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0580/42 May/Jun 2026, Question 8: Plotting Curves from a Table of Values

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 11 marks

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The question, in short

Complete a table of values for y = x³ − 2x² − 2, plot the curve for −1 ≤ x ≤ 3, then use a straight line to solve x³ − 2x² − x + 1 = 0.

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

Part (a) · 3 marks

Fill in the missing y-values in the table for y = x³ − 2x² − 2 at the given x values.

  1. 1x = −1: (−1)³ − 2(1) − 2 = −5
  2. 2x = 1: 1 − 2 − 2 = −3
  3. 3x = 2.5: 15.625 − 12.5 − 2 = 1.125 ≈ 1.1

Answer−5, −3, 1.1

Part (b) · 4 marks

Draw the graph of y = x³ − 2x² − 2 on the grid for x from −1 to 3.

  1. 1Plot the 9 points from the completed table
  2. 2(−1, −5), (0, −2), (1, −3), (2, −2), (3, 7) and the half-values
  3. 3Join with a smooth cubic curve (no ruled segments)

AnswerCorrect smooth curve through (−1, −5), (−0.5, −2.6), (0, −2), (0.5, −2.4), (1, −3), (1.5, −3.1), (2, −2), (2.5, 1.1), (3, 7)

Part (c) · 4 marks

Sketch the curve, draw a suitable straight line, and use it to solve x³ − 2x² − x + 1 = 0.

  1. 1x³ − 2x² − 2 = x − 3 is the same as x³ − 2x² − x + 1 = 0
  2. 2Draw the line y = x − 3 through (0, −3) and (2, −1)
  3. 3Read x at the three crossings with the curve
  4. 4x = −0.8, x = 0.55, x = 2.25

Answer−0.9 to −0.7, 0.5 to 0.6, 2.2 to 2.35

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Also searched as: 0580/42/M/J/26 · 0580_s26_qp_42 · 0580 s26 42 · question 8.
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