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