Solve Oblique Asymptote & Graph: f(x)=x^3+3x^2-x-5/x^2-1

[Johnyi]

Homework Statement

Find the oblique asymptote and sketch the graph

Homework Equations

f(x)=x^3+3x^2-x-5/x^2-1

The Attempt at a Solution

My question is that shouldn't there be a hole in the graph when x=-1 or 1? The denominator would end up being zero.

When should i be using an hole or when do i know it is an asymptote?

 

[Mentallic]

If you have a rational function of the form $$f(x)=\frac{x-b}{x-a}$$ then there doesn't exist a hole in the function at x=a, instead there is a vertical asymptote at that point. You will have a hole in the function when it is of the form $$f(x)=\frac{(x-a)(x-b)}{x-a}$$ since this simplifies into the linear equation $$f(x)=x-b$$ but at x=a there is a hole.

So basically, if the fraction is of the form 0/0 at some point x=a, then it could be a hole (since x=a is a zero of both the numerator and denominator, you can factor out x-a from both and then cancel). If it is of the form a/0 for some non-zero a, then it is a vertical asymptote.