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4MA1/1HR Jan 2019, Question 13: Stationary Points and Turning Points

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 1HR (Higher) · Calculator · 7 marks

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The question, in short

For the cubic curve y = x³ − x² − 8x + 12, differentiate it, locate both stationary x values, and prove the x-axis touches the curve.

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

Part (a) · 2 marks

Differentiate the given cubic expression to obtain the gradient function.

  1. 1Rule: multiply by the power, then lower the power by 1.
  2. 2x³ → 3x², x² → 2x, 8x → 8, 12 → 0
  3. 3dy/dx = 3x² − 2x − 8

Answer3x² − 2x − 8

Part (b) · 3 marks

Solve the gradient equal to zero to get the x values of both turning points, showing steps.

  1. 1Differentiate: dy/dx = 3x² − 2x − 8
  2. 2At turning points dy/dx = 0: 3x² − 2x − 8 = 0
  3. 3Factorise: (3x + 4)(x − 2) = 0
  4. 4x = 2, −4/3

Answer2, −4/3

Part (c) · 2 marks

Demonstrate that the x-axis meets the curve tangentially.

  1. 1Substitute x = 2 into y = x³ − x² − 8x + 12
  2. 2y = 8 − 4 - 16 + 12 = 0
  3. 3So (2, 0) is a point on the curve
  4. 4From (b), x = 2 is a turning point
  5. 5So the curve touches the x-axis at a turning point
  6. 6Therefore the x-axis is a tangent to the curve

AnswerAt x = 2, y = 8 − 4 − 16 + 12 = 0, and x = 2 is a turning point, so the curve touches the x-axis at (2, 0)

Stuck on a step? Read the free lesson: Stationary Points and Turning Points

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