# How do you find all the asymptotes for function y=3(2)^(x-1) ?

Aug 3, 2018

$y = \frac{3}{2} \left(x \ln \left(2\right) + 1\right)$ for $x \to 0$

#### Explanation:

In order to find eventual asymptotes, you have to find all critical points.
Also, $y = {e}^{\ln} \left(x\right) \iff 3 \cdot {2}^{x - 1} = {e}^{\ln} \left(3 \cdot {2}^{x - 1}\right) = 3 {e}^{\left(x - 1\right) \ln \left(2\right)} = \frac{3}{2} {e}^{x \ln 2}$

${y}^{'} = \frac{3 \ln 2}{2} {e}^{x \ln 2}$

Now, critical points exist if at any points, ${y}^{'} = 0$, but ${e}^{x}$ never decrease to 0 for any $x \in \mathbb{R}$, so there is no asymptotes in $\pm \infty$ for function $y$, however, when $x \to 0$, you can use a limited development to find an oblic asymptote :

$y = \frac{3}{2} \left(x \ln \left(2\right) + 1\right)$