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0580/42 Oct/Nov 2024, Question 2: Plotting Curves from a Table of Values

Cambridge IGCSE Mathematics 0580 · Paper 42 (Extended) · Calculator · 13 marks

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The question, in short

Using a table of values for y = x³ + 4x² − 4, fill in missing entries, plot the curve from x = −4.5 to 1.5, find the gradient at (1, 1) from a tangent, and solve x³ + 4x² − x − 6 = 0 graphically.

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

Part (a) · 2 marks

Work out the two missing y-values in the table for y = x³ + 4x² − 4.

  1. 1x = −4: y = (−4)³ + 4(−4)² − 4 = −64 + 64 − 4 = −4
  2. 2x = −1: y = (−1)³ + 4(−1)² − 4 = −1 + 4 − 4 = −1

Answer| x | −4 | −1 | | y | −4 | −1 |

Part (b) · 4 marks

Plot the curve y = x³ + 4x² − 4 on the grid for x values from −4.5 to 1.5.

  1. 1Plot (−4.5, −14.1), (−4, −4), (−3, 5), (−2, 4), (−1, −1), (0, −4), (1, 1), (1.5, 8.4)
  2. 2Join the points with a smooth curve (maximum near x = −2.7, minimum at x = 0)

AnswerCorrect graph

Part (ci) · 1 mark

Add a tangent line touching the curve at the point (1, 1).

  1. 1Place a ruler touching the curve only at (1, 1), with the same slope as the curve there, and draw a ruled line

AnswerRuled tangent at x = 1

Part (cii) · 2 marks

From your tangent, work out an approximate value for the curve's gradient at (1, 1).

  1. 1Gradient = rise ÷ run between two points on your tangent
  2. 2The true gradient is 3x² + 8x at x = 1: 3 + 8 = 11
  3. 3Gradient ≈ 11

Answer6 to 14

Part (d) · 4 marks

Draw a sketch of the curve, then add a straight line on the grid and use it to solve x³ + 4x² − x − 6 = 0.

  1. 1x³ + 4x² − x − 6 = 0 rearranges to x³ + 4x² − 4 = x + 2
  2. 2Draw the line y = x + 2 (through (0, 2) and (−4, −2))
  3. 3Read the x-values where the line meets the curve
  4. 4x ≈ −3.86, x ≈ −1.32, x ≈ 1.18

Answer−3.95 to −3.75, −1.4 to −1.25, 1.1 to 1.25

Stuck on a step? Read the free lesson: Plotting Curves from a Table of Values

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Also searched as: 0580/42/O/N/24 · 0580_w24_qp_42 · 0580 w24 42 · question 2.
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