Answer
$w(z)=-3\left(z^2-\frac{3}{2}\right)^2+\frac{19}{4}$
Vertex: $(3/2,19/4)$
Axis of symmetry: $z=3/2$.
Work Step by Step
Factor out the coefficient of $x^2$ and square half of the coefficient of the $x$-term: $(-3 / 2)^2=9/4$. Adding and subtracting this number after the $x$-term gives
$$
\begin{aligned}
& w(z)=-3 z^2+9 z-2\\
& w(z)=-3\left(z^2-3z+\frac{2}{3}\right) \\
& w(z)=-3\left(z^2-3z+\frac{9}{4}+\frac{2}{3}-\frac{9}{4}\right) \\
& w(z) = -3\left(z^2-3z+\frac{9}{4}\right)-3\left(-\frac{19}{12}\right) \\
&w(z)=-3\left(z^2-\frac{3}{2}\right)^2+\frac{19}{4}
\end{aligned}
$$ The vertex is $(3/2,19/4)$ and the axis of symmetry is $z=3/2$.
