| Date | November 2014 | Marks available | 3 | Reference code | 14N.1.sl.TZ0.1 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Solve | Question number | 1 | Adapted from | N/A |
Question
Let \(f(x) = {x^2} + x - 6\).
Write down the \(y\)-intercept of the graph of \(f\).
Solve \(f(x) = 0\).
On the following grid, sketch the graph of \(f\), for \( - 4 \le x \le 3\).
Markscheme
\(y\)-intercept is \( - 6,{\text{ }}(0,{\text{ }} - 6),{\text{ }}y = - 6\) A1
[1 mark]
valid attempt to solve (M1)
eg\(\;\;\;(x - 2)(x + 3) = 0,{\text{ }}x = \frac{{ - 1 \pm \sqrt {1 + 24} }}{2}\), one correct answer
\(x = 2,{\text{ }}x = - 3\) A1A1 N3
[3 marks]
A1A1A1
Note: The shape must be an approximately correct concave up parabola. Only if the shape is correct, award the following:
A1 for the \(y\)-intercept in circle and the vertex approximately on \(x = - \frac{1}{2}\), below \(y = - 6\),
A1 for both the \(x\)-intercepts in circles,
A1 for both end points in ovals.
[3 marks]
Total [7 marks]
Examiners report
Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).
Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).
In part (c), although most candidates were familiar with the general parabolic shape of the graph, many placed the vertex at the \(y\)-intercept \((0,{\text{ }} - 6)\), and very few candidates considered the endpoints of the function with the given domain.
