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WMA11/01 Oct 2021, Question 6: Equation of the Tangent Line

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

The question, in short

For the curve y = 2(x + 1)(x − 3)², draw its graph with axis intercepts marked, expand the cubic, then use that expansion to find the tangent line at x = 1/3.

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

Part (a) · 3 marks

Draw the curve and label where it crosses or touches the x-axis and y-axis, giving coordinates.

  1. 1Roots: x = −1 (single, crosses) and x = 3 (repeated, touches)
  2. 2x = 0: y = 2(1)(9) = 18
  3. 3Leading term 2x³ is positive, so the curve rises from bottom left to top right
  4. 4Sketch: crosses at (−1, 0), maximum, minimum touching at (3, 0), y-intercept (0, 18)

AnswerPositive cubic: cuts the x-axis at (−1, 0), touches the x-axis at (3, 0) (a minimum there), crosses the y-axis at (0, 18), with a maximum between x = −1 and x = 3

Part (b) · 3 marks

Expand the product into the cubic form ax³ + bx² + cx + d and state the constants.

  1. 1(x − 3)² = x² − 6x + 9
  2. 2(2x + 2)(x² − 6x + 9) = 2x³ − 12x² + 18x + 2x² − 12x + 18
  3. 3= 2x³ − 10x² + 6x + 18

Answerf(x) = 2x³ − 10x² + 6x + 18 (a = 2, b = −10, c = 6, d = 18)

Part (c) · 4 marks

Using your expanded form, determine the tangent line's equation at x = 1/3.

  1. 1f′(x) = 6x² − 20x + 6
  2. 2f′(1/3) = 6/9 − 20/3 + 6 = 0
  3. 3f(1/3) = 2(4/3)(−8/3)² = 2 × 4/3 × 64/9 = 512/27
  4. 4Gradient 0, so the tangent is y = 512/27

Answery = 512/27

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Also searched as: WMA11 October 2021 · WMA11/01 · wma11 oct21 · IAS Maths Pure Mathematics 1 October 2021 · question 6.
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