Answer
$\text{Vertex: }
\left(-\dfrac{1}{2}, -\dfrac{49}{4} \right)
\\\text{$y-$intercept: }
-12
\\\text{$x-$intercepts: }
\{-4,3\}$
Work Step by Step
In standard form, the given function, $
f(x)=x^2+x-12
$, is equivalent to
\begin{array}{l}\require{cancel}
f(x)=(x^2+x)-12
\\\\
f(x)=\left(x^2+x+\left(\dfrac{1}{2}\right)^2 \right)-12-\left(\dfrac{1}{2}\right)^2
\\\\
f(x)=\left(x^2+x+\dfrac{1}{4} \right)-12-\dfrac{1}{4}
\\\\
f(x)=\left(x+\dfrac{1}{2} \right)^2-\dfrac{48}{4}-\dfrac{1}{4}
\\\\
f(x)=\left(x+\dfrac{1}{2} \right)^2-\dfrac{49}{4}
.\end{array}
Substituting $x=0$, then the $y-$intercept is
\begin{array}{l}\require{cancel}
f(0)=\left(0+\dfrac{1}{2} \right)^2-\dfrac{49}{4}
\\\\
f(0)=\dfrac{1}{4}-\dfrac{49}{4}
\\\\
f(0)=-\dfrac{48}{4}
\\\\
f(0)=-12
.\end{array}
Substituting $f(x)=0$, then the $x-$intercepts are
\begin{array}{l}\require{cancel}
0=\left(x+\dfrac{1}{2} \right)^2-\dfrac{49}{4}
\\\\
\dfrac{49}{4}=\left(x+\dfrac{1}{2} \right)^2
\\\\
\pm\sqrt{\dfrac{49}{4}}=x+\dfrac{1}{2}
\\\\
\pm\dfrac{7}{2}=x+\dfrac{1}{2}
\\\\
-\dfrac{1}{2}\pm\dfrac{7}{2}=x
\\\\
x=\{ -4, 3 \}
.\end{array}
Hence, the given equation has the following characteristics:
\begin{array}{l}\require{cancel}
\text{Vertex: }
\left(-\dfrac{1}{2}, -\dfrac{49}{4} \right)
\\\text{$y-$intercept: }
-12
\\\text{$x-$intercepts: }
\{-4,3\}
.\end{array}
