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Thread: explain how it is the asymptote plz

  1. #1
    Junior Member
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    explain how it is the asymptote plz

    Hello.

    Can someone explain to me how the Horizontal Asymptote for this function is -1 (or the y-int. in an exponential function)?

    so the question is ...

    y= 2^x - 1

    I graphed it, and the asymptote is one, I just would appreciate an explanation as to how it is so. Thanks!
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  2. #2
    Senior Member DivideBy0's Avatar
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    For $\displaystyle y=2^x-1$, the asymptote should in fact be $\displaystyle y=-1$.

    To find out why, let's first examine your standard exponential graph $\displaystyle y=a^x$

    If a is positive, then any power of a can never make it negative. Even if you have a negative power, this will just give you a smaller and smaller fraction.

    e.g

    $\displaystyle 2^{-2} = \frac{1}{4}$

    $\displaystyle 2^{-4} = \frac{1}{16}$ etc.

    This is why $\displaystyle y=a^x, a >0$ can never be negative.

    With $\displaystyle y=2^x-1$, the asymptote has simply been translated down, since now $\displaystyle 2^x$ can go 1 lower than it used to.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    to simplify what he wrote, the "standard" equation $\displaystyle y=a^x$ is asymptotic to y=0 for all possible values of a, for all x in R.
    and so when you add/subtract a constant to $\displaystyle y=a^x$, the graph will shift up/down, and so does the asymptote..
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