Answer
Graph:
Work Step by Step
The degree of the numerator is 0, which is less than the degree of the denominator (1), so we can write
$y=f(x)=0+\displaystyle \frac{-3}{x-3}$
$\displaystyle \lim_{x\rightarrow+\infty}f(x)=0, \quad \lim_{x\rightarrow-\infty}f(x)=0,$
so, the horizontal asymptote is $y=0.$
f(x) is not defined for $x=3$. Vertical asymptote : $x=3$, with
$\displaystyle \lim_{x\rightarrow 3^{-}}f(x)=+\infty, \quad \displaystyle \lim_{x\rightarrow 3^{+}}f(x)=-\infty$
Plot some points:
y-intercept: $f(0)=1,$ plot ($0,1)$.
Also, use
$f(2)=3$ to plot $(2,3).$
$f(4)=-3$ to plot $(4,-3).$
Graph the asymptotes, and use the behavior in the vicinity of the asymptotes to sketch the graph.