The rational function is
[tex]\frac{2x+2}{x^2-1}[/tex]
For calculate vertical asymptote we need to find which value of x make zero the denominator
[tex]\begin{gathered} x^2-1=0 \\ (x+1)(x-1)=0 \\ x=1 \\ x=-1 \end{gathered}[/tex]
And for the numerator
[tex]\begin{gathered} 2x+2 \\ 2(x+1) \\ \end{gathered}[/tex]
The whole expression
[tex]\begin{gathered} \frac{2(x+1)}{(x+1)(x-1)} \\ \frac{2}{(x-1)} \end{gathered}[/tex]
In this case, the vertical asymptote is at -1
