Question

x^2+x-2/(x-1)(x+3)
determine the vertical asymtotes

Answer 1

set denominator = 0 ,
find respective values of x.
Those values of will represent your vertical asymptotes.

Answer 2

As vertical asymptotes are the y-coordinates reaching positive or negative infinity, we would expect \[y\rightarrow \pm \infty\] so \[\frac{x^2+x-2}{(x-1)(x+3)}\rightarrow \pm \infty\] hence \[\frac{1}{(x-1)(x+3)}\rightarrow\pm\infty\]
This implies
\[(x-1)(x+3)=0\]
Then solve for the value for x and you would have
\[x=1\]or\[x=-3\]