l (cm) 1 2 3 4 5 Perimeter P(cm) 6.28 12.57 18.86 21.14 31.43
1. On the grid provided, draw a graph of perimeter P, against l
1. the perimeter of a similar figure of side 2.5cm
2. the length of a similar figure whose perimeter is 9.43cm
3. the law connecting perimeter p and the length l
3. If the law is of the form P = 2k+ c where k and c are constants, find the value of k
15. 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
16. 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..
1. 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 2 2 2 2 2 2 2 2 2 y 26 2 -6 -46
2. On the grid provided below draw the graph of -x3 + 2x2 – 4x + 2 for - 3 x 4.
3. Use the graph to solve the equation -x3 + 2x2 – 4x + 2 = 0.
4. By drawing a suitable line on the graph solve the equation. –x3 + 2x2 – 5x + 3 = 0.
17. Determine the turning point of the curve y = 4x3 - 12x + 1. State whether the turning point is a maximum or a minimum point.
1. 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
2. Use the table to draw the graph of the function y = 2x3 – 3x2 + 1
3. Use your graph to find the values of x for :-
1. y > 0
2. The roots of the equation 2x3 – 3x2 + 1 = 0
3.  2x3 – 3x2 = 9
18. Find the radius and the centre of a circle whose equation is :
2x
2 + 2y2 – 6x + 10y + 9 = 0

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)
Completing of sq. for expression in x and y.
√ Expression.
Centre
3. CBE = 40o ( alt.segiment theoren)
∠BCE = 120o (Suppl. To BCD = 600 alt. seg.)
∠(40 + 120 + E) = 180o (Angle sum of Δ)
BEC = 20o
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
1. x = -3.85 ± 0.1 and x = - 2.15± 0.1
2. 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
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
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.
1.
 V 0 2 4 6 8 10 1 T 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
1. 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
2. T = 0.38
I   = 2.63 shown on graph
T
V = 1
-1
3.  I  = 4.45
T
T = (4.45)
= 0.2247
0.22
11. y = 2x3 + x2 + 3x -1
dy = 6x2 + 2x + 3
dx
= 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)
1.
 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
2.

3.
1. y = x3+ 4x2 + x -6
0 = x3 + 4x2 + x -4
y = -2
2. y=0
3. y = x3 + 4x2 + x - 6
0 = x3 + 4x2 + 0 – 2
y = x – 4
 x 1 0 -2 y -3 -4 -8
solution 0.8
-1.5
And -3.2
1, -2, -3
14.
1.
2.
1. P = 15.75cm
2. l=1.5cm
3. m = 35- 25 = 10 = 6.667
5.5 – 4.0 1.5
3. 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
15. 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

16. 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
17.
1.
 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
2. Check on the graph paper.
3. x = 0.5 + 0.1
4. –x3 + 2x2 – 5x + 3 = 0
Line to allow: y = x – 1
 x 0 1 y -1 0
x = 0.65
18. 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
1. table
2. plotting
scale
smooth curve
1. -0.5 < x < 1 and x>1
2.  x = 2.5 ±0.1
19.  2x2 + 2y2 – 6x + 10y + 9 = 0
x2+ y2 – 3x + 5y + 9/2 = 0
x2+ y2 – 3x + 5y = -9/2
x2– 3x + 9/4 + y2 + 5y + 25/4 = 8.5 – 4.5
(x – 3/2)2 + (y + 5/2)2 = 4
Centre = (1.5, -2.5)

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