Answer
$y=-2\left(x-2\right)^2+11$
Work Step by Step
Recall:
(1) The vertex form of $y=ax^2+bx+c$ is $y=a(x-h)^2+k$ where $(h, k)$ is the vertex.
(2) The vertex of the graph of $y=ax^2+bx+c$ is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.
The given function has $a=-2$ and $b=8$.
Thus, the $x$-coordinate of the vertex is:
\begin{align*}
x&=-\frac{b}{2a}\\\\
x&=-\frac{8}{2(-2)}\\\\
x&=-\frac{8}{-4}\\\\
x&=2
\end{align*}
Find the $y$-coordinate of the vertex by substituting $2$ to $x$ to obtain:
\begin{align*}
y&=-2x^2+8x+3\\\\
y&=-2\left(2\right)^2+8\left(2\right)+3\\\\
y&=-2\left(4\right)+16+3\\\\
y&=-8+16+3\\\\
y&=11
\end{align*}
The vertex is at $\left(2, 11\right)$ so $h=2$ and $k=11$.
Substitute the values of $a, h,$ and $k$ into the vertex form and then recall part (1) above to obtain:
$$y=-2\left(x-2\right)^2+11$$
