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Date November 2019 Marks available 2 Reference code 19N.2.SL.TZ0.T_4
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number T_4 Adapted from N/A

Question

The graph of the quadratic function fx=12x-2x+8 intersects the y-axis at 0, c.

The vertex of the function is -3, -12.5.

The equation fx=12 has two solutions. The first solution is x=-10.

Let T be the tangent at x=-3.

Find the value of c.

[2]
a.

Write down the equation for the axis of symmetry of the graph.

[2]
b.

Use the symmetry of the graph to show that the second solution is x=4.

[1]
c.

Write down the x-intercepts of the graph.

[2]
d.

On graph paper, draw the graph of y=fx for  -10x4  and  -14y14. Use a scale of 1cm to represent 1 unit on the x-axis and 1cm to represent 2 units on the y-axis.

[4]
e.

Write down the equation of T.

[2]
f.i.

Draw the tangent T on your graph.

[1]
f.ii.

Given fa=5.5 and f'a=-6, state whether the function, f, is increasing or decreasing at x=a. Give a reason for your answer.

[2]
g.

Markscheme

120-20+8  OR  1202+60-16  (or equivalent)      (M1)

Note: Award (M1) for evaluating f0.

c= -8          (A1)(G2)

Note: Award (G2) if -8 or 0, -8 seen.

[2 marks]

a.

x=-3      (A1)(A1)

Note: Award (A1) for “x= constant”, (A1) for the constant being -3. The answer must be an equation.

[2 marks]

b.

-3--10+-3      (M1)

OR

-8--10+2      (M1)

OR

-10+x2=-3      (M1)

OR

diagram showing axis of symmetry and given points (x-values labels, -10, -3 and 4, are sufficient) and an indication that the horizontal distances between the axis of symmetry and the given points are 7.      (M1)

Note: Award (M1) for correct working using the symmetry between x=-10 and x=-3. Award (M0) if candidate has used x=-10 and x=4 to show the axis of symmetry is x=-3. Award (M0) if candidate solved fx=12 or evaluated f-10 and f4.

x= 4      (AG)

[1 mark]

c.

-8 and 2      (A1)(A1)

Note: Accept x=-8, y=0 and x=2, y=0 or -8, 0 and 2, 0, award at most (A0)(A1) if parentheses are omitted.

[2 marks]

d.

      (A1)(A1)(A1)(A1)(ft)

Note: Award (A1) for labelled axes with correct scale, correct window. Award (A1) for the vertex, -3, -12.5, in correct location.
Award (A1) for a smooth continuous curve symmetric about their vertex. Award (A1)(ft) for the curve passing through their x and y intercepts in correct location. Follow through from their parts (a) and (d).

If graph paper is not used:
Award at most (A0)(A0)(A1)(A1)(ft). Their graph should go through their -8 and 2 for the last (A1)(ft) to be awarded.

 

[4 marks]

e.

y=-12.5  OR  y=0x-12.5      (A1)(A1)

Note: Award (A1) for "y= constant", (A1) for the constant being -12.5. The answer must be an equation.

 [2 marks]

f.i.

tangent to the graph drawn at x=-3        (A1)(ft)

Note: Award (A1) for a horizontal straight-line tangent to curve at approximately x=-3. Award (A0) if a ruler is not used. Follow through from their part (e).

 [1 mark]

f.ii.

decreasing       (A1)

gradient (of tangent line) is negative (at x=a)  OR  f'a<0        (R1)

Note: Do not accept "gradient (of tangent line) is -6". Do not award (A1)(R0).

[2 marks]

g.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.i.
[N/A]
f.ii.
[N/A]
g.

Syllabus sections

Topic 5—Calculus » SL 5.2—Increasing and decreasing functions
Show 45 related questions
Topic 2—Functions » SL 2.3—Graphing
Topic 5—Calculus » SL 5.4—Tangents and normal
Topic 2—Functions » SL 2.5—Modelling functions
Topic 2—Functions
Topic 5—Calculus

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