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

2 Answers
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+5x-6 = (x+6)(x-1)#.
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 + 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#.