| Date | May 2009 | Marks available | 4 | Reference code | 09M.1.sl.TZ2.14 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 14 | Adapted from | N/A |
Question
A quadratic curve with equation y = ax (x − b) is shown in the following diagram.
The x-intercepts are at (0, 0) and (6, 0), and the vertex V is at (h, 8).
Find the value of h.
Find the equation of the curve.
Markscheme
\(\frac{{0 + 6}}{2} = 3\) \(h = 3\) (M1)(A1) (C2)
Note: Award (M1) for any correct method.
[2 marks]
\(y = ax(x - 6)\) (A1)
\(8 = 3a(- 3)\) (A1)(ft)
\(a = - \frac{8}{9}\) (A1)(ft)
\(y = - \frac{8}{9} x (x - 6)\) (A1)(ft)
Notes: Award (A1) for correct substitution of \(b = 6\) into equation.
Award (A1)(ft) for substitution of their point V into the equation.
OR
\(y = a(x - 3)^2 + 8\) (A1)(ft)
Note: Award (A1)(ft) for correct substitution of their h into the equation.
\(0 = a(6 - 3)^2 + 8\) OR \(0 = a(0 - 3)^2 + 8\) (A1)
Note: Award (A1) for correct substitution of an x intercept.
\(a = - \frac{8}{9}\) (A1)(ft)
\(y = - \frac{8}{9}(x - 3)^2 + 8\) (A1)(ft) (C4)
[4 marks]
Examiners report
Most candidates successfully found h but very few could find the equation of the curve.
This question appeared to be the most difficult question on the paper.
