Express $3 - 3x - 2x^2$ in the form $a(x + b)^2 + c$ and hence state the coordinates of the turning point of the curve $y = 3 - 3x - 2x^2$.

1. Factor out the coefficient of $x^2$ from the $x$ terms

$3 - 3x - 2x^2 = -2\left(x^2 + \frac{3}{2}x\right) + 3$

2. Complete the square for the expression inside the bracket

$= -2\left[ \left(x + \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 \right] + 3$
$= -2\left[ \left(x + \frac{3}{4}\right)^2 - \frac{9}{16} \right] + 3$

3. Expand and simplify to reach the required form

$= -2\left(x + \frac{3}{4}\right)^2 + 2\left(\frac{9}{16}\right) + 3$
$= -2\left(x + \frac{3}{4}\right)^2 + \frac{9}{8} + \frac{24}{8}$
$= -2\left(x + \frac{3}{4}\right)^2 + \frac{33}{8}$

4. State the coordinates of the turning point

The turning point occurs when $x + \frac{3}{4} = 0 \implies x = -\frac{3}{4}$.
The corresponding $y$-value is $\frac{33}{8}$.
Coordinates of the turning point are $\left(-\frac{3}{4}, \frac{33}{8}\right)$.