Answer
Standard form: $y=\frac{1}{2} x^2-\frac{1}{2} x-6$
Vertex form: $ y=\frac{1}{2}\left(x-\frac{1}{2}\right)^2-\frac{49}{8}$
Factored form: $y=\frac{1}{2}(x-4)(x+3) $.
Work Step by Step
The standard form is obtained by writing the right-hand side as three terms and rearranging the terms:
$$
\begin{aligned}
& y=-6+\frac{x^2-x}{2}\\
&y=\frac{1}{2} x^2-\frac{1}{2} x-6 \\
\end{aligned}
$$
Complete the square to get the vertex form.
$$
\begin{aligned}
& y=\frac{1}{2} x^2-\frac{1}{2} x-6\\
& =\frac{1}{2}\left(x^2-x-12\right) \\
& =\frac{1}{2}\left(x^2-x+\frac{1}{4}-\frac{1}{4}-12\right) \\
& =\frac{1}{2}\left(x^2-x+\frac{1}{4}\right)+\frac{1}{2}\left(-\frac{1}{4}-12\right) \\
& =\frac{1}{2}\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\left(-\frac{49}{4}\right) \\
&y =\frac{1}{2}\left(x-\frac{1}{2}\right)^2-\frac{49}{8} .
\end{aligned}
$$
The factored form is
$$
\begin{aligned}
y & =\frac{1}{2}\left(x^2-x-12\right) \\
& =\frac{1}{2}(x-4)(x+3) .
\end{aligned}
$$
