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0580/11 May/Jun 2026, Question 18: Quadratic Graphs and the Parabola

Cambridge IGCSE Mathematics 0580 · Paper 11 (Core) · Non-calculator · 9 marks

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The question, in short

For y = x² + x − 3, fill missing table values, plot the curve for −4 ≤ x ≤ 4, state its line of symmetry, and use the graph to solve x² + x − 3 = 0.

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

Part (a) · 2 marks

Work out the missing y-values in the table for y = x² + x − 3 at the blank x entries.

  1. 1x = −3: 9 − 3 − 3 = 3
  2. 2x = 2: 4 + 2 − 3 = 3
  3. 3x = 4: 16 + 4 − 3 = 17

Answer3, 3, 17

Part (b) · 4 marks

Plot the points and draw the curve y = x² + x − 3 on the grid for x from −4 to 4.

  1. 1Plot the 9 points from the table
  2. 2Join with a smooth U-shaped curve
  3. 3The curve dips just below y = −3 between x = −1 and x = 0

AnswerCorrect smooth curve through (−4, 9), (−3, 3), (−2, −1), (−1, −3), (0, −3), (1, −1), (2, 3), (3, 9), (4, 17)

Part (c) · 1 mark

Give the equation of the vertical line of symmetry of the parabola.

  1. 1y = −3 at x = −1 and at x = 0
  2. 2The line of symmetry is halfway: x = −1/2

Answerx = −1/2

Part (d) · 2 marks

From your graph, find the x-values where x² + x − 3 equals 0.

  1. 1Read where the curve crosses the x-axis (y = 0)
  2. 2x ≈ −2.3 or x ≈ 1.3

Answerx = −2.4 to −2.2 and x = 1.2 to 1.4

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Also searched as: 0580/11/M/J/26 · 0580_s26_qp_11 · 0580 s26 11 · question 18.
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