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]