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0580/42 May/Jun 2024, Question 11: Gradient of a Curve at a Point

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 7 marks

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The question, in short

Given a point on a curve and its derivative, verify the point is stationary; then for y = 2x³ − 6x + 8 find a tangent gradient and stationary point x-values.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 2 marks

Substitute x = −1 into the given derivative and show the result equals zero at the point (−1, 6).

  1. 1dy/dx at x = −1: −4(−1)³ − 9(−1)² + 5 = 4 − 9 + 5 = 0
  2. 2Gradient is 0, so (−1, 6) is a stationary point

Answer−4(−1)³ − 9(−1)² + 5 = 0, so stationary point

Part (bi) · 3 marks

Differentiate 2x³ − 6x + 8 and evaluate the derivative at x = −2 to get the tangent gradient.

  1. 1The gradient of the tangent is the value of dy/dx at that point
  2. 22x³ → 6x², −6x → −6, 8 → 0
  3. 3dy/dx = 6x² − 6
  4. 4At x = −2: 6 × (−2)² − 6 = 24 − 6 = 18

Answer18

Part (bii) · 2 marks

Set the derivative of 2x³ − 6x + 8 to zero and solve for the x-values of stationary points.

  1. 16x² − 6 = 0
  2. 2x² = 1
  3. 3x = 1 and x = −1

Answer1 and −1

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