Does cancellation impact vertical asymptotes? Question: Let $r(x) = \frac{(x^2 + x)}{(x + 1)(2x - 4)}$. Does the graph has $x = 1$ as one of its asymptotes? 
Answer: No.
My reasoning: $\frac{(x^2 + x)}{(x + 1)(2x - 4)} = \frac{x(x + 1)}{(x + 1)(2x - 4)} = \frac{x}{(2x - 4)}$ 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?

You are correct saying that the function 
$$
r(x) = \frac{(x^2 + x)}{(x + 1)(2x - 4)}
$$
has not an asymptote for $x=-1$ since:
$$
\lim_{x \to -1}\frac{(x^2 + x)}{(x + 1)(2x - 4)}=\lim_{x \to -1}\frac{x(x + 1)}{(x + 1)(2x - 4)}=\frac{1}{6}
$$
but this function is not defined for $x=-1$ so its graph has an ''hole'' at the point $(-1 ,\frac{1}{6})$ and the functios is not the same as
$$
y=\frac{x}{(2x - 4)}
$$
