What is the vertical asymptote of x^2-16/x-8

Question

ipwnbunnies · Answer

The vertical asymptote occurs where the denominator is equal to 0. Because anything is defined when the denominator is 0.
Also, there won't be a vertical asymptote if the denominator factors into the numerator.
:3
That's not the case here, so solve the bottom for 0. That'll give you x = ?

ipwnbunnies · Answer

.-.  No.
The denominator of your function is (x-8). The vertical asymptote occurs at the x value that makes this 0.
x-8 = 0
Therefore, x = 8. There will be a vertical asymptote going through x value 8.

anonymous · Answer

>.< im not good at this stuff at all i have been looking for ways to do it right now.... all over google and using the online book they gave me but nothing seemed to help

ipwnbunnies · Answer

Oh, I'm sorry to hear that. .-.

anonymous · Answer

Thank you for helping me out i need to really learn how to do all of this again ._.

anonymous · Answer

OOOOOoooooooooh ok i get how to do that one now!!! thank you for the help lol

ipwnbunnies · Answer

:)