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Finding oblique asymptotes

  1. Jul 11, 2005 #1
    I'm having trouble figuring the oblique asymptote for this problem...

    I haven't ever done one of these before so I really don't know where to start.

    Here's the problem:

  2. jcsd
  3. Jul 11, 2005 #2


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    http://home.att.net/~srschmitt/precalc/precalc-12-03-00.html [Broken]
    Last edited by a moderator: May 2, 2017
  4. Jul 11, 2005 #3
    Hmm... I'm still not quite getting it.

    I need to factor the numerator then, correct? But it comes out as an imaginary number?
  5. Jul 11, 2005 #4


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    You only need to do the polynomial division, so you can write it in the form of Quotient + Remainder/Divisor, because of Euclides Algorithm Numerator = Quotient*Denominator + Remainder.

    The Oblique Asymptote will be the quotient.
  6. Jul 12, 2005 #5


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    The whole point of oblique (or horizontal) asymptotes is "what happens as x goes to infinity or negative infinity?"

    With something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] the simplest thing to do is to divide both numerator and denominator by the largest power of x in the denominator- here just x. [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]. Since those fractions with x in the denominator will go to 0 as x goes to either positive or negative infinity, it's easy to see that y approaches 2x+ 5.
  7. Jul 12, 2005 #6


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    I had an oblique asymptote question on my Calc I exam and I totally forgot how to do that. What I eventually did was to get an equation for the tangent at the point (a,f(a)) and then let a approach infinity, figuring that the tangent will coincide with the asymptote in that limit. Although I felt pretty good after coming up with that idea at the time and the answer was correct, I still received a big red cross through my solution :(
  8. Jun 16, 2007 #7
    Just pointing out something...I know it's a really old thread !

    Someone pointed out earlier that for something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] you can just divide by "x" and get that the asymptote is: [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]

    However, that is not the case I believe. The asymptote is actually 2x+3, and it is found by:

    = Lim x -> infinity, f(x) - (mx+b) = 0

    So I would do long division, get : 2x + 3 + 8/(x+1) and then use the formula mentioned above:

    = lim x-> infinity of 2x + 3 + 8/(x+1) - (2x+3)

    that's equal = 0, hence the mx+b part is an oblique asymptote. Which is, 2x+3 = oblique asymptote in this case.

    Hope this helps!
  9. Nov 10, 2009 #8
    Hi, dale 123....you can check your answer by graphing it. Use a program if you're having trouble with the graph. Once it's graphed you'll see that your answer is not right. The correct answer is still 2x+5.
  10. Jan 1, 2010 #9
    You're both wrong, dale 123 is correct. To find the oblique asymptote, you must use polynomial long division, and then analyze the function as it approaches infinity.

    Taking the limit first, like HallsofIvy did, is wrong because 11/x and 1/x approach infinity at different rates, and therefore add to the numerator and denominator in slightly different ways.

    If you are not finding the limit at infinity of the entire function, as in this problem, then you MUST write it in the form "term + term + ... + term", and analyze each individual term as x approaches infinity. This is because fractions can do unexpected things when it comes to limits.
  11. Jan 1, 2010 #10


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    Oh, I am so embarassed! When I read my post I thought "wait, that's not right, you divide the polynomial to find the asymptote!". Don't know what I was thinking 5 and a half years ago. Thanks, dale123 and erh5060, for fixing that.
  12. Jan 1, 2010 #11
    So dividing by the largest power of x works for horizontal asymptotes but not generally for slant asymptotes. That's good to know.

    Forgive me; you mean years ago. :wink:
  13. Jan 2, 2010 #12


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    Ahh, yes, of course! I didn't make a mistake- what I wrote clearly must have been true 4½ years ago but not now! :rofl:
  14. Jan 3, 2010 #13
    There is also a nice way to find oblique assymptotes for any function in general, not only rational functions, since the method described here works only for rational functions.

    Let f(x) be sucha function and assume that L=ax+b is the oblique assymptote. Let P be any point in the graph of f and Q be a point in the line L, such that d(P,Q) is the shortest distance. Also let denote by K the point on the line L when we draw a perpendicular segment from point P to the x-axis. Now, after a little bit of geometry we find that L is an oblique assymptote iff

    [tex]\lim_{x\rightarrow \infty}(f(x)-L)=lim_{x\rightarrow \infty}(f(x)-ax-b)=0[/tex]

    If we divide the whole thing by x, and take the limit we find that:



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