Answer
Sample answer: $\displaystyle \quad g(x)=\frac{1}{x-3}$
Work Step by Step
The graph approaches $y=0$, the horizontal asymptote, on both sides of the graph.
$f(x)=0+$(rational expression with degree of numerator less than that of the denominator)
The rational expression should have $x-3$ as a factor of the denominator.
$\displaystyle \frac{1}{x-3}$ is adequate, because the one-sided limits are
$\displaystyle \lim_{x\rightarrow 3^{-}}\frac{1}{x-3}=-\infty,\qquad\lim_{x\rightarrow 3^{+}}\frac{1}{x-3}=+\infty,$
as needed by the problem.
Sample answer: $\displaystyle \quad g(x)=\frac{1}{x-3}$
