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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 − 4xc. 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]
b.

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]

b.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.

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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