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0580/42 Feb/Mar 2024, Question 8: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 12 marks

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The question, in short

Sketch two straight-line and quadratic graphs, then locate and classify the stationary points of a cubic curve.

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

Part (a) · 2 marks

Draw the line y = 4 − 3x on the given axes.

  1. 1Straight line crossing the y-axis at 4 (positive intercept)
  2. 2Negative gradient −3: it slopes down, crossing the x-axis at 4/3
  3. 3Draw it with a ruler

AnswerRuled line with negative gradient and positive y-intercept

Part (b) · 2 marks

Draw the curve y = −x² on the given axes.

  1. 1Upside-down parabola (negative x²)
  2. 2Its maximum (vertex) is at the origin, so the curve lies below the x-axis

AnswerNegative quadratic, with vertex at origin

Part (ci) · 5 marks

Work out the coordinates of the stationary points of y = 10 + 9x² − 2x³, showing every step.

  1. 1dy/dx = 18x − 6x²
  2. 218x − 6x² = 0
  3. 36x(3 − x) = 0, so x = 0 or x = 3
  4. 4x = 0: y = 10; x = 3: y = 10 + 81 − 54 = 37
  5. 5(0, 10) and (3, 37)

Answer(0, 10) and (3, 37)

Part (cii) · 3 marks

Using the points (0, 10) and (3, 37), decide which is a maximum and which a minimum, justifying your choice.

  1. 1d²y/dx² = 18 − 12x
  2. 2At x = 0: 18 > 0, so (0, 10) is a minimum
  3. 3At x = 3: 18 − 36 = −18 < 0, so (3, 37) is a maximum

Answer(0, 10) minimum with correct reason AND (3, 37) maximum with correct reason

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Also searched as: 0580/42/F/M/24 · 0580_m24_qp_42 · 0580 m24 42 · question 8.
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