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0580/21 May/Jun 2025, Question 22: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 21 (Extended) · Non-calculator · 6 marks

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The question, in short

A curve is given by y = xⁿ + qx² + 9x, with gradient 3x² − 12x + 9. Determine n and q, then locate the curve's turning points.

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Part (a) · 2 marks

Use the given gradient expression to work out the values of the constants n and q.

  1. 1Derivative of xⁿ is nxⁿ⁻¹ = 3x², so n = 3
  2. 2Derivative of qx² is 2qx = −12x, so q = −6

Answern = 3, q = −6

Part (b) · 4 marks

Find the x- and y-coordinates of the points where the curve's gradient is zero.

  1. 1From dy/dx = 3x² − 12x + 9: n = 3 and q = −6, so y = x³ − 6x² + 9x
  2. 2Turning points: 3x² − 12x + 9 = 0, so 3(x − 1)(x − 3) = 0
  3. 3x = 1: y = 1 − 6 + 9 = 4
  4. 4x = 3: y = 27 − 54 + 27 = 0
  5. 5Turning points (1, 4) and (3, 0)

Answer(1, 4) and (3, 0)

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Also searched as: 0580/21/M/J/25 · 0580_s25_qp_21 · 0580 s25 21 · question 22.
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