Recognising Relationships
Learn to recognise rates of change from graphs and tables for Queensland Year 11 Mathematical Methods (QCAA). A rate tells you how quickly one quantity changes as another does.
You will learn to spot increasing and decreasing intervals, tell a constant rate from a changing one, find where the rate is greatest, least or zero, and calculate an average rate of change — the intuition behind calculus in the QCAA course.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), before any differentiation rules, you learn to recognise relationships from a graph or table: read the sign of the rate of change (rising, falling or flat), find where a quantity is increasing or decreasing, locate where the rate is greatest, least or zero, and tell a constant (straight-line) rate from a changing (curved) one. This page shows how to read the gradient of a graph as a rate of change.
The rate of change of a quantity tells you how fast it is changing as another quantity varies. On a graph of \(y\) against \(x\), the rate of change is the gradient: where the graph rises the rate is positive, where it falls the rate is negative, and where it is horizontal the rate is zero.
A constant rate produces a straight-line graph — the steepness is the same everywhere. A changing rate produces a curve: the graph gets steeper where the rate grows and flatter where it shrinks. The point where a curve is steepest has the greatest rate of change; a turning point (a peak or trough) has a rate of change of zero.
The average rate of change between two points is the gradient of the straight line joining them — the secant — found as change in \(y\) over change in \(x\).
The rate of change read from a graph is a gradient. Over an interval from \(x=a\) to \(x=b\), the average rate of change is
For a straight-line graph this gradient is the same over every interval — the rate is constant:
How to read a rate of change from a graph
- Direction: decide whether the graph is rising (positive rate), falling (negative rate) or horizontal (zero rate).
- Steepness: compare how steep the graph is — the steepest part has the greatest rate; a straight line has a constant rate.
- Describe: state the rate with its sign and, in a practical context, its units and meaning (for example metres per second).
Read the direction of the graph on each stage.
Stage 1 — the graph rises from \((0,0)\) to \((8,400)\):
| \(\text{rate}\) | \(=\) | \(\dfrac{400-0}{8-0}\) |
| \(=\) | \(+50\text{ m/min (positive)}\) |
Stage 2 — the graph is horizontal from \((8,400)\) to \((12,400)\):
| \(\text{rate}\) | \(=\) | \(\dfrac{400-400}{12-8}\) |
| \(=\) | \(0\text{ m/min (zero)}\) |
Stage 3 — the graph falls from \((12,400)\) to \((20,200)\):
| \(\text{rate}\) | \(=\) | \(\dfrac{200-400}{20-12}\) |
| \(=\) | \(-25\text{ m/min (negative)}\) |
Walking away at \(+50\) m/min, resting at \(0\) m/min, then returning at \(-25\) m/min.
The rate of change is the steepness of the curve. On \(y=x^2\) the curve gets steeper as \(x\) increases, so compare the short intervals just before each point.
Steepness near \(P\), on \([0.5,\,1]\):
| \(\text{gradient}\) | \(=\) | \(\dfrac{1-0.25}{1-0.5}\) |
| \(=\) | \(1.5\) |
Steepness near \(Q\), on \([1.5,\,2]\):
| \(\text{gradient}\) | \(=\) | \(\dfrac{4-2.25}{2-1.5}\) |
| \(=\) | \(3.5\) |
Steepness near \(R\), on \([2.5,\,3]\):
| \(\text{gradient}\) | \(=\) | \(\dfrac{9-6.25}{3-2.5}\) |
| \(=\) | \(5.5\) |
The curve is steepest at \(R(3,9)\), so the rate of change is greatest there.
| \(t\) (min) | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| \(V\) (L) | 40 | 30 | 24 | 20 |
Average rate over \([0,2]\):
| \(\dfrac{\Delta V}{\Delta t}\) | \(=\) | \(\dfrac{30-40}{2-0}\) |
| \(=\) | \(\dfrac{-10}{2}\) | |
| \(=\) | \(-5\text{ L/min}\) |
Average rate over \([4,6]\):
| \(\dfrac{\Delta V}{\Delta t}\) | \(=\) | \(\dfrac{20-24}{6-4}\) |
| \(=\) | \(\dfrac{-4}{2}\) | |
| \(=\) | \(-2\text{ L/min}\) |
Both rates are negative (draining). \(|-5|>|-2|\), so the tank empties faster at the start — it is decreasing at a decreasing rate.
\([0,2]\): \(-5\) L/min; \([4,6]\): \(-2\) L/min; it drains fastest in the first two minutes.
Track the direction of the height as time increases. The graph is a parabola opening downwards with a peak.
Average rate on the way up, \([0,2]\):
| \(\text{rate}\) | \(=\) | \(\dfrac{h(2)-h(0)}{2-0}\) |
| \(=\) | \(\dfrac{20-0}{2}\) | |
| \(=\) | \(+10\text{ m/s (rising)}\) |
Average rate on the way down, \([2,4]\):
| \(\text{rate}\) | \(=\) | \(\dfrac{h(4)-h(2)}{4-2}\) |
| \(=\) | \(\dfrac{0-20}{2}\) | |
| \(=\) | \(-10\text{ m/s (falling)}\) |
The height rises to a peak of \(20\) m at \(t=2\) s, where the graph is momentarily horizontal, then falls.
The rate is positive before \(t=2\) s, zero at the peak \((2,20)\), and negative after.
Common pitfalls
Frequently asked questions
How do you tell the sign of a rate of change from a graph?
If the graph rises as you move right the rate is positive; if it falls the rate is negative; if it is horizontal the rate is zero.
Where is the rate of change greatest on a curve?
Where the curve is steepest. The rate of change is the gradient, so the steepest part of the graph has the greatest rate.
What does a straight-line graph tell you about the rate?
That the rate is constant — the same over every interval — because a straight line has the same gradient everywhere.
What is the average rate of change between two points?
The gradient of the straight line (the secant) joining them: \(\dfrac{f(b)-f(a)}{b-a}\), the change in \(y\) divided by the change in \(x\).
What does a rate of change of zero mean?
The quantity is momentarily not changing. On a graph this is a horizontal part, such as a peak, a trough or a flat section.
What is the difference between increasing at an increasing rate and at a decreasing rate?
Both are rising, so the rate is positive. Increasing at an increasing rate gets steeper (rate growing); increasing at a decreasing rate flattens out (rate shrinking).