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Math Help - All asymptotes for this function.

  1. #1
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    All asymptotes for this function.

    How can i calculate all asymptotes (horizontal and vertical) for this function.

    (6) / (1+2e-5x​)
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  2. #2
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    Re: All asymptotes for this function.

    To find vertical asymptotes, equate the denominator to zero, and solve for x.

    To find horizontal asymptotes, compute:

    \lim_{x\to\pm\infty}\frac{6}{1+2e^{-5x}}

    What do you find?
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  3. #3
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    Re: All asymptotes for this function.

    Horizontal:

    6 /(1 + (2/∞))= 6

    vertical:

    1+2e-5x=0

    2*(-5x)=-ln1

    x= ln1/10 = 0.

    These are the results i got when i tried earlier, however with f(x) = 0 i get x as 2, so the vertical cant be 0. Was wondering if there might be two horizontal
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: All asymptotes for this function.

    You have found one horizontal asymptote, try the limit as x goes to negative infinity as well.

    You are incorrectly solving for x to find any possible vertical asymptotes:

    1+2e^{-5x}=0

    e^{-5x}=-\frac{1}{2}

    Without taking logs, we should know that:

    0<e^{-5x}

    for all real x, hence there are no vertical asymptotes.
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  5. #5
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    Re: All asymptotes for this function.

    I dont really see how using -∞ will change the results, but again i'm asking here as i dont know how to do this.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: All asymptotes for this function.

    Since the exponent is negative, the denominator goes to 1 as goes go to infinity and goes to infinity as x goes to negative infinity.

    6/1 = 6 so y = 6 is the horizontal asymptote on the right.

    6/∞ = 0 so y = 0 is the horizontal asymptote to the left.
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