# Graphical Methods - Mathematics Form 3 Notes

## Introduction

• These are ways or methods of solving mathematical functions using graphs.

## Graphing Solutions of Cubic Equations

• A cubic equation has the form
ax3 + bx2 + cx + d = 0
where a, b , c and d are constants
• It must have the term in x3 or it would not be cubic (and so a 0), but any or all of b, c and d can be zero. For instance,
x3 6x2 + 11x 6 = 04x3 + 57 = 0x3 + 9= 0
are all cubic equations.
• The graphs of cubic equations always take the following shapes.

y =x3 6x2 + 11 x 6 = 0.
• Notice that it starts low down on the left, because as x gets large and negative so does x3 and it finishes higher to the right because as x gets large and positive so does x3. The curve crosses the x-axis three times, once where x = 1 , once where x = 2 and once where x = 3. This gives us our three separate solutions.

Example

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 16 4 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. −6 + x + 4x2 + x3 = 0
2. x3 + 4x2 + x – 4 = 0
3. −2 + 4x2 + x3 = 0

Solution

The table shows corresponding values of x and y for y = −6 + + 4xx3

 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 1 6 4 0 4 1 6 X3 −64 −27 −8 −1 0 1 8 Y=−6+x+4x2 +x3 −10 0 0 −4 −6 0 20

From the graph the solutions for x are x = −3 , x = −2, x = 1
To solve equation y = −6 + + 4xx3 we draw a straight line from the
diffrence of the two equations and then we read the coordinates at the point of the intersetion of the curve and the straight line

y = x3 + 4x2 + x − 6
0 = x3 + 4x2 + x − 4
y = –2                           solutions 0.8, −1.5 and −3.2

 x 1 0 −2 y −3 −4 −8

y = x3 + 4x2 + x – 6
0 = x
3 + 4x2 + 0 – 2
y = x – 4

## Average Rate of Change

### Defining the Average Rate of Change

• The notion of average rate of change can be used to describe the change in any variable with respect to another. If you have a graph that represents a plot of data points of the form (x, y), then the average rate of change between any two points is the change in the y value divided by the change in the x value.

Note;
• The rate of change of a straight ( the slop)line is the same between all points along the line
• The rate of change of a quadratic function is not constant (does not remain the same)
• The average rate of change of with respect to x = change in y
change in x

Example

The graph below shows the rate of growth of a plant,from the graph, the change in height between day 1 and day 3 is given by 7.5 cm – 3.8 cm = 3.7 cm.
Average rate of change is 3.7 cm/2 days = 1.85 cm/day
The average rate of change for the next two days is 1.3 cm/2 days = 0.65cm/day

Note;

• The rate of growth in the first 2 days was 1.85 cm/day while that in the next two days is only 0.65 cm/day. These rates of change are represented by the gradients of the lines PQ and QR respectively.

• The gradient of the straight line is 20 ,which is constant. The gradient represents the rate of distance with time (speed) which is 20 m/s.

### Rate of Change at an Instant

We have seen that to find the rate of change at an instant (particular point), we:

• Draw a tangent to the curve at that point
• Determine the gradient of the tangent

The gradient of the tangent to the curve at the point is the rate of change at that point.

## Empirical Graphs

An Empirical graph is a graph that you can use to evaluate the fit of a distribution to your data by drawing the line of best fit. This is because raw data usually have some errors.

Example

The table below shows how length l cm of a metal rod varies with increase in temperature T (0C).

 Temperature oC 0 1 2 3 5 6 7 8 Length cm 4 4.3 4.7 4.9 5 5.9 6 6.4

Solution

NOTE;

• There is a linear relation between length and temperature.
• We therefore draw a line of best fit that passes through as many points as possible.
• The remaining points should be distributed evenly below and above the line

The line cuts the y – axis at (0, 4) and passes through the point (5, 5.5).Therefore, the gradient of the line is 1.5/5 = 0.3.The equation of the line is l =0.3T + 4.

## Reduction of Non-linear Laws to Linear Form.

• When we plot the graph of xy=k, we get a curve.But when we plot y against 1/x , w get a straight line whose gradient is k.The same approach is used to obtain linear relations from non-linear relations of the form y= kxn.

Example

The table below shows the relationship between A and r

 r 1 2 3 4 5 A 3.1 1 2.6 28.3 50.3 78.5

It is suspected that the relation is of the form A=Kr2.By drawing a suitable graph,verify the law connecting A and r and determine the value of K.

Solution

If we plot A against r2,we should get a straight line.

 r 1 2 3 4 5 A 3.1 12.6 28.3 50.3 78.5 r2 1 4 9 16 25

Since the graph of A against r2 is a straight line, the law A =kr2 holds. The gradient of this line is 3.1 to one decimal place. This is the value of k.

Example

From 1960 onwards, the population P of Kisumu is believed to obey a law of the form P =kAt,Where k and A are constants and t is the time in years reckoned from 1960.The table below shows the population of the town since 1960.

 r 1960 1965 1970 1975 1980 1985 1990 P 5000 6080 7400 9010 10960 13330 16200

By plotting a suitable graph, check whether the population growth obeys the given law. Use the graph to estimate the value of A.

Solution

The law to be tested is P=kAt. Taking logs of both sides we get log P =log⁡(kAt). Log P = log K + t log A, which is in the form y = mx + c. Thus we plot log P against t.

(Note that log A is a constant).

The table below shows the corresponding values of t and log p.

 r 1960 1965 1970 1975 1980 1985 1990 Log P 3.699 3.784 3.869 3.955 4.04 4.125 4.21

Since the graph is a straight line, the law P =kAt holds.

Log A is given by the gradient of the straight line.Therefore, log A = 0.017.
Hence, A = 1.04
Log k is the vertical intercept.
Hence log k =3.69
Therefore k = 4898
Thus, the relationship is P = 4898(1.04
)t

Note;

• Laws of the form y=kAx can be written in the linear form as: log y = log k + x log A (by taking logs of both sides)
• When log y is plotted against x , a straight line is obtained.Its gradient is log A and the intercept is log k.
• The law of the form y =kXn,where k and n are constants can be written in linear form as; Log y = log k + n log x.
• We therefore plot log y is plotted against log x.
• The gradient of the line gives n while the vertical intercept is log k

Summary

For the law y = d + cx2 to be verified it is necessary to plot a graph of the variables in a modified form as follows
y = d +cx2 is compared with y = mx + c that is y =cx2 + d

1. Y is plotted on the y axis
2. x2 is plotted on the x axis
3. The gradient is c
4. The vertical axis intercept is d

For the law y – a = b x to be verified it is necessary to plot a graph of the variables in amodified form as follows
y a = b√x , i.e. y = b√x + a which is compared with y = mx + c

1. should be plotted on the y axis
2. should be plotted on the x axis
3. The gradient is b
4. The vertical axis intercept is a

For the law y e = f/x to be verified it is necessary to plot a graph of the variables in a modified form as follows.
The law y e = f/x is f (1/x+ e compared with y = mx + c.

1. should be plotted on the vertical axis
2. 1/should be plotted on the horizontal axis
3. The gradient is f
4. The vertical axis intercept is e

For the law y cx = bxto be verified it is necessary to plot a graph of the variables in a modified form as follows.
The law y cx = bx2 is y/x= bx + c compared with y = mx + c,

1. y/x should be plotted on y axis
2. X should be plotted on x axis
3. The gradient is b
4. The vertical axis intercept is c

For the law y = a/x + bx to be verified it is necessary to plot a graph of the variables in a modified form as follows.
The law
y = a + b compared with y = mx + c

1. y/x should be plotted on the vertical axis
2. 1/x2 should be plotted on the horizontal axis
3. The gradient is a
4. The vertical intercept is b

## Equation of a Circle

• A circle is a set of all points that are of the same distance r from a fixed point. The figure below is a circle centre (0,0) and radius 3 units

• P (x ,y ) is a point on the circle. Triangle PON is right – angled at N.
By Pythagoras’ theorem;
ON2 + PN2 = OP2
But ON = x, PN = y and OP = 3 .Therefore, x2 + y2 = 32

Note;
• The general equation of a circle centre (0, 0) and radius r is x2 + y2 = r2

Example

Find the equation of a circle centre (0, 0) passing through (3, 4)

Solution

Let the radius of the circle be r
From Pythagoras theorem;
r = √(32 x 42)
r = 5

Example

Consider a circle centre (5 , 4 ) and radius 3 units.

Solution

In the figure below triangle CNP is right angled at N.By pythagoras theorem;
CN2 + NP2 = CP2
But CN= (x – 5), NP = (y – 4) and CP =3 units.
Therefore,
(x – 5)2 + (y – 4 )2 = 32 this is the equation of a circle.

Note;
The equation of a circle centre ( a,b) and radius r units is given by;
(x–a)2 + (y–b)2 = (r)2

Example

Find the equation of a circle centre (–2 ,3) and radius 4 units

Solution

General equation of the circle is (x–a)2 + (y–b)2 = r2 .Therefore a = –2, b =3 and r = 4
(x–(–2))2 + (y–(3))2 = 42
(x+2)2 + (y–3)2 = 16

Example

Line AB is the diameter of a circle such that the co-ordinates of A and B are (–1, 1 ) and(5, 1 ) respectively.

1. Determine the centre and the radius of the circle
2. Hence, find the equation of the circle

Solution

1. (1+5, 1+21) = (2,1)
2       2
=√32 = 3
2. Equation of the circle is ;
(x–2)2 + (y–1)2 = 32
(x–2)2 + (y–1)2 = 9

Example

The equation of a circle is given by x 6x + y2 + 4y – 3 = 0.Determine the centre and radius of the circle.

Solution

x2 6x +y2 + 4y = 3
Completing the square on the left hand side;
x2 6x + 9+y2 + 4y + 4 = 3 + 9 + 4
(x – 3)
2 + (y + 2)2 = 4 – 3 = 0
Therefore centre of the circle is (3,–2) and radius is 4 units. Note that the sign changes to opposite positive sign becomes negative while negative sign changes to positive.

Example

Write the equation of the circle that has A(1, – 6) and B(5, 2) as endpoints of a diameter.

Solution

Method 1:

Determine the center using the Midpoint Formula:

C(1+5 , –6+2)→C(3, –2)
2        2
Determine the radius using the distance formula (center and end of diameter):
r = (3−1)2 + (−2+6)2 = √(4 + 16) = 20 = 2√5
Equation of circle is: (−3)2 + (y+2)2 = 20

Method 2:

Determine center using Midpoint Formula (as before): C(3,−2).
Thus, the circle equation will have the form (x−3)2 + (y+2)2 = r2
Find r2 by plugging the coordinates of a point on the circle in for x and y.
Let’s use B(5, 2):r2 = (5−3)2 + (2+2)2 = 22 + 42 = 4 + 16 = 20
Again, we get this equation for the circle: (x−3)2 + (y+2)2 = 20

## Past KCSE Questions on the Topic.

1. The table shows the height metres of an object thrown vertically upwards varies with the time t seconds
 T 0 1 2 3 4 5 6 7 8 9 10 S 45.1 49.9 −80
The relationship between s and t is represented by the equations s = at2 + bt + 10 where b are constants.
1.
1. Using the information in the table, determine the values of a and b
2. Complete the table
2.
1. Draw a graph to represent the relationship between s and t
2. Using the graph determine the velocity of the object when t = 5 seconds
2. Data collected form an experiment involving two variables X and Y was recorded as shown in the table below
 x 1.1 1.2 1.3 1.4 1.5 1.6 y −0.3 0.5 1 .4 2.5 3.8 5.2
The variables are known to satisfy a relation of the form y = ax3 + b where a and b are constants
1. For each value of x in the table above, write down the value of x3
1. By drawing a suitable straight line graph, estimate the values of a and b
2. Write down the relationship connecting y and x
3. Two quantities P and r are connected by the equation p = krn. The table of values of P and r is given below.
 P 1.2 1.5 2 2.5 3.5 4.5 R 1.58 2.25 3.39 4.74 7.86 11 .5
1. State a liner equation connecting P and r.
2. Using the scale 2 cm to represent 0.1 units on both axes, draw a suitable line graph on the grid provided. Hence estimate the values of K and n.
4. The points which coordinates (5,5) and (−3,−1 ) are the ends of a diameter of a circle centre A
Determine:
1. The coordinates of A
2. The equation of the circle, expressing it in form x2 + y2 + ax + by + c = 0 where a, b, and c are constants each computer sold
5. The figure below is a sketch of the graph of the quadratic function y = k(x+1 ) (x−2)
Find the value of k
6. The table below shows the values of the length X (in metres ) of a pendulum and the corresponding values of the period T (in seconds) of its oscillations obtained in an experiment.
 X (metres) 0.4 1 .0 1 .2 1 .4 1 .6 T (seconds) 1 .25 2.01 2.1 9 2.37 2.53
1. Construct a table of values of log X and corresponding values of log T, correcting each value to 2 decimal places
2. Given that the relation between the values of log X and log T approximate to a linear law of the form m log X + log a where a and b are constants
1. Use the axes on the grid provided to draw the line of best fit for the graph of log T against log X.
2. Use the graph to estimate the values of a and b
3. Find, to decimal places the length of the pendulum whose period is 1 second.
7. Data collection from an experiment involving two variables x and y was recorded as shown in the table below
 X 1.1 1.2 1.3 1.4 1.5 1.6 Y −0.3 0.5 1.4 2.5 3.8 5.2
The variables are known to satisfy a relation of the form y = ax3 + b where a and b are constants
1. For each value of x in the table above. Write down the value of x3
2.
1. By drawing s suitable straight line graph, estimate the values of a and b
2. Write down the relationship connecting y and x
8. Two variables x and y, are linked by the relation y = axn. The figure below shows part of the straight line graph obtained when log y is plotted against log x.
Calculate the value of a and n
9. The luminous intensity I of a lamp was measured for various values of voltage v across it. The results were as shown below
 V(volts) 30 36 40 44 48 50 54 L (Lux ) 708 1248 1726 2320 3038 3848 4380
It is believed that V and l are related by an equation of the form l = aVnwhere a and n are constant.
1. Draw a suitable linear graph and determine the values of a and n
2. From the graph find
1. The value of I when V = 52
2. The value of V when I = 2800
10. In a certain relation, the value of A and B observe a relation B= CA + KA2 where C and K are constants. Below is a table of values of A and B
 A 1 2 3 4 5 6 B 3.2 6.75 10.8 15.1 20 25.2
1. By drawing a suitable straight line graphs, determine the values of C and K.
2. Hence write down the relationship between A and B
3. Determine the value of B when A = 7
11. The variables P and Q are connected by the equation P = abq where a and b are constants. The value of p and q are given below
 P 6.56 1 7.7 47.8 1 29 349 941 2540 6860 Q 0 1 2 3 4 5 6 7
1. State the equation in terms of p and q which gives a straight line graph
2. By drawing a straight line graph, estimate the value of constants a and b and give your answer correct to 1 decimal place

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