Question

How to find the asymptotes of `y=1-2^x` ?

Answer 1

`y=1`

Explanation:

We have

`lim_{x to infty)(1-2^x)=-infty`

since the function

`f(x)=1-2^x` is strictly monotonousl decreasing,

`f'(x)=-2^x\ln(2)`

`lim_(x to -\infty}(1-2^x)=1`
since
`lim_(x \to -infty)2^x=0`