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Math Help - Finding an asymptote

  1. #1
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    Finding an asymptote

    how do you find the asymptotes of a curve like:

    (x+3)/(x^2+4)

    where the denominator is a power for eg.

    and why do you integrate 3/x to get 3lnx and not -9x^{-3}??

    help very appreciated thanks!!
    Last edited by mr fantastic; May 13th 2010 at 08:43 PM. Reason: Fixed some latex
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  2. #2
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by Yehia View Post
    how do you find the asymptotes of a curve like:

    (x+3)/(x^2+4)

    where the denominator is a power for eg.

    and why do you integrate 3/x to get 3lnx and not -9x^-3??

    help very appreciated thanks!!
    To get the asymptotes, test its end behaviors by taking both limits as x\rightarrow \infty and as x\rightarrow -\infty. Both of which are 0, therefore there is only one asymptote which is y = 0.

    To answer your second question, the reason it turns into y = ln |x| is because when you are integrating using the reverse power rule which says

     \int {x}^{r}dx = \frac{{x}^{r+1}}{r+1}

    however this only works if  r \neq -1, because then denominator becomes 0. In your case, we have  y = 3{x}^{-1}, so r = -1 so it doesn't work. So we use  y = ln |x| . Make sure that it's absolute value of x and not just regular x. If you don't put the absolute value signs, you'll get it wrong.

    Cheers!
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Yehia View Post
    and why do you integrate 3/x to get 3lnx and not -9x^{-3}?
    A better question is, why do you think that the integral of 3/x is -9x^{-3}?
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