4PM1/02R Jun 2023, Question 9: Tangents, Normals and Stationary Points (Further)

Part (a) · 2 marks

State the equations of the horizontal and vertical asymptotes of the curve.

1. 1As x → ±∞, y → −2x/x = −2
2. 2Vertical asymptote where x + 6 = 0: x = −6

Answer(i) y = −2 (ii) x = −6

Part (b) · 2 marks

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

1. 1y = 0: 3 − 2x = 0, x = 3/2
2. 2x = 0: y = 3/6 = 1/2

Answer(i) (3/2, 0) (ii) (0, 1/2)

Part (c) · 3 marks

Sketch the curve, clearly showing both asymptotes and the axis crossing points.

1. 1Draw dashed asymptotes x = −6 and y = −2
2. 2Right branch: above y = −2, right of x = −6, through (0, 1/2) and (3/2, 0)
3. 3Left branch: below y = −2, left of x = −6

AnswerTwo branches with asymptotes x = −6 and y = −2: the right-hand branch comes down from near x = −6, passes through (0, 1/2) and (3/2, 0) and approaches y = −2; the left-hand branch lies below y = −2 to the left of x = −6

Part (d) · 3 marks

Prove that the gradient of the tangent to the curve is always negative.

1. 1dy/dx = [(x + 6)(−2) − (3 − 2x)(1)]/(x + 6)²
2. 2= (−2x − 12 − 3 + 2x)/(x + 6)² = −15/(x + 6)²
3. 3(x + 6)² > 0 and −15 < 0, so the gradient is always negative

Answerdy/dx = −15/(x + 6)², which is always negative — shown

Part (e) · 5 marks

Given the tangent y = −(3/5)x + k with k > 0, determine k.

1. 1−15/(x + 6)² = −3/5, so (x + 6)² = 25
2. 2x + 6 = ±5, so x = −1 or x = −11
3. 3At x = −1: y = (3 + 2)/5 = 1
4. 4Tangent: y − 1 = −(3/5)(x + 1), so y = −(3/5)x + 2/5
5. 5k = 2/5 (x = −11 gives y = −5 and k = −58/5 < 0, rejected)

Answerk = 2/5

More past-paper questions on tangents, normals and stationary points (further)

• 4PM1/01R Jun 2025 Q9 · 9 marks
• 4PM1/02 Jun 2025 Q6 · 10 marks
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• 4PM1/01 Jun 2024 Q7 · 13 marks