Determine whether or not $f$is continuous.

The graph of the function $g$ is obtained by applying the following transformations to the graph of $f$:

a reflection in the $y$–axis followed by a translation by the vector $\left( \begin{array}{l}2\\0\end{array} \right)$.

Find $g(x)$.

$1 - 2(2) = -  3$ and $\frac{3}{4}{(2 - 2)^2} - 3 =  - 3$     A1

both answers are the same, hence f is continuous (at $x = 2$)     R1

Note:     R1 may be awarded for justification using a graph or referring to limits. Do not award A0R1.

[2 marks]

reflection in the y-axis

$f( - x) = \left\{ \begin{array}{r}1 + 2x,\\{\textstyle{3 \over 4}}{(x + 2)^2} - 3,\end{array} \right.\begin{array}{*{20}{c}}{x \ge  - 2}\\{x <  - 2}\end{array}$     (M1)

Note:     Award M1 for evidence of reflecting a graph in y-axis.

translation $\left( \begin{array}{l}2\\0\end{array} \right)$

$g(x) = \left\{ \begin{array}{r}2x - 3,\\{\textstyle{3 \over 4}}{x^2} - 3,\end{array} \right.\begin{array}{*{20}{c}}{x \ge 0}\\{x < 0}\end{array}$     (M1)A1A1

Note:     Award (M1) for attempting to substitute $(x - 2)$ for x, or translating a graph along positive x-axis.

     Award A1 for the correct domains (this mark can be awarded independent of the M1).

     Award A1 for the correct expressions.

[4 marks]