Graph$\frac{{\left(y+3\right)}^2}{16}-\frac{{\left(x+2\right)}^2}{9}=1$.

We have:

$\frac{{\left(y+3\right)}^2}{16}-\frac{{\left(x+2\right)}^2}{9}=1$

The equation of hyperbola is $\frac{{\left(y+3\right)}^2}{16}-\frac{{\left(x+2\right)}^2}{9}=1$.

Rewrite the equation in the standard form:

$\frac{{\left(y-\left(-3\right)\right)}^2}{16}-\frac{{\left(x-\left(-2\right)\right)}^2}{9}=1$

The equation a horizontal hyperbola in standard form, with the center $\left(h,k\right)$is given by:

localid="1646029649224" $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$

Comparing with the equation of a horizontal hyperbola, it follows:

localid="1646028550888" $$\begin{gathered} a=-2, \\ b=-3 \\ \mathrm{Center}:\left(-2,-3\right) \end{gathered}$$

Find the value of $a:$

$$\begin{gathered} a^2=16 \\ a=\sqrt{16} \\ a=\pm 4 \\ {b}^2=9 \\ b=\sqrt{9} \\ b=\pm 3 \end{gathered}$$

Since the length cannot be negative, discard the negative solutions.

To find the vertices of the hyperbola, insert the values:

$$\begin{gathered} \left(h,k+a\right),\left(h,k-a\right) \\ =\left(-2,-3+4\right),\left(-2,-3-4\right) \\ =\left(-2,1\right),\left(-2,-7\right) \end{gathered}$$

Substitute the value into the formula $y=\pm \frac{a}{b}\left(x-h\right)+k$:

$$\begin{gathered} y=\pm \frac{4}{3}\left(x-\left(-2\right)\right)-3 \\ y=\frac{4}{3}x-\frac{1}{3}, \\ y=-\frac{4}{3}x-\frac{17}{3} \end{gathered}$$

Connect the point with a line. The graph of the vertices $y=\frac{4}{3}x-\frac{1}{3},y=-\frac{4}{3}x-\frac{1}{7}$.

Draw the hyperbola passing through the vertices and considering the asymptotes. The graph of the function is shown below: