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Math Help - Finding asymptotes.

  1. #1
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    Finding asymptotes.

    firstly, how do you get from a function like: x-1 + (1/x-3) to the original function before long division?

    and secondly i know how you find the asymptotes of a funtion like:
    (x^2 + 3)/(x+2)
    but how do you get them if it is:
    (x+2)/(x^2+3)

    ??

    cheers and thanks very appreciated!
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Yehia View Post
    firstly, how do you get from a function like: x-1 + (1/x-3) to the original function before long division?

    \color{red}{ x-1 + \frac{1}{x-3} = \frac{(x-3)(x-1)+1}{x-3} = \frac{x^2-4x+4}{x-3}}

    and secondly i know how you find the asymptotes of a funtion like:
    (x^2 + 3)/(x+2)
    but how do you get them if it is:
    (x+2)/(x^2+3)

    \color{red}{\text{No vertical asymptotes, since the denominator never = 0.}} \color{red}{\text{Find horizontal asymptotes the same way you would for the other problem.}}

    ??

    cheers and thanks very appreciated!
    aNon1
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  3. #3
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    Quote Originally Posted by Anonymous1 View Post
    aNon1

    thanks a lot!!! ok i get it now, but i still dont see how you find the asymptote...?
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  4. #4
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Yehia View Post
    thanks a lot!!! ok i get it now, but i still dont see how you find the asymptote...?
    For vertical asymptotes set the denominator =0 and solve for x.

    For horizontal:

    To find horizontal asymptotes:

    If the degree (the largest exponent) of the denominator is bigger than the degree of the numerator, the horizontal asymptote is the x-axis (y = 0).
    If the degree of the numerator is bigger than the denominator, there is no horizontal asymptote.
    If the degrees of the numerator and denominator are the same, the horizontal asymptote equals the leading coefficient (the coefficient of the largest exponent) of the numerator divided by the leading coefficient of the denominator
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