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Homework Help: Horizontal asymtotes

  1. Apr 24, 2010 #1
    When the power of the leading coefficient of the numerator of a rational function is a lot greater than the power of the leading coefficient of the denominator , ie

    [tex]f(x)=\frac{x^3+1}{x-1}[/tex]

    The horizontal asymtote is y=x^2 according to the book . Is that true ? Is there any proof for this . I only know when its one power difference , that would be oblique asymtote .
     
  2. jcsd
  3. Apr 24, 2010 #2
    I think that is right , look at the highest power on top and the highest power on the bottom and divide them .
     
  4. Apr 24, 2010 #3

    HallsofIvy

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    I really doubt that your text book says horizontal asymptote. A horizontal asymptote is a straight horizontal line and only occurs in a rational function when the numerator and denominator have the same degree. A "slant" or "skew" asymptote, a non-horizontal line, occurs when the numerator has degree one higher than the denominator.

    I've never seen a curve called an "asymptote" before but since
    [tex]\frac{x^3}{x- 1}= x^2+ x+ 1+ \frac{2}{x-1}[/tex]
    so for large x, the graph comes close to [itex]x^2+ x+ 1[/itex] which itself will be close to [itex]x^2[/itex]. (Though I would say the "asymptote" was [itex]x^2+ x+ 1[/itex], not just [itex]x^2[/itex].)
     
  5. Apr 24, 2010 #4
    thank you . It's not horizontal asymtote , i just didn't know what to call it , is there a name for that kind of asymtote ?

    Is it necessary to include this asymtote in my sketching ? Usually , i would just include the vertical asymtote in my sketching if there is no horizontal asym.
     
  6. Apr 24, 2010 #5
    I think you might be able to call it a slant asymptote.
     
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