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Date November 2018 Marks available 2 Reference code 18N.1.SL.TZ0.T_11
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number T_11 Adapted from N/A

Question

Consider the curve y = 5x3 − 3x.

The curve has a tangent at the point P(−1, −2).

Find d y d x .

[2]
a.

Find the gradient of this tangent at point P.

[2]
b.

Find the equation of this tangent. Give your answer in the form y = mx + c.

[2]
c.

Markscheme

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

15x2 − 3      (A1)(A1) (C2)

Note: Award (A1) for 15x2, (A1) for −3. Award at most (A1)(A0) if additional terms are seen.

 

[2 marks]

a.

15 (−1)2 − 3      (M1)

Note: Award  (M1) for substituting −1 into their  d y d x .

 

= 12     (A1)(ft) (C2)

Note: Follow through from part (a).

 

[2 marks]

b.

(y − (−2)) = 12 (x − (−1))     (M1)

OR

−2 = 12(−1) + c     (M1)

Note: Award  (M1) for point and their gradient substituted into the equation of a line.

 

y = 12x + 10     (A1)(ft) (C2)

Note: Follow through from part (b).

 

[2 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5—Calculus » SL 5.4—Tangents and normal
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Topic 2—Functions » SL 2.5—Modelling functions
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Topic 5—Calculus

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