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Date	November 2018	Marks available	4	Reference code	18N.2.SL.TZ0.T_4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Sketch	Question number	T_4	Adapted from	N/A

Question

Consider the function $f\left( x \right) = \frac{{27}}{{{x^2}}} - 16x,\,\,\,x \ne 0$ .

Sketch the graph of y = f (x), for −4 ≤ x ≤ 3 and −50 ≤ y ≤ 100.

[4]

Use your graphic display calculator to find the zero of f (x).

[1]

b.i.

Use your graphic display calculator to find the coordinates of the local minimum point.

[2]

b.ii.

Use your graphic display calculator to find the equation of the tangent to the graph of y = f (x) at the point (–2, 38.75).

Give your answer in the form y = mx + c.

[2]

b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(A1)(A1)(A1)(A1)

Note: Award (A1) for axis labels and some indication of scale; accept y or f(x).

Use of graph paper is not required. If no scale is given, assume the given window for zero and minimum point.

Award (A1) for smooth curve with correct general shape.

Award (A1) for x-intercept closer to y-axis than to end of sketch.

Award (A1) for correct local minimum with x-coordinate closer to y-axis than end of sketch and y-coordinate less than half way to top of sketch.

Award at most (A1)(A0)(A1)(A1) if the sketch intersects the y-axis or if the sketch curves away from the y-axis as x approaches zero.

[4 marks]

1.19 (1.19055…) (A1)

Note: Accept an answer of (1.19, 0).

Do not follow through from an incorrect sketch.

[1 mark]

b.i.

(−1.5, 36) (A1)(A1)

Note: Award (A0)(A1) if parentheses are omitted.

Accept x = −1.5, y = 36.

[2 marks]

b.ii.

y = −9.25x + 20.3 (y = −9.25x + 20.25) (A1)(A1)

Note: Award (A1) for −9.25x, award (A1) for +20.25, award a maximum of (A0)(A1) if answer is not an equation.

[2 marks]

b.iii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

Syllabus sections

Topic 2—Functions » SL 2.1—Equations of straight lines, parallel and perpendicular

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22M.2.AHL.TZ2.4d:
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18N.2.SL.TZ0.T_4b.iii:
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Give your answer in the form y = mx + c.
21N.1.SL.TZ0.4c:
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22M.1.SL.TZ1.4b:
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18N.2.SL.TZ0.T_4b.i:
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21N.1.SL.TZ0.4d:
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