91Ó°ÊÓ

We have:

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

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

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

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

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

$\left(h,k\right)=\left(-2,-2\right)$

Find the value of $a:$

$$\begin{gathered} a^2=9 \\ a=\sqrt{9} \\ a=Â\pm 9 \\ {b}^2=\sqrt{9} \\ b=Â\pm 9 \end{gathered}$$

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

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

$\left(h,k+c\right),\left(h,k-c\right)$

$$\begin{gathered} c^2=a^2+b^2 \\ {c}^2=9^2+9^2 \\ c=\sqrt{162} \\ c=Â\pm 3\sqrt{2} \end{gathered}$$

Substitute the values:

$\left(h,k+c\right).\left(h,k-c\right)=\left(-2,-2+3\sqrt{2}\right),\left(-2,-2-3\sqrt{2}\right)$

Substitute $x=-2$into the hyperbola equation:

$$\begin{gathered} \frac{{\left(y+2\right)}^2}{9}-\frac{{\left(-2+2\right)}^2}{9}=1 \\ y=-5,1 \end{gathered}$$

Therefore, the points $\left(-2,-5\right)$and$\left(-2,1\right)$lies on the hyperbola.

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

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

Connect the point with a line. The graph of the vertices $y=x,y=-x-4$.

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