Answer
The vertical asymptote moves to $x=1$.
The horizontal asymptote remains $y=0.$
The domain changes to $(-\infty,1)\cup(1,\infty)$
The range remains $(-\infty,0)\cup(0,\infty)$
![](https://gradesaver.s3.amazonaws.com/uploads/solution/b1fd3f81-893c-416e-a683-acc61ac0859d/result_image/1573944081.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240617%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240617T194337Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=91056854040400913b772ccc2a156b9932da0dd26884cf762b2ecc57cb0967d4)
Work Step by Step
Let $f(x)=\displaystyle \frac{1}{x}$
The asymptotes are $x=0$ (vertical) and $y=0$ (horizontal).
The domain and range are both $\mathbb{R}/\{0\}=(-\infty,0)\cup(0,\infty)$
The graph and the asymptotes of f are graphed with red dashed lines.
$r(x)=\displaystyle \frac{1}{x-1}=f(x-1),$
so its graph (blue solid line) is obtained from the graph of f by shifting it to the right by 1 unit.
The vertical asymptote moves to $x=1$.
The horizontal asymptote remains $y=0.$
The domain changes to $(-\infty,1)\cup(1,\infty)$
The range remains $(-\infty,0)\cup(0,\infty)$