vertical asymptote question

for $\displaystyle f(x)=(6-3x-3x^2)/(x^2-1)$

I got vertical asymptotes at x=1 and x= -1 but the case where x=1 is wrong

yet $\displaystyle (x^2-1)=0$ factors to (x+1)(x-1)=0 which implies that $\displaystyle x^2-1)=0$ is zero for both x+1 and x-1

Where is my error?

thanks!

Re: vertical asymptote question

$\displaystyle f(x)=\frac{6-3x-3x^2}{x^2-1}=\frac{3(2+x)(1-x)}{(x+1)(x-1)}=-\frac{3(x+2)}{x+1}$

So, your original function has a vertical asymptote at $\displaystyle x=-1$, but a hole in the graph at $\displaystyle x=1$.