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Vertical asymptotes

  1. Sep 17, 2004 #1
    The question is
    fine a equation f(x) that has a vertical asymptotes of X=1and X=3 and a horizontal asymptotes of Y=1

    A) (x-1)(x-3) / y-1 = x^2-4x+3 / x-1


    B) (x-1) /(x-1)(x-3) = just the oppsite of above

    is any of these correct?
    I think B is correct but right now I can't find any info on it in my book
  2. jcsd
  3. Sep 17, 2004 #2


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    Homework Helper

    I think it won't be correct, the Horizontal asymtopte for B will be y=0.

    If i remember correctly the horizontal asymptote is

    [tex] \lim_{x \rightarrow \infty} \frac{(x-1)}{(x-1)(x-3)} [/tex]

    The above case will give y=0. The numerator must of the form
    [tex] x^2 + Bx +C [/tex] where B and C don't allow factors equal to those of the denominator.

    To make myself clearer:

    [tex] \lim_{x \rightarrow \infty} \frac{Ax^2 +Bx +C}{x^2-4x+3} [/tex]

    [tex] \lim_{x \rightarrow \infty} \frac{A +\frac{B}{x} +\frac{C}{x^2}}{1-\frac{4}{x}+\frac{3}{x^2}} = \frac{A}{1}[/tex]

    [tex] \frac{A}{1} = 1 [/tex]

    because horizontal asymptote should be y = 1
    Last edited: Sep 17, 2004
  4. Sep 18, 2004 #3
    B gives you a removable discontinuity at x=1.

    Try this:

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