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

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Answer 1 · 2018-07-07T19:30:59

$y=1$

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
$lim_(x \to -infty)2^x=0$

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