Average Gradient Between Two Points

The average gradient measures how steep a curve is on average between two points. It prepares you directly for differential calculus in Grade 12, where the two points move closer and closer together.

The average gradient of $f$ between $x = a$ and $x = b$ is the gradient of the straight line joining the two points on the curve:

$average gradient = \frac{f ( b ) - f ( a )}{b - a}$

Example. $f (x) = x^{2} + 2 x$ between $x = 1$ and $x = 3$ :

$\frac{f ( 3 ) - f ( 1 )}{3 - 1} = \frac{15 - 3}{2} = 6$

With a variable gap. Between $x$ and $x + h$ on $f (x) = x^{2} - 3 x$ :

$\frac{( x + h ) ^{2} - 3 ( x + h ) - ( x ^{2} - 3 x )}{h} = \frac{2 x h + h ^{2} - 3 h}{h} = 2 x + h - 3$

As $h$ gets very small, this approaches $2 x - 3$ , the gradient at a single point. That idea is the start of Grade 12 calculus.

Average gradient as a rate of change

If $y$ depends on time, the average gradient is the average rate of change over the time interval.

Example. A ball rolls down a ramp so that its distance after $t$ seconds is $s (t) = 5 t^{2}$ metres.

Between $t = 1$ and $t = 3$ : $\frac{45 - 5}{2} = 20$ m/s
Between $t = 2$ and $t = 2, 1$ : $\frac{22 , 05 - 20}{0 , 1} = 20, 5$ m/s

The shorter the interval, the closer the average speed is to the speed at that instant.

Always give units for a rate: units of $y$ per unit of $x$ .

Key terms

Average gradient: The gradient of the line joining two points on a curve.
Secant: A line joining two points on a curve.
Rate of change: How fast one quantity changes compared with another.

Common mistakes

Subtracting the $y$ -values and $x$ -values in different orders.
Using $f (b) - f (a)$ without dividing by $b - a$ .
Expanding $(x + h)^{2}$ as $x^{2} + h^{2}$ .
Leaving out units for a rate of change.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Calculate the average gradient of $f (x) = x^{2} + 2 x$ between $x = 1$ and $x = 3$ .

Worked solution

Hidden - try the question first

Solution

$f (3) = 15$ and $f (1) = 3$
$\frac{15 - 3}{3 - 1}$
$= 6$

Question 2

5 marks total

routine

Calculate the average gradient of $f (x) = - 2 x^{2} + 3$ between $x = - 1$ and $x = 2$ .

Worked solution

Hidden - try the question first

Solution

$f (3) = 15$
$f (- 1) = - 2 + 3 = 1$
$\frac{- 5 - 1}{2 - ( - 1 )}$
$= - 2$

Question 3

6 marks total

complex

Given $f (x) = x^{2} - 3 x$ .

(a) Show that the average gradient between $x$ and $x + h$ is $2 x + h - 3$ . (4)

(b) Calculate this average gradient when $x = 2$ and $h = 0, 01$ . (1)

Worked solution

Hidden - try the question first

Solution

(a) $f (x + h) = x^{2} + 2 x h + h^{2} - 3 x - 3 h$
$f (x + h) - f (x) = 2 x h + h^{2} - 3 h$
$\frac{2 x h + h ^{2} - 3 h}{h}$
$= 2 x + h - 3$
(b) $4 + 0, 01 - 3 = 1, 01$
(c) $1$

Question 4

8 marks total

problem-solving

A ball rolls down a ramp. The distance it has travelled after $t$ seconds is $s (t) = 5 t^{2}$ metres.

(a) Calculate the average speed between $t = 1$ and $t = 3$ . (3)

(b) Calculate the average speed between $t = 2$ and $t = 2, 1$ . (3)

Worked solution

Hidden - try the question first

Solution

(a) $f (x + h) = x^{2} + 2 x h + h^{2} - 3 x - 3 h$ and $s (1) = 5$
$\frac{45 - 5}{3 - 1}$
$= 20$ m/s
(b) $s (2, 1) = 22, 05$ and $s (2) = 20$
$\frac{22 , 05 - 20}{0 , 1}$
$= 20, 5$ m/s
(c) About 20 m/s
The average speed over shorter intervals ending or starting at $t = 2$ approaches 20 m/s

FAQs

Why do we learn average gradient in Grade 11?

It leads directly to the derivative in Grade 12, which is the gradient at a single point.

Is average gradient the same as the gradient of a straight line?

It is the gradient of the straight line joining two points on a curve. For a straight-line graph, it equals the line's gradient everywhere.

What units does an average rate of change have?

Units of the output per unit of the input, such as metres per second or rand per month.

Average gradient