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