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

Dec 18, 2014

$x = - 6$ 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: ${x}^{2} + 5 x - 6 = \left(x + 6\right) \left(x - 1\right)$.
We set this equal to zero, and find $x = - 6$ and $x = 1$.

Dec 18, 2014

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.

$0 = {x}^{2} + 5 x - 6$

Now we can factor this piece of the function.

$0 = \left(x + 6\right) \cdot \left(x - 1\right)$

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 $- 6 , 1$.