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4024/12 May/Jun 2025, Question 22: Quadratic Graphs and the Parabola

Cambridge O Level Mathematics 4024 · Paper 12 (Non-calculator) · Non-calculator · 7 marks

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The question, in short

Complete the square for x² + 4x − 12, use the result to locate the turning point, and sketch the curve showing axis intercepts.

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

Part (a) · 2 marks

Rewrite x² + 4x − 12 in completed-square form (x + a)² + b, stating values of a and b.

  1. 1Halve the x coefficient: 4 ÷ 2 = 2, so use (x + 2)²
  2. 2(x + 2)² = x² + 4x + 4, which is 4 too many
  3. 3x² + 4x − 12 = (x + 2)² − 4 − 12 = (x + 2)² − 16

Answer(x + 2)² − 16

Part (b) · 1 mark

Using the completed-square form, give the coordinates of the turning point of y = x² + 4x − 12.

  1. 1Minimum when x + 2 = 0: x = −2, y = −16
  2. 2Turning point (−2, −16)

Answer(−2, −16)

Part (c) · 4 marks

Sketch y = x² + 4x − 12 on axes, marking where it crosses the x-axis and y-axis.

  1. 1x-intercepts: (x + 6)(x − 2) = 0, so x = −6 and x = 2
  2. 2y-intercept: x = 0 gives y = −12
  3. 3Draw a U-shaped curve through −6 and 2 on the x-axis and −12 on the y-axis, minimum at (−2, −16) below the x-axis, left of the y-axis

AnswerCorrect sketch with minimum in correct quadrant and values −6, 2, −12 labelled

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Also searched as: 4024/12/M/J/25 · 4024_s25_qp_12 · 4024 s25 12 · question 22.
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