Answer
See graph and explanations.
Work Step by Step
Step 1. Given $y=\frac{x+1}{x-3}=\frac{x-3+4}{x-3}=1+\frac{4}{x-3}$
Step 2. It is easier to graph the function by translating the know function $f(x)=\frac{1}{x}$ where $f'(x)=-\frac{1}{x^2}\lt0$ and thus decreasing over all its open intervals. The asymptotes are $x=0$ and $y=0$ for this function.
Step 3. To graph the given function from $f(x)=\frac{1}{x}$, first stretch vertically by a factor of $4$ to get $\frac{4}{x}$, then shift $3$ unit to the right to get $\frac{4}{x-3}$, and finally shift up $1$ unit to get $y=1+\frac{4}{x-3}$ as shown in the figure.
Step 4. The asymptotes will also shift to become $x=3$ and $y=1$.
