Show that $f(x) = 3{x^2} + 6x - 9$ .

For the graph of f

(i)     write down the coordinates of the vertex;

(ii)    write down the y-intercept;

(iii)   find both x-intercepts.

Hence sketch the graph of f .

Let $g(x) = {x^2}$ . The graph of f may be obtained from the graph of g by the following two transformations

a stretch of scale factor t in the y-direction,

followed by a translation of $\left( \begin{array}{l} p\\ q \end{array} \right)$ .

Write down $\left( \begin{array}{l} p\\ q \end{array} \right)$ and the value of t .

$f(x) = 3({x^2} + 2x + 1) - 12$     A1

$= 3{x^2} + 6x + 3 - 12$     A1

$= 3{x^2} + 6x - 9$     AG     N0

[2 marks]

(i) vertex is $( - 1, - 12)$     A1A1     N2

(ii) $y = - 9$ , or $(0, - 9)$     A1     N1

(iii) evidence of solving $f(x) = 0$     M1

e.g. factorizing, formula

correct working     A1

e.g. $3(x + 3)(x - 1) = 0$ , $x = \frac{{ - 6 \pm \sqrt {36 + 108} }}{6}$

$x = - 3$ , $x = 1$ , or $( - 3{\text{, }}0){\text{, }}(1{\text{, }}0)$     A1A1     N2

[7 marks]

     A1A1A1     N3

Note: Award A1 for a parabola opening upward, A1 for vertex in approximately correct position, A1 for intercepts in approximately correct positions. Scale and labelling not required.

[3 marks]

$\left( \begin{array}{l} p\\ q \end{array} \right) = \left( \begin{array}{l} - 1\\ - 12 \end{array} \right)$ , $t = 3$     A1A1A1     N3

[3 marks]