# Does cancellation impact vertical asymptotes? Question: Let r(x)=(x2+x)/((x+1)(2x−4)). Does the graph has x=1 as one of its asymptotes? My reasoning: (x2+x)/((x+1)(2x−4))=x(x+1)/((x+1)(2x−4))=x/(2x−4) and so, it cannot have x=−1 as one of its asymptotes. However, what if I don't calcel and then say that −1 is a vertical asymptote? Will I be wrong?

Does cancellation impact vertical asymptotes?
Question: Let $r\left(x\right)=\frac{\left({x}^{2}+x\right)}{\left(x+1\right)\left(2x-4\right)}$. Does the graph has x=1 as one of its asymptotes?
My reasoning: $\frac{\left({x}^{2}+x\right)}{\left(x+1\right)\left(2x-4\right)}=\frac{x\left(x+1\right)}{\left(x+1\right)\left(2x-4\right)}=\frac{x}{\left(2x-4\right)}$ and so, it cannot have x=−1 as one of its asymptotes.
However, what if I don't calcel and then say that −1 is a vertical asymptote? Will I be wrong?
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Julianna Bell
You are correct saying that the function
$r\left(x\right)=\frac{\left({x}^{2}+x\right)}{\left(x+1\right)\left(2x-4\right)}$
has not an asymptote for x=−1 since:
$\underset{x\to -1}{lim}\frac{\left({x}^{2}+x\right)}{\left(x+1\right)\left(2x-4\right)}=\underset{x\to -1}{lim}\frac{x\left(x+1\right)}{\left(x+1\right)\left(2x-4\right)}=\frac{1}{6}$
but this function is not defined for x=−1 so its graph has an ''hole'' at the point $\left(-1,\frac{1}{6}\right)$ and the functios is not the same as
$y=\frac{x}{\left(2x-4\right)}$