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4PM1/02 Jun 2025, Question 10: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 17 marks

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The question, in short

For the curve y = (7 − x)/(2x + 3), find asymptotes, axis crossings, sketch it, show the gradient at A(1/2, 13/8), and find the tangent at P parallel to that at A.

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

Part (a) · 2 marks

State the equations of the two asymptotes: the one parallel to the y-axis and the one parallel to the x-axis.

  1. 12x + 3 = 0 gives x = −3/2
  2. 2As x → ±∞, y → −x/(2x) = −1/2

Answer(i) x = −3/2 (ii) y = −1/2

Part (b) · 2 marks

Find the coordinates where the curve meets the x-axis and the y-axis.

  1. 1y = 0: 7 − x = 0, so (7, 0)
  2. 2x = 0: y = 7/3, so (0, 7/3)

Answer(7, 0) and (0, 7/3)

Part (c) · 3 marks

Sketch the curve, clearly marking both asymptotes and the coordinates of the axis crossing points.

  1. 1Asymptotes x = −3/2 and y = −1/2
  2. 2Right branch: from +∞ near x = −3/2 through (0, 7/3) and (7, 0), then towards y = −1/2 from above
  3. 3Left branch: below y = −1/2, tending to −∞ near x = −3/2

AnswerTwo-branch curve with asymptotes x = −3/2 and y = −1/2; the right branch passes through (0, 7/3) and (7, 0) and approaches y = −1/2 from above; the left branch lies below y = −1/2

Part (d) · 3 marks

Show that the gradient of the curve at the point A(1/2, 13/8) equals −17/16.

  1. 1dy/dx = [−1(2x + 3) − 2(7 − x)]/(2x + 3)² = −17/(2x + 3)²
  2. 2At x = 1/2: −17/(4)² = −17/16

AnswerGradient at A = −17/16 shown

Part (e) · 7 marks

Given the tangent at P is parallel to the tangent at A, find the tangent equation at P in the form ax + by + c = 0 with integer coefficients.

  1. 1−17/(2x + 3)² = −17/16, so (2x + 3)² = 16
  2. 22x + 3 = ±4, x = 1/2 or x = −7/2
  3. 3At P, x = −7/2: y = (7 + 7/2)/(−7 + 3) = −21/8
  4. 4y + 21/8 = −(17/16)(x + 7/2)
  5. 534x + 32y + 203 = 0

Answer34x + 32y + 203 = 0

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