0580/42 Oct/Nov 2025, Question 11: Plotting Curves from a Table of Values
Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 10 marks
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The question, in short
Table of values for y = x³ − 2x + 3; complete missing y values, plot the curve for −2 ≤ x ≤ 2, then use a straight line to solve x³ − 2.5x + 1 = 0.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Part (a) · 2 marks
Find the missing y values in the table for x = −1.5 and x = 1.5 using y = x³ − 2x + 3.
- 1x = −1.5: y = (−1.5)³ − 2(−1.5) + 3 = −3.375 + 3 + 3 = 2.625 ≈ 2.63
- 2x = 1.5: y = (1.5)³ − 2(1.5) + 3 = 3.375 − 3 + 3 = 3.375 ≈ 3.38
Answer2.63 and 3.38
Part (b) · 4 marks
Plot the graph of y = x³ − 2x + 3 on the grid for x from −2 to 2.
- 1Plot the 9 points: (−2, −1), (−1.5, 2.63), (−1, 4), (−0.5, 3.88), (0, 3), (0.5, 2.13), (1, 2), (1.5, 3.38), (2, 7)
- 2Join them with a smooth curve: it rises to a maximum near (−0.8, 4.1), falls to a minimum near (0.8, 1.9), then rises steeply to (2, 7)
AnswerCorrect graph
Part (c) · 4 marks
Sketch the curve, draw a suitable straight line on the grid, and use it to solve x³ − 2.5x + 1 = 0.
- 1x³ − 2.5x + 1 = 0 → x³ − 2x + 3 = 0.5x + 2
- 2Draw the ruled line y = 0.5x + 2 (through (−2, 1) and (2, 3))
- 3Read the x-values where the line crosses the curve
- 4x ≈ −1.75, x ≈ 0.43, x ≈ 1.32
Answer−1.85 to −1.7, 0.4 to 0.5, 1.25 to 1.4
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Also searched as: 0580/42/O/N/25 · 0580_w25_qp_42 · 0580 w25 42 · question 11.
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