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WMA11/01 Jun 2025, Question 7: Cubic Equations and Curve Meetings

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 1 (WMA11) · Calculator · 10 marks

The question, in short

Given a cubic curve touching the x-axis at the origin and cutting it at x = 4, passing through (10, 120), find its equation, sketch a related quadratic, and find intersections.

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

Part (a) · 3 marks

Find the cubic function f(x) using the given x-axis behaviour and the point (10, 120).

  1. 1Touches at O and cuts at x = 4, so f(x) = ax²(x − 4)
  2. 2Through (10, 120): 120 = a × 100 × 6 = 600a
  3. 3a = 0.2, so f(x) = 0.2x²(x − 4)

Answerf(x) = 0.2x²(x − 4) (= 0.2x³ − 0.8x²)

Part (b) · 2 marks

Sketch the curve y = 1.2x(8 − x) on a copy of the given graph.

  1. 1y = 1.2x(8 − x) has a negative x² coefficient, so it is ∩-shaped
  2. 2It crosses the x-axis at x = 0 and x = 8 (to the right of x = 4)
  3. 3Vertex at x = 4: y = 1.2 × 4 × 4 = 19.2; the curve continues below the x-axis on both sides

AnswerAn upside-down (∩-shaped) parabola through the origin O and through (8, 0) on the positive x-axis to the right of x = 4, with maximum at (4, 19.2), extending below the x-axis into quadrants 3 and 4

Part (c) · 5 marks

Use algebra to find the coordinates where the cubic and quadratic curves intersect, showing all working.

  1. 11.2x(8 − x) = 0.2x²(x − 4)
  2. 2Multiply by 5: 6x(8 − x) = x²(x − 4)
  3. 348x − 6x² = x³ − 4x², so x³ + 2x² − 48x = 0
  4. 4x(x² + 2x − 48) = 0, so x(x + 8)(x − 6) = 0
  5. 5x = 0, 6, −8
  6. 6y = 1.2x(8 − x): (0, 0), (6, 14.4), (−8, −153.6)

Answer(0, 0), (6, 14.4) and (−8, −153.6)

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