- Graphing Solutions of Cubic Equations
- Average Rate of Change
- Empirical Graphs
- Reduction of Non-linear Laws to Linear Form.
- Equation of a Circle
- Past KCSE Questions on the Topic.
- These are ways or methods of solving mathematical functions using graphs.
- 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 = 0, 4x3 + 57 = 0, x3 + 9x = 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.
- 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
- Using the grid provided draw the graph for y = −6 + x + 4x2 + x3 for −4≤x ≤2
- Use the graph to solve the equations:-
- −6 + x + 4x2 + x3 = 0
- x3 + 4x2 + x – 4 = 0
- −2 + 4x2 + x3 = 0
The table shows corresponding values of x and y for y = −6 + x + 4x2 + x3
|4x2||64||36||1 6||4||0||4||1 6|
From the graph the solutions for x are x = −3 , x = −2, x = 1
To solve equation y = −6 + x + 4x2 + x3 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
y = x3 + 4x2 + x – 6
0 = x3 + 4x2 + 0 – 2
y = x – 4
- 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.
- 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 y with respect to x = change in y
change in x
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
- 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.
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.
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.
The table below shows how length l cm of a metal rod varies with increase in temperature T (0C).
- 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.
- 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.
The table below shows the relationship between A and r
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.
If we plot A against r2,we should get a straight line.
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.
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.
By plotting a suitable graph, check whether the population growth obeys the given law. Use the graph to estimate the value of A.
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.
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
- 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
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
- Y is plotted on the y axis
- x2 is plotted on the x axis
- The gradient is c
- 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
- y should be plotted on the y axis
- x should be plotted on the x axis
- The gradient is b
- 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.
- y should be plotted on the vertical axis
- 1/x should be plotted on the horizontal axis
- The gradient is f
- The vertical axis intercept is e
For the law y – cx = bx2 to 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,
- y/x should be plotted on y axis
- X should be plotted on x axis
- The gradient is b
- 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
- y/x should be plotted on the vertical axis
- 1/x2 should be plotted on the horizontal axis
- The gradient is a
- The vertical intercept is b
- 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
- The general equation of a circle centre (0, 0) and radius r is x2 + y2 = r2
Find the equation of a circle centre (0, 0) passing through (3, 4)
Let the radius of the circle be r
From Pythagoras theorem;
r = √(32 x 42)
r = 5
Consider a circle centre (5 , 4 ) and radius 3 units.
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.
The equation of a circle centre ( a,b) and radius r units is given by;
(x–a)2 + (y–b)2 = (r)2
Find the equation of a circle centre (–2 ,3) and radius 4 units
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
Line AB is the diameter of a circle such that the co-ordinates of A and B are (–1, 1 ) and(5, 1 ) respectively.
- Determine the centre and the radius of the circle
- Hence, find the equation of the circle
- (–1+5, 1+21) = (2,1)
Radius =√[(5–2)2 + (1–1)2]
=√32 = 3
- Equation of the circle is ;
(x–2)2 + (y–1)2 = 32
(x–2)2 + (y–1)2 = 9
The equation of a circle is given by x2 – 6x + y2 + 4y – 3 = 0.Determine the centre and radius of the circle.
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.
Write the equation of the circle that has A(1, – 6) and B(5, 2) as endpoints of a diameter.
Determine the center using the Midpoint Formula:
C(1+5 , –6+2)→C(3, –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
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
- 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
- Using the information in the table, determine the values of a and b
- Complete the table
- Draw a graph to represent the relationship between s and t
- Using the graph determine the velocity of the object when t = 5 seconds
- 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
- For each value of x in the table above, write down the value of x3
- By drawing a suitable straight line graph, estimate the values of a and b
- Write down the relationship connecting y and x
- 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 2.0 2.5 3.5 4.5 R 1.58 2.25 3.39 4.74 7.86 11 .5
- State a liner equation connecting P and r.
- 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.
- The points which coordinates (5,5) and (−3,−1 ) are the ends of a diameter of a circle centre A
- The coordinates of A
- 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
- The figure below is a sketch of the graph of the quadratic function y = k(x+1 ) (x−2)
Find the value of k
- 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
- Construct a table of values of log X and corresponding values of log T, correcting each value to 2 decimal places
- 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
- Use the axes on the grid provided to draw the line of best fit for the graph of log T against log X.
- Use the graph to estimate the values of a and b
- Find, to decimal places the length of the pendulum whose period is 1 second.
- 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
- For each value of x in the table above. Write down the value of x3
- By drawing s suitable straight line graph, estimate the values of a and b
- Write down the relationship connecting y and x
- 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
- 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
- Draw a suitable linear graph and determine the values of a and n
- From the graph find
- The value of I when V = 52
- The value of V when I = 2800
- 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
- By drawing a suitable straight line graphs, determine the values of C and K.
- Hence write down the relationship between A and B
- Determine the value of B when A = 7
- 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
- State the equation in terms of p and q which gives a straight line graph
- 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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