| 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
0+62=30+62=3 h=3h=3 (M1)(A1) (C2)
Note: Award (M1) for any correct method.
[2 marks]
y=ax(x−6)y=ax(x−6) (A1)
8=3a(−3)8=3a(−3) (A1)(ft)
a=−89a=−89 (A1)(ft)
y=−89x(x−6)y=−89x(x−6) (A1)(ft)
Notes: Award (A1) for correct substitution of b=6b=6 into equation.
Award (A1)(ft) for substitution of their point V into the equation.
OR
y=a(x−3)2+8y=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+80=a(6−3)2+8 OR 0=a(0−3)2+80=a(0−3)2+8 (A1)
Note: Award (A1) for correct substitution of an x intercept.
a=−89a=−89 (A1)(ft)
y=−89(x−3)2+8y=−89(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.
