Answer
$y=2\left(x-\frac{5}{4}\right)^2+\frac{71}{8}$
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 $b=-5$.
Thus, the $x$-coordinate of the vertex is:
\begin{align*}
x&=-\frac{b}{2a}\\\\
x&=-\frac{-5}{2(2)}\\\\
x&=-\frac{-5}{4}\\\\
x&=\frac{5}{4}
\end{align*}
Find the $y$-coordinate of the vertex by substituting $\frac{5}{4}$ to $x$ to obtain:
\begin{align*}
y&=2x^2-5x+12\\\\
y&=2\left(\frac{5}{4}\right)^2-5\left(\frac{5}{4}\right)+12\\\\
y&=2\left(\frac{25}{16}\right)-\left(\frac{25}{4}\right)+12\\\\
y&=\frac{25}{8}-\frac{25}{4}+12\\\\
y&=\frac{25}{8}-\frac{50}{8}+\frac{96}{8}\\\\
y&=\frac{71}{8}
\end{align*}
The vertex is at $(\frac{5}{4}, \frac{71}{8})$ so $h=\frac{5}{4}$ and $k=\frac{71}{8}$.
Substitute the values of $a, h,$ and $k$ into the vertex form in the recall part (1) above to obtain:
$$y=2\left(x-\frac{5}{4}\right)^2+\frac{71}{8}$$
