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    This module and collection are included inLens: Siyavula: Mathematics (Gr. 10-12)
    By: Siyavula

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Introduction

The gradient of a straight line graph is calculated as:

y 2 - y 1 x 2 - x 1 y 2 - y 1 x 2 - x 1
(1)

for two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) on the graph.

We can now define the average gradient between two points even if they are defined by a function which is not a straight line, (x1,y1)(x1,y1) and (x2,y2)(x2,y2) as:

y 2 - y 1 x 2 - x 1 y 2 - y 1 x 2 - x 1
(2)

This is the same as Equation 1.

Straight-Line Functions

Investigation : Average Gradient - Straight Line Function

Fill in the table by calculating the average gradient over the indicated intervals for the function f(x)=2x-2f(x)=2x-2. Note that (x1x1;y1y1) is the co-ordinates of the first point and (x2x2;y2y2) is the co-ordinates of the second point. So for AB, (x1x1;y1y1) is the co-ordinates of point A and (x2x2;y2y2) is the co-ordinates of point B.

Table 1
  x 1 x 1 x 2 x 2 y 1 y 1 y 2 y 2 y 2 - y 1 x 2 - x 1 y 2 - y 1 x 2 - x 1
A-B          
A-C          
B-C          
Figure 1
Figure 1 (MG10C12_001.png)

What do you notice about the gradients over each interval?

The average gradient of a straight-line function is the same over any two intervals on the function.

Parabolic Functions

Investigation : Average Gradient - Parabolic Function

Fill in the table by calculating the average gradient over the indicated intervals for the function f(x)=2x-2f(x)=2x-2:

Table 2
  x 1 x 1 x 2 x 2 y 1 y 1 y 2 y 2 y 2 - y 1 x 2 - x 1 y 2 - y 1 x 2 - x 1
A-B          
B-C          
C-D          
D-E          
E-F          
F-G          

What do you notice about the average gradient over each interval? What can you say about the average gradients between A and D compared to the average gradients between D and G?

Figure 2
Figure 2 (MG10C12_002.png)

The average gradient of a parabolic function depends on the interval and is the gradient of a straight line that passes through the points on the interval.

For example, in Figure 3 the various points have been joined by straight-lines. The average gradients between the joined points are then the gradients of the straight lines that pass through the points.

Figure 3: The average gradient between two points on a curve is the gradient of the straight line that passes through the points.
Figure 3 (MG10C12_003.png)

Method: Average Gradient

Given the equation of a curve and two points (x1x1, x2x2):

  1. Write the equation of the curve in the form y=...y=....
  2. Calculate y1y1 by substituting x1x1 into the equation for the curve.
  3. Calculate y2y2 by substituting x2x2 into the equation for the curve.
  4. Calculate the average gradient using:
    y2-y1x2-x1y2-y1x2-x1
    (3)

Exercise 1: Average Gradient

Find the average gradient of the curve y=5x2-4y=5x2-4 between the points x=-3x=-3 and x=3x=3

Solution
  1. Step 1. Label points :

    Label the points as follows:

    x 1 = - 3 x 1 = - 3
    (4)
    x 2 = 3 x 2 = 3
    (5)

    to make it easier to calculate the gradient.

  2. Step 2. Calculate the yy coordinates :

    We use the equation for the curve to calculate the yy-value at x1x1 and x2x2.

    y 1 = 5 x 1 2 - 4 = 5 ( - 3 ) 2 - 4 = 5 ( 9 ) - 4 = 41 y 1 = 5 x 1 2 - 4 = 5 ( - 3 ) 2 - 4 = 5 ( 9 ) - 4 = 41
    (6)
    y 2 = 5 x 2 2 - 4 = 5 ( 3 ) 2 - 4 = 5 ( 9 ) - 4 = 41 y 2 = 5 x 2 2 - 4 = 5 ( 3 ) 2 - 4 = 5 ( 9 ) - 4 = 41
    (7)
  3. Step 3. Calculate the average gradient :
    y 2 - y 1 x 2 - x 1 = 41 - 41 3 - ( - 3 ) = 0 3 + 3 = 0 6 = 0 y 2 - y 1 x 2 - x 1 = 41 - 41 3 - ( - 3 ) = 0 3 + 3 = 0 6 = 0
    (8)
  4. Step 4. Write the final answer :

    The average gradient between x=-3x=-3 and x=3x=3 on the curve y=5x2-4y=5x2-4 is 0.

Average gradient for other functions

We can extend the concept of average gradient to any function. The average gradient for any function also depends on the interval chosen and is the gradient of a straight line that passes through the two points. So we can use the formula that we found for the average gradient of parabolic functions and apply it to any function. We will consider the average gradient of just two functions here: exponential functions and hyperbolic functions.

Average gradient of exponential functions

For example, if we were asked to find the average gradient of the function g(x)=3.2x+2g(x)=3.2x+2 between the points (-4;2,2)(-4;2,2) and (-0,6;4)(-0,6;4). This is shown in Figure 4.

Figure 4: The average gradient for an exponential function.
Figure 4 (exponent.png)
Using the formula we find:
y 2 - y 1 x 2 - x 1 = 4 - 2,2 ( - 0,6 ) - ( - 4 ) = 1,8 - 0,6 + 4 = 1,8 5,2 = 0,35 y 2 - y 1 x 2 - x 1 = 4 - 2,2 ( - 0,6 ) - ( - 4 ) = 1,8 - 0,6 + 4 = 1,8 5,2 = 0,35
(9)

Average gradient of hyperbolic functions

For example, if we were asked to find the average gradient of the function g(x)=2x+2g(x)=2x+2 between the points (-4;-2,5)(-4;-2,5) and (0,5;6)(0,5;6) and (-4;2,2)(-4;2,2) and (-0,6;4)(-0,6;4). This is shown in Figure 5.

Figure 5: The average gradient for a hyperbolic function.
Figure 5 (hyperbola.png)

For the first point we would get:

y 2 - y 1 x 2 - x 1 = ( - 2,5 ) - 1 ( - 4 ) - 0,5 = - 3,5 - 4,5 = 0,78 y 2 - y 1 x 2 - x 1 = ( - 2,5 ) - 1 ( - 4 ) - 0,5 = - 3,5 - 4,5 = 0,78
(10)
Similarly for the second points we would find that the average gradient is: 0,530,53

Summary

  • The average gradient between two points is: y 2 - y 1 x 2 - x 1 y 2 - y 1 x 2 - x 1
  • The average gradient of a straight-line function is the same over any two intervals on the function
  • The average gradient of a parabolic function depends on the interval and is the gradient of a straight line that passes through the points on the interval
  • We can extend the concept of average gradient to any function

End of Chapter Exercises

  1. An object moves according to the function d=2t2+1d=2t2+1 , where dd is the distance in metres and tt the time in seconds. Calculate the average speed of the object between 2 and 3 seconds. The speed is the gradient of the function dd
     Click here for the solution
  2. Given: f(x)=x3-6xf(x)=x3-6x. Determine the average gradient between the points where x=1x=1 and x=4x=4.
     Click here for the solution
  3. Find the average gradient of each of the following functions between the points where x=2x=2 and x=3x=3
    1. f(x)=x2+3f(x)=x2+3
    2. f(x)=4x+1f(x)=4x+1
    3. f(x)=2x-3f(x)=2x-3
    Click here for the solution

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