Average rates and secant gradients | Year 11

01 / LEARN

The idea, step by step

Change divided by interval

An average rate compares the total change in an output with the change in its input. Subtract outputs in the same order as inputs. Its sign tells whether the output has risen or fallen overall, not what happens at every moment.

A secant joins two distinct points on a graph. Its gradient equals the average rate between them. Find both output values from the original function before forming the quotient.

average rate = \frac{f (b) - f (a)}{b - a}, a \neq b .

Interpret the result

Distance against time gives metres per second; cost against items gives dollars per item. State the output units divided by input units and explain the sign in context.

Equal endpoint values give zero average change even when the graph rises and falls between them. A zero secant gradient does not mean all tangent gradients are zero. Estimate the sign from the graph as a check.

A common trap

Subtracting the inputs in the opposite order from the outputs.

Instead: Keep matching endpoint order in both subtractions; reversing only one changes the sign.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Find the average rate of $f (x) = x^{2}$ from $x = 1$ to $x = 3$ .

$f (1) = 1$ and $f (3) = 9$ .
The changes are $Δ y = 8$ and $Δ x = 2$ .
The secant gradient is $8 / 2 = 4$ .

4 output units per input unit.

Example 2

Worked example 2

A tank has $V (t) = 120 - 6 t$ litres after $t$ minutes. Find its average rate from 2 to 7 minutes.

$V (2) = 108$ and $V (7) = 78$ litres.
$Δ V = - 30$ litres over $Δ t = 5$ minutes.
The rate is $- 30 / 5 = - 6$ .

−6 L/min: the volume falls by 6 litres per minute.

Check your understanding

What is the average rate of $f (x) = 3 x + 2$ from 0 to 4?

Show answer

$(14 - 2) / (4 - 0) = 3$ . A line has the same secant gradient on every interval.

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the average rate of $f (x) = x^{2}$ from 0 to 4.
3 marks
Find the secant gradient of $g (x) = x^{3}$ between 1 and 3.
3 marks
A population rises from 320 to 395 over 5 years. Find its average yearly increase.
3 marks
$C (t) = 100 - 4 t$ . Find its average rate from 2 to 7.
3 marks
A secant passes through (1,5) and (4,17). Find its gradient.
3 marks
$h (t) = t^{2} - 4 t$ on $0 \leq t \leq 4$ . Its average rate is 0. Must its instantaneous rate always be 0? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the average rate of $f (x) = x^{2}$ from 0 to 4.
3 marks
$f (0) = 0, f (4) = 16$ .
$(16 - 0) / (4 - 0) = 4$ .
4
Find the secant gradient of $g (x) = x^{3}$ between 1 and 3.
3 marks
$g (1) = 1, g (3) = 27$ .
$(27 - 1) / (3 - 1) = 13$ .
13
A population rises from 320 to 395 over 5 years. Find its average yearly increase.
3 marks
The change is 395−320=75 people.
Divide by 5 years: 75/5=15.
15 people per year
$C (t) = 100 - 4 t$ . Find its average rate from 2 to 7.
3 marks
$C (2) = 92, C (7) = 72$ .
$(72 - 92) / (7 - 2) = - 4$ .
−4 units per time unit
A secant passes through (1,5) and (4,17). Find its gradient.
3 marks
$Δ y = 12, Δ x = 3$ .
$m = 12 / 3 = 4$ .
4
$h (t) = t^{2} - 4 t$ on $0 \leq t \leq 4$ . Its average rate is 0. Must its instantaneous rate always be 0? Explain.
3 marks
$h (0) = h (4) = 0$ , but $h (2) = - 4$ .
Equal endpoints do not imply a constant graph.
No. It falls and rises back to the starting value.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topic 6: rates of change, derivatives and applications. Introductory polynomial calculus with explicit modelling and interpretation.

Official curriculum source ↗

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