# Math Help - explain how it is the asymptote plz

1. ## 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!

2. For $y=2^x-1$, the asymptote should in fact be $y=-1$.

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

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

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

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

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

With $y=2^x-1$, the asymptote has simply been translated down, since now $2^x$ can go 1 lower than it used to.

3. to simplify what he wrote, the "standard" equation $y=a^x$ 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 $y=a^x$, the graph will shift up/down, and so does the asymptote..