#### Answer

The graph is shown below:

#### Work Step by Step

We know that a rational function $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ can consist of the vertical and horizontal asymptotes.
For vertical asymptotes, we keep the denominator function $q\left( x \right)$ equal to 0, and thereby find its roots; that is, a vertical asymptote is the root of the rational function.
So, for the vertical asymptote to be given by $x=3$ , we can write the denominator function of the rational function as,
$q\left( x \right)=x-3$
We have a horizontal asymptote as:
$\begin{align}
& y=\frac{0}{1} \\
& =0
\end{align}$
And for the horizontal asymptote to be the x-axis or $y=0$, the degree of the numerator function $p(x)$ has to be less than the degree of the denominator function, $q\left( x \right)$.
Let $f\left( x \right)=\frac{p\left( x \right)}{x-3}$.
Since the degree of $p\left( x \right)$ is supposed to be one less than that of $q\left( x \right)$ for the function to have a horizontal asymptote, thus the degree of $p\left( x \right)$ is 0, i.e., it is a constant.
For calculating $p\left( x \right)$ , we calculate the y intercept. Since the y intercept is given to be -1,
$\begin{align}
& f\left( 0 \right)=-1 \\
& \frac{p\left( x \right)}{0-3}=-1 \\
& p\left( x \right)=3
\end{align}$
Therefore, the required function is given by,
$f\left( x \right)=\frac{3}{x-3}$
Now, for plotting,
Press $y=$ to write the equation $y=\frac{3}{x-3}$.
Press WINDOW to set the window $\left( -10,10,1 \right)$ and $\left( -10,10,1 \right)$.
Press GRAPH to plot the graph of the function.
Thus, the graph of the function $f\left( x \right)=\frac{3}{x-3}$ has a vertical asymptote at $x=3$ , horizontal asymptote $y=0$ , $y$ -intercept at $-1$ ,and no $x$ -intercept.