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0580/43 Oct/Nov 2021, Question 9: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 43 (Extended) · Calculator · 14 marks

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The question, in short

For f(x) = x(x − 1)(x − 2), find axis crossings, expand the expression, locate turning points to 1 decimal place, and sketch the curve.

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

Part (a) · 2 marks

Find the coordinates where y = f(x) meets the x-axis.

  1. 1f(x) = 0 when x = 0, x − 1 = 0 or x − 2 = 0
  2. 2(0, 0), (1, 0), (2, 0)

Answer(0, 0), (1, 0), (2, 0)

Part (b) · 2 marks

Show that expanding f(x) gives x³ − 3x² + 2x.

  1. 1x(x − 1)(x − 2) = x(x² − x − 2x + 2)
  2. 2= x(x² − 3x + 2) = x³ − 3x² + 2x

Answerx(x² − x − 2x + 2) or (x² − x)(x − 2) or (x − 1)(x² − 2x) leading to x³ − 3x² + 2x with no errors or omissions

Part (c) · 8 marks

Find turning point coordinates, showing working, rounded to 1 decimal place.

  1. 1dy/dx = 3x² − 6x + 2
  2. 23x² − 6x + 2 = 0: x = (6 ± √((−6)² − 4 × 3 × 2))/(2 × 3) = (6 ± √12)/6
  3. 3x = 0.422… or x = 1.577…
  4. 4y = f(0.422…) = 0.384…, y = f(1.577…) = −0.384…
  5. 5Turning points (0.4, 0.4) and (1.6, −0.4)

Answer(0.4, 0.4) and (1.6, −0.4)

Part (d) · 2 marks

Sketch the graph of y = f(x) on the given axes.

  1. 1Draw a cubic rising from bottom left through (0, 0)
  2. 2Maximum at about (0.4, 0.4) above the x-axis, down through (1, 0)
  3. 3Minimum at about (1.6, −0.4) below the x-axis, up through (2, 0) and on to top right

AnswerCorrect sketch: cubic through the origin with maximum and minimum in the correct quadrants

Stuck on a step? Read the free lesson: Stationary Points and Turning Points

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Also searched as: 0580/43/O/N/21 · 0580_w21_qp_43 · 0580 w21 43 · question 9.
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