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-   -   An exponential equation (https://www.askmehelpdesk.com/showthread.php?t=66831)

  • Feb 26, 2007, 10:46 PM
    Aerlinn
    An exponential equation
    This equation: f(x)= (x-2)2^(x-1)
    Says to state the equation of the horizontal asymptote.

    How do I do I do this without using a calculator?
  • Feb 27, 2007, 12:45 AM
    Capuchin
    Well the asymptote is horizontal, so it's of the form y = n, where you need to find n.

    You need to find the point where x --> infinity
  • Feb 27, 2007, 01:13 AM
    Aerlinn
    And... how is that done... algebraically?
  • Feb 27, 2007, 02:20 PM
    asterisk_man
    we need to take limits when x goes to + and - infinity



    Therefore, the horizontal asymptote is y=0.

    A plot of this function confirms this result.
  • Feb 27, 2007, 09:01 PM
    Aerlinn
    I see... Thanks :)
    I don't understand the lim x2^x=0 bit however... :S
  • Feb 28, 2007, 07:38 AM
    asterisk_man
    you don't understand limits or you don't understand why x2^x is going to 0?
  • Mar 14, 2007, 02:00 AM
    Aerlinn
    Don't understand why x2^x is going to 0...
  • Mar 14, 2007, 08:08 AM
    asterisk_man
    The basic idea is that as x becomes negative 2^x is going to be getting closer and closer to 0 and x is going to be getting more and more negative. However 2^x is going to be going to 0 faster than x is going to - infinity so 2^x wins.

    Try a few examples:
    -1 * 2^-1 = -1 * 1/2 = -1/2 = -0.5
    -2 * 2^-2 = -2 * 1/4 = -1/2 = -0.5
    -3 * 2^-3 = -3 * 1/8 = -3/8 = -0.375
    -4 * 2^-4 = -4 * 1/16 = -1/4 = -0.25
    -5 * 2^-5 = -5 * 1/32 = -5/32 = -0.15625
    -6 * 2^-6 = -6 * 1/64 = -3/32 = -0.09375

    you can see its getting smaller and smaller. The magnitude of x is increasing by 1 each time but the magnitude of 2^x is halving each time.

    do you follow now?

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