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Thread: A horizontal asymptote I'm not seeing...

  1. #1
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    A horizontal asymptote I'm not seeing...

    Alright, in the back of the book it says the horizontal asymptote is $\displaystyle y = 2$. I know that's true because when I went and graphed it, the lines were scared of that number.

    Anyway, here is the equation:

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

    $\displaystyle n < m$ makes $\displaystyle y= 0 $ is what I thought but there's a two.

    No idea what to do here.

    Wouldn't getting rid of the bottom make it:
    $\displaystyle f(x) = 2+ \frac{1}{x}$
    $\displaystyle f(x) = 2x+1$

    Unless...

    $\displaystyle \frac{2}{1}+\frac{1}{x}$

    $\displaystyle \frac{2x}{x}+\frac{1}{x}$

    $\displaystyle = \frac{2x+1}{x}$

    which makes the horizontal asymptote 2...

    Never saw an example period of that though.
    Last edited by A Beautiful Mind; Oct 2nd 2009 at 11:10 AM.
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  2. #2
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    Talking

    I don't understand how you're "getting rid of the bottom"...? This isn't an equation that you're solving, so you can multiply through by x or something; it's a functional statement containing a fraction.

    To find the horizontal asymptote, restate the rational function as one combined fraction:

    $\displaystyle 2\, +\, \frac{1}{x}\, =\, \frac{2x}{x}\, +\, \frac{1}{x}\, =\, \frac{2x\, +\, 1}{x}$

    Then apply the asymptote rules they've given you.
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  3. #3
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    $\displaystyle y=f(x)$
    $\displaystyle f(x)=\frac{1}{x}$

    picture the graph of this in your mind. you know it has two asymptotes x=0 and y=0.

    $\displaystyle f(x)+2$ is a vertical translation so move all points up. the Y axis is still an asymptote but the x axis has moved up 2.

    $\displaystyle \begin{array}
    \\\lim
    \\x\rightarrow+\infty
    \end{array}
    $
    Last edited by sammy28; Oct 4th 2009 at 01:41 AM. Reason: added latex
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