Question

What is/are the vertical asymptote(s) for `y=(x^2+2x)/(x^2+5x-6)`?

Answer 1

`x=-6` and `x=1`

The vertical asymptotes are where the denominator is equal to zero, because you can't divide by that. Therefore you should use factorisation here for the denominator: `x^2+5x-6 = (x+6)(x-1)`.
We set this equal to zero, and find `x=-6` and `x=1`.

Answer 2

To find the vertical asymptote of any function all we need to do is find where this function is undefined.
If a function is undefined when its' denominatore equals 0 (dividing by 0) then we can find the vertical asymptote by taking the denominator and setting it equal to 0.

`0=x^2 + 5x -6`

Now we can factor this piece of the function.

`0=(x+6)*(x-1)`

We have 2 vertical asymptotes because we have 2 numbers that make the denominator=0

The denominator is 0 if `x=-6` or if `x=1` so your 2 vertical asymptotes are `-6,1`.