91Ó°ÊÓ

We have:

$16x^2-25y^2+96x-50y-281=0$

The equation is $16x^2-25y^2+96x-50y-281=0$.

Move the constant to the right:

$16x^2-25y^2+96x-50y=281$

Reorder the terms:

$$\begin{gathered} 16x^2+96x-25y^2-50y=281 \\ 16\left(x^2+6x\right)-25y^2-50y=281 \end{gathered}$$

Add $16×9$to the both sides:

localid="1646050052764" $$\begin{gathered} 16\left(x^2+6x+9\right)-25y^2-50y=281+16×9 \\ 16\left(x^2+6x+9\right)-25y^2-50y=425 \\ 16{\left(x+3\right)}^2-25\left(y^2+2y\right)=425 \end{gathered}$$

Add $-25×1$to the both sides:

localid="1646050063867" $$\begin{gathered} 16{\left(x+3\right)}^2-25\left(y^2+2y+1\right)=425-25×1 \\ 16{\left(x+3\right)}^2-25\left(y^2+2y+1\right)=425-25 \\ 16{\left(x+3\right)}^2-25{\left(y+1\right)}^2=400 \\ \frac{16{\left(x+3\right)}^2}{400}-\frac{25{\left(y+1\right)}^2}{16}=1 \\ \frac{{\left(x+3\right)}^2}{25}-\frac{{\left(y+1\right)}^2}{16}=1 \end{gathered}$$

Rewrite the equation:

$\frac{{\left(x+\left(-3\right)\right)}^2}{5^2}-\frac{{\left(y-\left(-1\right)\right)}^2}{4^2}=1$

Observe that the equation is in the form of a horizontal hyperbola with the center located at $\left(-3,-1\right)$.

Therefore, the two axes of symmetry of the given hyperbola are the transverse axis $y=-1$and the conjugate axis$x=-3$

The vertices of the hyperbola are:

$$\begin{gathered} \left(h-a,k\right),\left(h+a,k\right)=\left(-3-5,-1\right),\left(-3+5,-1\right) \\ =\left(-8,-1\right),\left(+2,-1\right) \end{gathered}$$

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

$$\begin{gathered} y=Â\pm \frac{4}{5}\left(x-\left(-3\right)\right)-1 \\ y=\frac{4}{5}x+\frac{7}{5}, \\ y=-\frac{4}{5}x-\frac{17}{5} \end{gathered}$$

Connect the point with a line. The graph of the vertices $y=\frac{4}{5}x+\frac{7}{5},y=-\frac{4}{5}x-\frac{17}{5}$.

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