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Asymptotes and holes

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the oblique asymptote and sketch the graph

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


    3. The attempt at a solution

    My question is that shouldnt 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?
     
  2. jcsd
  3. Oct 1, 2011 #2

    Mentallic

    User Avatar
    Homework Helper

    If you have a rational function of the form [tex]f(x)=\frac{x-b}{x-a}[/tex] 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 [tex]f(x)=\frac{(x-a)(x-b)}{x-a}[/tex] since this simplifies into the linear equation [tex]f(x)=x-b[/tex] 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.
     
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