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Date May 2018 Marks available 2 Reference code 18M.1.SL.TZ1.T_12
Level Standard Level Paper Paper 1 Time zone Time zone 1
Command term Find Question number T_12 Adapted from N/A

Question

Consider the quadratic function f(x)=ax2+bx+22.

The equation of the line of symmetry of the graph y=f(x) is x=1.75.

The graph intersects the x-axis at the point (−2 , 0).

Using only this information, write down an equation in terms of a and b.

[1]
a.

Using this information, write down a second equation in terms of a and b.

[1]
b.

Hence find the value of a and of b.

[2]
c.

The graph intersects the x-axis at a second point, P.

Find the x-coordinate of P.

[2]
d.

Markscheme

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

1.75=b2a (or equivalent)      (A1) (C1)

Note: Award (A1) for f(x)=(1.75)2a+1.75b or for y=(1.75)2a+1.75b+22 or for f(1.75)=(1.75)2a+1.75b+22.

[1 mark]

a.

(2)2×a+(2)×b+22=0 (or equivalent)      (A1) (C1)

Note: Award (A1) for (2)2×a+(2)×b+22=0 seen.

Award (A0) for y=(2)2×a+(2)×b+22.

[1 mark]

b.

a = −2, b = 7     (A1)(ft)(A1)(ft) (C2)

Note: Follow through from parts (a) and (b).
Accept answers(s) embedded as a coordinate pair.

[2 marks]

c.

−2x2 + 7x + 22 = 0     (M1)

Note: Award (M1) for correct substitution of a and b into equation and setting to zero. Follow through from part (c).

(=) 5.5     (A1)(ft) (C2)

Note: Follow through from parts (a) and (b).

OR

x-coordinate = 1.75 + (1.75 − (−2))     (M1)

Note: Award (M1) for correct use of axis of symmetry and given intercept.

(=) 5.5     (A1) (C2)

[2 marks]

d.

Examiners report

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

Syllabus sections

Topic 2—Functions » SL 2.5—Modelling functions
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Topic 2—Functions

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