Question

Give the location of the vertical asymptote(s) if they exist, and state the functions domain.
I believe the textbook's solution may be incorrect

Answer 1

\[h(x)=\frac{x^2-1}{2x^2+3x-5}\]

Answer 2

\[h(x)=\frac{(x-1)(x+1)}{(x-1)(2x+5)}\]
it has 1 vertical asymptote at x=-5/2

Answer 3

I factored them and got
\[h(x)= \frac{(x+1)(x-1)}{(x+\frac{5}{2})(x-1)}\]

Answer 4

I wrote that the same v.a value but my book states x=1 also.
I believe that is a hole.
For the domain