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Oblique Asymptotes: What happens to the Remainder?

  1. Oct 24, 2011 #1
    Let's say I'm trying to find the oblique asymptote of the function:

    f(x)=
    -3x2 + 2
    x-1

    Forgive my poor formatting.

    So because the denominator isn't linear, we do polynomial long division of the function and ultimately get -3x - 3 as our quotient, with a remainder of -1. For the sake of the oblique asymptote, we disregard the remainder. But why? I haven't found a website that explains it; they all simply say to ignore the remainder. It bothers me that we just forget about it, regardless of what it is.
    Can you explain to me why we disregard the remainder, and where it "goes?"
     
  2. jcsd
  3. Oct 24, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    An "asymptote" is a line that a curve approaches as x goes to, in this case, negative infinity and infinity. Yes, long division gives a quotient of -3x- 3 with a remaider of -1. That means that
    [tex]\frac{-3x^2+ 2}{x- 1}= -3x- 3- \frac{1}{x- 1}[/tex]
    As x goes to either infinity or negative infinity, that last fraction goes to 0.
     
  4. Oct 24, 2011 #3
    That makes perfect sense! Thank you so much!
     
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