0580/41 Oct/Nov 2024, Question 8: Equation of the Tangent Line

Part (ai) · 2 marks

Find the values of x for which g(x) = 20.

1. 1x² − 16 = 20
2. 2x² = 36
3. 3x = 6 and x = −6

Answer6 and −6

Part (aii) · 2 marks

Find the inverse function f⁻¹(x).

1. 1y = 7 − 3x, swap: x = 7 − 3y
2. 23y = 7 − x
3. 3f⁻¹(x) = (7 − x)/3

Answer(7 − x)/3

Part (aiii) · 3 marks

Find gf(x) + 1 in its simplest form.

1. 1gf(x) + 1 = (7 − 3x)² − 16 + 1
2. 2= 49 − 21x − 21x + 9x² − 15
3. 3= 9x² − 42x + 34

Answer9x² − 42x + 34

Part (aiv) · 4 marks

Sketch y = g(x) on the axes, marking where it crosses the axes.

1. 1g(x) = (x − 4)(x + 4), so the graph crosses the x-axis at −4 and 4
2. 2When x = 0, y = −16: the y-intercept and lowest point (turning point) is (0, −16)
3. 3Sketch a U-shaped parabola through (−4, 0), (0, −16) and (4, 0), marking these values

AnswerCorrect sketch with roots marked at −4 and 4 and y-intercept and turning point at y = −16

Part (av) · 5 marks

Find the tangent to y = g(x) at x = −3, in the form y = mx + c.

1. 1Gradient = g′(x) = 2x
2. 2At x = −3: gradient = 2 × (−3) = −6
3. 3y = (−3)² − 16 = −7, so the point is (−3, −7)
4. 4−7 = −6 × (−3) + c, so c = −25
5. 5y = −6x − 25

Answery = −6x − 25

Part (bi) · 2 marks

Sketch the graph of y = h(x) on the axes.

1. 1h(0) = 3⁰ = 1, so the curve crosses the y-axis at (0, 1)
2. 2Sketch an increasing curve that gets closer and closer to the x-axis for negative x and rises steeply for positive x

AnswerCorrect sketch with y-intercept above x-axis

Part (bii) · 1 mark

State the equation of the asymptote to y = h(x).

1. 1As x → −∞, 3ˣ → 0, so the asymptote is y = 0

Answery = 0

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