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/4f3a91b8-88fc-492f-89e5-8a245be4826d/result_image/1573944976.png?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Credential=AKIAJVAXHCSURVZEX5QQ%2F20240617%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Date=20240617T172015Z&X-Amz-Expires=900&X-Amz-SignedHeaders=host&X-Amz-Signature=869dcbb441aeaecd780823a452f58c2a3cf0a188053b52255f42014bceb59247)
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.
$r(x)=\displaystyle \frac{3}{x+1}=3\cdot f(x+1),$
so its graph (blue solid line) is obtained from the graph of f by
-shifting it to the left by $1$ unit, to obtain $f(x+1)$
and then,
- vertically stretching by a factor of 3.
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 $(-\infty,0)\cup(0,\infty)$