Answer
The graph is shown below.
Work Step by Step
Consider the equation:
$\frac{{{y}^{2}}}{16}-\frac{{{x}^{2}}}{9}=1$
Identify a and b by comparing $\frac{{{x}^{2}}}{{{3}^{2}}}-\frac{{{y}^{2}}}{{{3}^{2}}}=1$ with the standard equation of a hyperbola $\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1$.
Here,
$a=3$ and $b=4$
The asymptotes are,
$\begin{align}
& y=\frac{b}{a}x \\
& =\frac{4}{3}x
\end{align}$
And,
$\begin{align}
& y=\frac{b}{a}x \\
& =-\frac{4}{3}x
\end{align}$
For $a=3$ and $b=4$, the possible pairs are as shown below,
$\left( -3,4 \right),\left( 3,4 \right),\left( 3,-4 \right),\left( -3,-4 \right)$
These pairs form a rectangle and both asymptotes pass through the corner points of the rectangle.
The required table to plot both asymptotes is shown below,
$\begin{matrix}
x & y=\frac{4}{3}x & y=-\frac{4}{3}x \\
-3 & -4 & 4 \\
0 & 0 & 0 \\
3 & 4 & -4 \\
\end{matrix}$
The vertices of the vertical hyperbola are the mid-points of the top and bottom sides of the rectangle; thus, the vertices of the vertical hyperbola are $\left( 0,-4 \right)$ and $\left( 0,4 \right)$.
Consider the hyperbola equation,
$\frac{{{y}^{2}}}{16}-\frac{{{x}^{2}}}{9}=1$
Rewrite the equation as shown below,
$\begin{align}
& \frac{{{y}^{2}}}{16}-\frac{{{x}^{2}}}{9}=1 \\
& \frac{{{y}^{2}}}{16}=1+\frac{{{x}^{2}}}{9} \\
& \frac{{{y}^{2}}}{16}=\frac{9+{{x}^{2}}}{9} \\
& y=\pm \frac{4}{3}\sqrt{9+{{x}^{2}}}
\end{align}$
Substitute $x=-2$ into the equation,
$\begin{align}
& y=\pm \frac{4}{3}\sqrt{9+{{\left( -2 \right)}^{2}}} \\
& =\pm \frac{4}{3}\sqrt{9+4} \\
& =\pm \frac{4}{3}\left( 3.6 \right) \\
& =\pm 4.807
\end{align}$
Substitute $x=0$ into the equation,
$\begin{align}
& y=\pm \frac{4}{3}\sqrt{9+{{\left( 0 \right)}^{2}}} \\
& =\pm \frac{4}{3}\sqrt{9} \\
& =\pm \frac{4}{3}\left( 3 \right) \\
& =\pm 4
\end{align}$
Substitute $x=2$ into the equation,
$\begin{align}
& y=\pm \frac{4}{3}\sqrt{9+{{\left( 2 \right)}^{2}}} \\
& =\pm \frac{4}{3}\sqrt{9+4} \\
& =\pm \frac{4}{3}\left( 3.6 \right) \\
& =\pm 4.807
\end{align}$
$\begin{matrix}
x & y=\pm \frac{4}{3}\sqrt{9+{{x}^{2}}} \\
-2 & \pm 4.807 \\
0 & \pm 4 \\
2 & \pm 4.807 \\
\end{matrix}$
Now plot the graph of the hyperbola as shown in Figure - 2.
