Express $-9 + 6x - 2x^2$ in the form $a(x+b)^2 + c$ and hence explain why $-9 + 6x - 2x^2$ is negative for all real values of $x$.

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

$-2x^2 + 6x - 9 = -2(x^2 - 3x) - 9$

2. Complete the square for the expression inside the bracket

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

3. Expand and simplify to reach the required form

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

4. Explain why the expression is always negative

For all real values of $x$, $\left(x - \frac{3}{2}\right)^2 \geqslant 0$.
Therefore, $-2\left(x - \frac{3}{2}\right)^2 \leqslant 0$.
Subtracting $\frac{9}{2}$ from both sides gives $-2\left(x - \frac{3}{2}\right)^2 - \frac{9}{2} \leqslant -\frac{9}{2}$.
Since $-\frac{9}{2} < 0$, the expression is always negative.