Question

what is the removable discontinuity, horizontal and vertical asymptotes for f(x)=(x+1)/(x^2-x-6)

Answer 1

Take the limit of the function as x approaches infinity to get the asymptotes.

Answer 2

but how do I do that?

Answer 3

First things first, that function can be simplified into a single linear term by factoring. That might make it a little easier to see.

For a vertical asymptote, you want to look at the function to see what value would cause the limit to approach infinity at that point, like this:
\[\lim_{x \rightarrow a}=\pm \infty\]For a horizontal asymptote, you want to take the limit as x itself approaches infinity. If it converges to a finite value, that is the location of your asymptote, like this:
\[\lim_{x \rightarrow \pm \infty}f(x)=L\]
Make sense?

Answer 4

I am just studying this in my class and I found my vertical asymptote by finding what makes the denominator 0.

Answer 5

so the vertical asymptote is x=3 and x=-2?

Answer 6

I think so. I am not an expert because I just started studying this.

Answer 7

thanks! do you know how to find the removable discontinuity?

Answer 8

I am trying to figure that out right now. Let me see if I can find the answer.

Answer 9

A removable discontinuity — that’s a fancy term for a hole

Answer 10

Factor both the numerator and denominator. If a term cancels out from both the numerator and denominator then that represents a hole.

Answer 11

So first factor your equation.

Answer 12

This is not you equation.

Answer 13

x-3 cancels out from the numerator and denominator so x=3 is the hole.

Answer 14

ok so then what's the asymptotes?

Answer 15

http://www.wolframalpha.com/widgets/view.jsp?id=21344fc46140bef0640b08781cbb4daf