Answer
Please see image.
Work Step by Step
The graph of $f(x)=a(x-h)^{2}+k$ has the same shape as the graph of $y=a(x-h)^{2}.$
If $k$ is positive, the graph of $y=a(x-h)^{2}$ is shifted $k$ units up.
If $k$ is negative, the graph of $y=a(x-h)^{2}$ is shifted $|k|$ units down.
The vertex is $(h, k)$, and the axis of symmetry is $x=h.$
For $a\gt 0$, the minimum function value is $k$.
For $a\lt 0$, the maximum function value is $k.$
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$k=-1$; shifted $1$ unit down.
$h=-2$
$a=\displaystyle \frac{3}{2}$; opens upward
The vertex is $(-2,-1)$.
The axis of symmetry is $x=2.$
Since $a\gt 0$, the minimum function value is $-1$.
Make a table of function values and plot the points,
and join with a smooth curve.