Answer
See below
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Work Step by Step
The standard form of the equation is: $y=a(x-h)^2+k$,
where $(h,k)$ is the vertex.
Since the vertex is $(0,-1)$, we get $h=0,k=-1$. Substitute:
$y=a(x-0)^2+(-1)\\
\rightarrow y=ax^2-1$
Substitute $(-2,3)$
$3=a(-2)^2-1\\
\rightarrow a=1$
Hence, the vertex form is $y=x^2-1$
The standard form: $y=ax^2+bx+c$
Since $a=1\\b=0\\c=-1$
We have: $y=x^2-1$
The intercept form of the equation: $y=a(x-p)(x-q)\\
\rightarrow y=(x-1)(x+1)$