91影视

We have:

$\frac{{(x-2)}^2}{4}-\frac{{\left(y-2\right)}^2}{9}=1$

The equation of hyperbola is $\frac{{(x-2)}^2}{4}-\frac{{\left(y-2\right)}^2}{9}=1$.

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

localid="1646028360708" $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

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

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

Vertices of a horizontal hyperbola with the standard equation, where $\left(h,k\right)$is the center, are given by:

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

Determine the vertices of the horizontal hyperbola:

$$\begin{gathered} =\left(2卤2,2\right) \\ =\left(2-2,2\right),\left(2+2,2\right) \\ =\left(0,2\right),\left(4,2\right) \end{gathered}$$

Asymptotes of a horizontal hyperbola with the standard equation are given below:

$y-k=卤\frac{b}{a}\left(x-h\right)$

$$\begin{gathered} y-2=卤\frac{3}{2}\left(x-2\right) \\ y=2卤\frac{3}{2}\left(x-2\right) \end{gathered}$$

Connect the point with a line. The graph of the vertices $y=2卤\frac{3}{2}\left(x-2\right)$.

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