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Math Help - Help trying to find horizontal Asymptotes.

  1. #1
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    Help trying to find horizontal Asymptotes.

    [itex]y=\frac{x}{x^2+5x+4x}[/itex]

    I am stuck on finding the horizontal asymptotes. I know for sure that there is one at y=0. I have graphed this on my calculator and it seems that there is another between 0 and 1.
    Am i on the right track?
    I did take the inverse of this function and solved it and it came up with 3 answers.
    Last edited by johnsy123; March 29th 2014 at 08:25 PM.
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  2. #2
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    Re: Help trying to find horizontal Asymptotes.

    Quote Originally Posted by johnsy123 View Post
    y=\frac{x}{x^2+5x+4x}

    I am stuck on finding the horizontal asymptotes. I know for sure that there is one at y=0. I have graphed this on my calculator and it seems that there is another between 0 and 1.
    Am i on the right track?
    I did take the inverse of this function and solved it and it came up with 3 answers.
    Is that x after the 4 in the denominator correct? Is this actually

    $y=\dfrac{x}{x^2+9x}$ or do you mean $y=\dfrac{x}{x^2+5x+4}$
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    Re: Help trying to find horizontal Asymptotes.

    Quote Originally Posted by romsek View Post
    Is that x after the 4 in the denominator correct? Is this actually

    $y=\dfrac{x}{x^2+9x}$ or do you mean $y=\dfrac{x}{x^2+5x+4}$
    Hey, yes the one on the right is correct......x/(x^2+5x+4).
    Apologies for the poorly inputed text.
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  4. #4
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    Re: Help trying to find horizontal Asymptotes.

    Quote Originally Posted by johnsy123 View Post
    Hey, yes the one on the right is correct......x/(x^2+5x+4).
    Apologies for the poorly inputed text.
    You find horizontal asymptotes by letting $x \to \pm \infty$. In this case 0 is the only horizontal asymptote.

    This is seen by noting that as $x \to \pm \infty$ the $x^2$ term in the denominator overwhelms the others and you are left with $\dfrac{x}{x^2}=\dfrac{1}{x}$

    Clearly $\dfrac{1}{x} \to 0$ as $x \to \pm \infty$

    Do you also need to find the vertical asymptotes?
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