# Thread: 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 $\displaystyle y=2^x-1$, 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.

3. 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..