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Date November 2014 Marks available 1 Reference code 14N.1.sl.TZ0.1
Level SL only Paper 1 Time zone TZ0
Command term Write down Question number 1 Adapted from N/A

Question

Let \(f(x) = {x^2} + x - 6\).

Write down the \(y\)-intercept of the graph of \(f\).

[1]
a.

Solve \(f(x) = 0\).

[3]
b.

On the following grid, sketch the graph of \(f\), for \( - 4 \le x \le 3\).

[3]
c.

Markscheme

\(y\)-intercept is \( - 6,{\text{ }}(0,{\text{ }} - 6),{\text{ }}y =  - 6\)     A1

[1 mark]

a.

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]

b.

    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]

c.

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).

a.

Parts (a) and (b) of this question were answered quite well by nearly all candidates, with only a few factoring errors in part (b).

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.

c.

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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