Average Gradient on a Curve — Grade 11 Maths (CAPS)

Why a curve has no single gradient, how to find the average gradient between two points, what happens as the two points slide together — and how that idea becomes the gradient at a single point, which is where calculus begins.

Also known as: average gradient · gradient of a curve · secant · rate of change grade 11

A straight line has one gradient, everywhere along it. A curve does not — it is steeper in some places than others. So the question “what is the gradient of y = x²?” has no single answer, and we ask a narrower question instead: what is the gradient between two particular points?

1The average gradient is just the gradient of the chord

Join the two points with a straight line — a chord, or secant — and find its gradient in the ordinary way.

average gradient = (y₂ − y₁)/(x₂ − x₁) = (f(b) − f(a))/(b − a)

Between A(1 ; 1) and B(3 ; 9) on y = x², the rise is 8 and the run is 2, so the average gradient is 4.

Worked ExampleDetermine the average gradient of f(x) = x² between x = 1 and x = 3.

1
f(1) = 1² = 1 and f(3) = 3² = 9
Find both y-values first.
2
average gradient = (9 − 1)/(3 − 1)
Change in y over change in x.
3
= 8/2 = 4

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It is exactly the gradient formula from analytical geometry. The only new idea is which two points: they must both lie on the curve, so you work their y-values out from the equation rather than reading them off.

2The average gradient changes depending on where you look

This is the point of the whole topic. Take the same curve and move the interval:

f(x) = x² between x = 1 and x = 3 → (9 − 1) ÷ 2 = 4
between x = 0 and x = 2 → (4 − 0) ÷ 2 = 2
between x = 3 and x = 5 → (25 − 9) ÷ 2 = 8

Three different answers on the same curve. The parabola gets steeper as x grows, and the average gradient measures that.

⚠️

Assuming a negative average gradient means the curve is always falling. It means the curve is lower at the end than at the start — the curve may well have risen and fallen in between. The average gradient only sees the two endpoints.

3Sliding the points together

Now the idea that leads to calculus. Keep one point fixed at x = 1 and bring the second one closer:

from x = 1 to x = 3 → gradient 4
from x = 1 to x = 2 → (4 − 1) ÷ 1 = 3
from x = 1 to x = 1,5 → (2,25 − 1) ÷ 0,5 = 2,5
from x = 1 to x = 1,1 → (1,21 − 1) ÷ 0,1 = 2,1
from x = 1 to x = 1,01 → (1,0201 − 1) ÷ 0,01 = 2,01

The answers are closing in on 2. That number is the gradient of the curve at the single point x = 1 — the gradient of the tangent there. You cannot reach it by putting both points at x = 1, because that gives 0 ÷ 0. But you can get as close as you like.

as the two points slide together, the CHORD becomes the TANGENT

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This is the whole idea of the derivative, a year early. In Grade 12 you will write it as a limit and get a formula for it; here you only need to see numerically that the average gradients settle on a value.

4In context: average rate of change

Outside of graphs, the same calculation is called an average rate of change.

Worked ExampleA ball's height after t seconds is h(t) = 20t − 5t². Determine the average rate of change of its height (its average velocity) between t = 1 and t = 3 seconds.

1
h(1) = 20(1) − 5(1)² = 15 m
2
h(3) = 20(3) − 5(3)² = 60 − 45 = 15 m
3
average rate = (15 − 15)/(3 − 1) = 0 m/s
4
The ball is at the same height at both times — it went up and came back down. Its average velocity is zero even though it travelled 10 m in those 2 seconds (5 m up and 5 m back down).
A good illustration of what the average gradient does and does not tell you.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Determine the average gradient of f(x) = x² between x = 2 and x = 5.
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f(2) = 4, f(5) = 25 ✓; (25 − 4) ÷ (5 − 2) = 21 ÷ 3 = 7 ✓
2
Determine the average gradient of f(x) = x² + 1 between x = −1 and x = 2.
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f(−1) = 2, f(2) = 5 ✓; (5 − 2) ÷ (2 − (−1)) = 3 ÷ 3 = 1 ✓
3
Write down the formula for the average gradient of f between x = a and x = b.
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(f(b) − f(a))/(b − a) ✓✓
4
For f(x) = x², calculate the average gradient between x = 2 and x = 2,1 (to two decimal places).
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f(2) = 4; f(2,1) = 4,41 ✓; (4,41 − 4) ÷ 0,1 ✓ = 4,10 ✓
5
Using the answer above, estimate the gradient of y = x² AT the point where x = 2.
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Approximately 4 ✓✓ (the average gradients close in on 4 as the interval shrinks)
6
Explain why the gradient of a curve cannot be given as a single number, but the gradient of a straight line can.
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A straight line has the same steepness everywhere ✓; a curve changes steepness from point to point ✓, so a gradient can only be quoted for a stated interval or at a stated point ✓
7
The average gradient of f between x = 0 and x = 4 is 0. Does this mean the graph is a horizontal line? Explain.
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No ✓. It means f(4) = f(0) ✓ — the graph returns to the same height, but it may have risen and fallen in between ✓
8
A ball's height is h(t) = 30t − 5t² metres. Determine its average velocity between t = 1 and t = 4 seconds.
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h(1) = 25, h(4) = 120 − 80 = 40 ✓; (40 − 25) ÷ 3 ✓ = 5 m/s ✓
9
What does the chord between two points on a curve become as the two points slide together?
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The tangent to the curve at that point ✓✓
10
For f(x) = 2x², determine the average gradient between x = 1 and x = 1 + h, and simplify your answer.
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f(1) = 2; f(1 + h) = 2(1 + h)² = 2 + 4h + 2h² ✓; average gradient = (4h + 2h²)/h ✓ = 4 + 2h, h ≠ 0 ✓