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Thread: Show the graph of a function intersects its horizontal asymptote at one point

  1. #1
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    Show the graph of a function intersects its horizontal asymptote at one point

    the function is $\displaystyle \frac{2(x^2+1)}{(x-1)(x+2)}$

    So it will have a vertical asymptote $\displaystyle x\rightarrow 1^+, \,\,\, x\rightarrow 1^-, \,\,\, x\rightarrow 2^+, \,\,\, x\rightarrow 2^-$ at $\displaystyle x =1\,\,\, and\,\,\, x = -2 $

    And it will have a horizontal asymptote at the line $\displaystyle y=2$

    and to determine whether the graph approaches the horizontal asymptote from above or below

    $\displaystyle f(x) - 2 = \frac{-2x+6}{(x-1)(x+2)}$

    as $\displaystyle x\rightarrow+\infty, \,\, f(x)<2$ and $\displaystyle x\rightarrow -\infty, \,\,\,f(x) >2$

    And because $\displaystyle \lim_{x\rightarrow1^+}f(x) = +\infty$ but as $\displaystyle x\rightarrow+\infty,\,\, f(x) < 2$

    Thus the graph would intersect the horizontal asymptote at $\displaystyle y=2$ being the point $\displaystyle (3,2)$

    Is this correct?
    Last edited by Jonroberts74; Oct 29th 2013 at 02:54 PM.
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  2. #2
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    Re: Show the graph of a function intersects its horizontal asymptote at one point

    Hey Jonroberts74.

    It all looks good.
    Thanks from Jonroberts74
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