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Date	May 2018	Marks available	2	Reference code	18M.1.sl.TZ1.4
Level	SL only	Paper	1	Time zone	TZ1
Command term	Find	Question number	4	Adapted from	N/A

Question

Let f(x) = ax² − 4x − c. A horizontal line, L , intersects the graph of f at x = −1 and x = 3.

The equation of the axis of symmetry is x = p. Find p.

[2]

a.i.

Hence, show that a = 2.

[2]

a.ii.

The equation of L is y = 5 . Find the value of c.

[3]

Markscheme

METHOD 1 (using symmetry to find p)

valid approach (M1)

eg $\frac{{ - 1 + 3}}{2}$ ,

p = 1 A1 N2

Note: Award no marks if they work backwards by substituting a = 2 into $- \frac{b}{{2a}}$ to find p.

Do not accept $p = \frac{2}{a}$ .

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a M1

eg a + 4 − c = a(3²) − 4(3) − c, f(−1) = f(3)

correct working A1

eg 8a = 16

a = 2 AG N0

valid approach to find p (M1)

eg $- \frac{b}{{2a}},\,\,\,\frac{4}{{2\left( 2 \right)}}$

p = 1 A1 N2

[2 marks]

a.i.

METHOD 1

valid approach M1

eg $- \frac{b}{{2a}},\,\,\,\frac{4}{{2a}}$ (might be seen in (i)), f' (1) = 0

correct equation A1

eg $\frac{4}{{2a}}$ = 1, 2a(1) − 4 = 0

a = 2 AG N0

METHOD 2 (calculating a first)
(i) & (ii) valid approach to calculate a M1

eg a + 4 − c = a(3²) − 4(3) − c, f(−1) = f(3)

correct working A1

eg 8a = 16

a = 2 AG N0

[2 marks]

a.ii.

valid approach (M1)
eg f(−1) = 5, f(3) =5

correct working (A1)
eg 2 + 4 − c = 5, 18 − 12 − c = 5

c = 1 A1 N2

[3 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Syllabus sections

Topic 2 - Functions and equations » 2.4 » The quadratic function

$x \mapsto a{x^2} + bx + c$ : its graph,

$y$ -intercept

$(0, c)$ . Axis of symmetry.

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