Average Gradient on a Curve (Grade 11) — Worksheet & Memo

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Grade 11 · Functions and Graphs

Average Gradient on a Curve (Grade 11)

MEMORANDUM

Name:

Class:

Date:

Mark

/ 25

Answer all questions. Show all your working — marks are awarded for method as well as the final answer. Teacher copy: accept any correct equivalent method.

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

TOTAL: 25 marks