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0580/32 May/Jun 2026, Question 18: Quadratic Graphs and the Parabola

Cambridge IGCSE Mathematics 0580 · Paper 32 (Core) · Calculator · 10 marks

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The question, in short

For y = −x² + 3x + 5, fill in missing table values, plot the curve for −2 ≤ x ≤ 5, find the highest point, and use the graph to solve the equation equal to zero.

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

Part (a) · 3 marks

Complete the y-values in the table for y = −x² + 3x + 5 at the given x values from −2 to 5.

  1. 1x = −1: −1 − 3 + 5 = 1
  2. 2x = 0: 5
  3. 3x = 1: −1 + 3 + 5 = 7
  4. 4x = 2: −4 + 6 + 5 = 7
  5. 5x = 4: −16 + 12 + 5 = 1

Answer1, 5, 7, 7, 1

Part (b) · 4 marks

Draw the graph of y = −x² + 3x + 5 on the grid for x from −2 to 5.

  1. 1Plot the 8 points from the table
  2. 2Join with a smooth curve, not ruled
  3. 3The curve peaks between x = 1 and x = 2, above y = 7 and below 7.5

AnswerCorrect smooth parabola through (−2, −5), (−1, 1), (0, 5), (1, 7), (2, 7), (3, 5), (4, 1), (5, −5)

Part (c) · 1 mark

Sketch the curve for −2 ≤ x ≤ 5 and state the coordinates of its highest point.

  1. 1By symmetry the highest point is at x = (1 + 2)/2 = 1.5
  2. 2y = −2.25 + 4.5 + 5 = 7.25
  3. 3Highest point (1.5, 7.25)

Answer(1.5, 7.1 to 7.5)

Part (d) · 2 marks

Using your graph, find the solutions of −x² + 3x + 5 = 0.

  1. 1Read where the curve crosses the x-axis (y = 0)
  2. 2x ≈ −1.2
  3. 3x ≈ 4.2

Answerx = −1.3 to −1.1 or x = 4.1 to 4.3

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Also searched as: 0580/32/M/J/26 · 0580_s26_qp_32 · 0580 s26 32 · question 18.
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