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0580/42 Oct/Nov 2021, Question 5: 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 − 3x² + 3, plot the curve for −1 ≤ x ≤ 3, then use a straight line to solve x³ − 3x² + 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 three missing y-values in the table for y = x³ − 3x² + 3 at the given x-values.

  1. 1x = −1: y = −1 − 3 + 3 = −1
  2. 2x = 1.5: y = 3.375 − 6.75 + 3 = −0.375
  3. 3x = 3: y = 27 − 27 + 3 = 3

Answer−1, −0.375, 3

Part (b) · 4 marks

Plot the points from the table and draw the curve y = x³ − 3x² + 3 on the grid for −1 ≤ x ≤ 3.

  1. 1Plot (−1, −1), (−0.5, 2.125), (0, 3), (0.5, 2.375), (1, 1), (1.5, −0.375), (2, −1), (2.5, −0.125), (3, 3)
  2. 2Join with a smooth curve: maximum at (0, 3), minimum at (2, −1)

AnswerCorrect graph

Part (c) · 4 marks

Add a suitable straight line to the grid and use its intersections with the curve to solve x³ − 3x² + x + 1 = 0.

  1. 1x³ − 3x² + x + 1 = 0 rearranges to x³ − 3x² + 3 = 2 − x
  2. 2Rule the line y = 2 − x, through (0, 2) and (2, 0)
  3. 3Read the x-coordinates where the line meets the curve
  4. 4x ≈ −0.4, x = 1, x ≈ 2.4

Answer−0.45 to −0.35, 0.95 to 1.05, 2.35 to 2.45

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Also searched as: 0580/42/O/N/21 · 0580_w21_qp_42 · 0580 w21 42 · question 5.
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