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 $y=f(x)=0+\displaystyle \frac{1}{2x+4}$
$\displaystyle \lim_{x\rightarrow+\infty}f(x)=0, \quad \lim_{x\rightarrow-\infty}f(x)=0,$
so, the horizontal asymptote is $y=0.$
f(x) is not defined for $x=-2$. Vertical asymptote : $x=-2$, with
$\displaystyle \lim_{x\rightarrow-2^{-}}f(x)=-\infty, \quad \displaystyle \lim_{x\rightarrow-2^{+}}f(x)=+\infty$
Plot some points:
y-intercept: $f(0)=1/4,$ plot ($0,1/4)$.
Also, use
$f(-3/2)=1$ to plot $(-3/2,1).$
$f(-5/2)=-1$ to plot $(-5/2,-1).$
Graph the asymptotes, and use the behavior in the vicinity of the asymptotes to sketch the graph.