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Year 11 Methods (Unit 1 & 2) Rates Of Change

Average Rate Of Change

20 practice questions 1 video lesson Theory + worked examples

Understand the average rate of change for Queensland Year 11 Mathematical Methods (QCAA). It measures how much a function rises or falls across an interval — the change in y divided by the change in x.

You will learn to compute this rate over any interval, read it as the gradient of a secant line, and work backwards from a known rate — a stepping stone to the instantaneous rate of change.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 2), the average rate of change of a function \(y=f(x)\) over an interval \([a,b]\) is \(\dfrac{f(b)-f(a)}{b-a}\) — the gradient of the secant line joining \((a,f(a))\) and \((b,f(b))\). This page shows how to compute it for polynomial and power functions, read it as a secant gradient, apply it in context (such as average velocity), and reverse it to find an endpoint.

The average rate of change of \(y=f(x)\) between \(x=a\) and \(x=b\) is the total change in \(y\) divided by the total change in \(x\). It measures the overall rate across the interval, even when the actual rate varies within it.

Geometrically it is the gradient of the secant — the straight line that cuts the curve at the two endpoints \((a,f(a))\) and \((b,f(b))\). Unlike the gradient of a tangent (which is the rate at a single instant), the secant gradient averages the behaviour between the two points.

For a straight-line (linear) function the rate is constant, so the average rate over any interval is the same. For a curve it generally differs from one interval to the next.

Average rate of change \(=\dfrac{f(b)-f(a)}{b-a}=\) gradient of the secant. Evaluate the function at both endpoints, subtract, and divide by the change in \(x\).
Average rate of change is the secant gradientA parabola with a straight secant line joining points A and B; the average rate of change equals the gradient of that secant. x y A B
The average rate of change of \(y=x^2\) on \([1,3]\) is the gradient of the secant \(AB\): \(\dfrac{9-1}{3-1}=4\).
A straight line has a constant rateA straight line where the average rate of change is the same over every interval, equal to the fixed gradient. x y
For a straight line the average rate is the same over every interval — it equals the fixed gradient.

The average rate of change of \(y=f(x)\) over \([a,b]\) is

\[\text{average rate}=\dfrac{f(b)-f(a)}{b-a}\]
f(b)-f(a)b-a

This is exactly the gradient of the secant through \(\bigl(a,f(a)\bigr)\) and \(\bigl(b,f(b)\bigr)\):

\[m_{\text{secant}}=\dfrac{y_2-y_1}{x_2-x_1}\]
msecant=y2-y1x2-x1
Average rate is a secant, not a tangent. It uses the two endpoints only; the rate at a single instant (the tangent gradient) comes later, in instantaneous rate of change.

How to find an average rate of change over \([a,b]\)

  1. Evaluate: substitute the endpoints to find \(f(a)\) and \(f(b)\).
  2. Subtract: form the change in \(y\), which is \(f(b)-f(a)\), and the change in \(x\), which is \(b-a\).
  3. Divide: compute \(\dfrac{f(b)-f(a)}{b-a}\) and, in context, attach the correct units.
Example 1 — Average rate of a curve
Find the average rate of change of \(f(x)=x^2+1\) over the interval \([1,\,4]\).
Solution

Evaluate at the endpoints:

\(f(1)\)\(=\)\((1)^2+1=2\)
\(f(4)\)\(=\)\((4)^2+1=17\)

Divide the change in \(y\) by the change in \(x\):

\(\text{average rate}\)\(=\)\(\dfrac{f(4)-f(1)}{4-1}\)
\(=\)\(\dfrac{17-2}{3}\)
\(=\)\(\dfrac{15}{3}\)
\(=\)\(5\)

Average rate of change \(=5\).

Secant of y=x^2+1 on the interval one to fourA parabola with a secant joining one comma two and four comma seventeen; its gradient is five. x y
5
Example 2 — Secant gradient
Find the gradient of the secant to \(y=x^2-3x\) between \(x=1\) and \(x=5\).
Solution

Evaluate at the endpoints:

\(f(1)\)\(=\)\((1)^2-3(1)=-2\)
\(f(5)\)\(=\)\((5)^2-3(5)=10\)

The secant gradient is the average rate of change:

\(m_{\text{secant}}\)\(=\)\(\dfrac{f(5)-f(1)}{5-1}\)
\(=\)\(\dfrac{10-(-2)}{4}\)
\(=\)\(\dfrac{12}{4}\)
\(=\)\(3\)

Secant gradient \(=3\).

Secant of y=x^2-3x on the interval one to fiveA parabola with a secant joining one comma minus two and five comma ten; its gradient is three. x y
3
Example 3 — Average velocity in context
A ball's height is \(s=40t-5t^2\) (metres), with \(t\) in seconds. Find its average velocity over the interval \([1,\,3]\).
Solution

Height at each endpoint:

\(s(1)\)\(=\)\(40(1)-5(1)^2\)
\(=\)\(40-5=35\)
\(s(3)\)\(=\)\(40(3)-5(3)^2\)
\(=\)\(120-45=75\)

Average velocity is displacement over time:

\(\bar v\)\(=\)\(\dfrac{s(3)-s(1)}{3-1}\)
\(=\)\(\dfrac{75-35}{2}\)
\(=\)\(\dfrac{40}{2}\)
\(=\)\(20\text{ m/s}\)

Average velocity \(=20\) m/s.

Average velocity as a secant gradientA projectile height curve with a secant joining one comma thirty-five and three comma seventy-five; the average velocity is twenty metres per second. x y
20 m/s
Example 4 — Find the endpoint (reverse)
For \(y=x^2\), the average rate of change over \([2,\,b]\) is \(7\). Find \(b\) (with \(b>2\)).
Solution

Write the average rate and factor the difference of squares:

\(\dfrac{b^2-2^2}{b-2}\)\(=\)\(7\)
\(\dfrac{(b-2)(b+2)}{b-2}\)\(=\)\(7\)
\(b+2\)\(=\)\(7\)

Solve for \(b\):

\(b\)\(=\)\(7-2\)
\(=\)\(5\)

The endpoint is \(b=5\).

Finding the endpoint from the average rateA parabola with a secant of gradient seven from two comma four; the far endpoint is five comma twenty-five. x y b=5
b=5

Common pitfalls

Subtracting in a different order top and bottom. Keep the order consistent: if the numerator is \(f(b)-f(a)\) then the denominator must be \(b-a\), not \(a-b\).
Averaging the two \(y\)-values. The average rate is \(\dfrac{f(b)-f(a)}{b-a}\), not \(\dfrac{f(a)+f(b)}{2}\); it is a gradient, not an average height.
Mistaking it for the instantaneous rate. The average rate uses two points (a secant). The rate at a single instant (a tangent gradient) is a different, later idea.
Sign slips with negatives. When an endpoint value is negative, keep the minus sign: \(10-(-2)=12\), not \(8\).

Frequently asked questions

What is the average rate of change of a function?

The change in \(y\) divided by the change in \(x\) over an interval, \(\dfrac{f(b)-f(a)}{b-a}\).

How is the average rate of change related to a secant?

It is exactly the gradient of the secant line, the straight line that joins the two endpoints \((a,f(a))\) and \((b,f(b))\) on the curve.

What is the difference between average and instantaneous rate of change?

The average rate uses two points (a secant gradient) over an interval; the instantaneous rate is the rate at a single point (a tangent gradient).

How do you find the average velocity of a moving object?

Divide the change in position by the change in time: \(\dfrac{s(t_2)-s(t_1)}{t_2-t_1}\). It is the average rate of change of position.

Why is the average rate of a straight line the same everywhere?

Because a straight line has a constant gradient, so the secant over any interval has the same gradient — the rate never changes.

Can an average rate of change be negative?

Yes. If \(f(b)

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