Answer
$$
y=-2 x^{2}+12 x-5
$$
The vertex of this parabola is $(3,13)$.
Work Step by Step
$$
y=-2 x^{2}+12 x-5
$$
To find the vertex of this parabola. First, we find $y^{\prime}(x) $,
$$
\begin{aligned}
y^{\prime}(x) &=-4 x+12\\
&=-4(x-3)\\
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $y^{\prime} =0 $. Now, solve the equation $y^{\prime} =0 $ to get
$$
\begin{aligned}
y^{\prime}(x) =-4(x-3) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x-3 &=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x & =3 \\
\end{aligned}
$$
The only critical number is : $x =3.$
When $x=3$
$$
y=-2(3)^{2}+12(3)-5=13
$$
So, the vertex of this parabola is $(3,13)$.
