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Help find the asymptote!

  1. Aug 20, 2009 #1
    Can anyone please explain to me how to find the asymptote of a function?




    Your help is much appreciated!
     
  2. jcsd
  3. Aug 20, 2009 #2
    There are two types of asymptotes horizontal and vertical.

    If there are breaking points of the function( ex. f(x)=1/x, breaking point x=0), there is vertical asymptote.

    If [tex]\lim_{x \to \infty} f(x) = a \,\mbox{ or } \lim_{x \to -\infty} f(x) = a[/tex] then there is horizontal asymptote.

    Its pretty easy to find them.

    For more info, read the Wikipedia Article.

    Regards.
     
  4. Aug 20, 2009 #3

    HallsofIvy

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    You can also have "slant" (or "oblique") asymptotes. For example,
    [tex]y= \frac{3x^2- 2}{x- 1}[/tex]
    has a vertical asymptote at x= 1 (because then the denominator x-1 is 0).Also, dividing the numerator by the denominator gives y= 3x+ 3+ 1/(x-1). For very large x, the fraction, 1/(x-1) goes to 0 so the graph approaches the "slant asymptote" y= 3x+ 3.
     
  5. Aug 20, 2009 #4

    HallsofIvy

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    Of course, you can have many different vertical asymptotes but at most two horizontal or slant asymptotes ([itex]x\to\infty[/itex] and [itex]x\to -\infty[/itex] may be different) and you cannot have both horizontal and slant asymptotes on the same "side" of the graph.

    Note that the denominator being 0 does not necessarily mean a vertical asymptote. For example, [tex]y= \frac{x^2- 1}{x- 1}[/itex] has denominator 0 but its graph is just the line y= x+ 1 with a "hole" at (1, 2).
     
  6. Aug 21, 2009 #5
    You can also have a parabolic asymptote! I discovered this in my math class last year and it was fascinating!

    Try graphing this function: [tex]\frac{x^3}{x - 1}[/tex]

    Indeed, in any rational function where the numerator is a polynomial of a higher degree than the denominator, the function will have an asymptote that is shaped like the remainder function! So if you have a cubic numerator and a linear denominator, the degree difference is 3 - 1 = 2 and it follows that your asymptote will be parabolic. Dividing a quintic by a parabola gives you a cubic-shaped asymptote.

    Try graphing this one, too: [tex]\frac{x^5 - x^4 - x^3 + 1}{x^2 - x - 1}[/tex]

    (This is so fun :smile:)
     
  7. Aug 21, 2009 #6

    Mentallic

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    Could you please explain this a little more clearly?

    It might be worthy to note that this is the case because both the numerator and denominator have equal factors (x-1)
    So, as [tex]x\rightarrow 1[/tex] both the numerator and denominator tend towards 0, and instead of having a vertical asymptote because of having the undefined [tex]\frac{a}{0}[/tex] where [tex]a\neq 0[/tex], we have [tex]\frac{0}{0}[/tex] which is also undefined, but different such that the value is some real, finite value. In this case, 2, which thus makes me wonder why the function should be stripped of its point (1,2).

    I like your enthusiasm! :biggrin:
     
  8. Aug 21, 2009 #7

    Oblique and horizontal asymptotes on the "right hand side" of the graph are determined by how f(x) behaves as x tends to infinity. If there exists an oblique asymptote on the right hand side, then f(x) is going to infinity or negative infinity (for large enough x the graph starts looking like a straight line with a nonzero slope). The horizontal asymptote will exist on the right hand side if and only if f(x) tends to a constant value as x tends to infinity, so you obviously can't have both on the same side.
     
    Last edited: Aug 21, 2009
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