| Date | May 2018 | Marks available | 2 | Reference code | 18M.1.sl.TZ1.4 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Show that | 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.
Hence, show that a = 2.
The equation of L is y = 5 . Find the value of c.
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]
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]
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]
