Question

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

Answer 1

`y=3/2(xln(2)+1)` for `x to 0`

Explanation:

In order to find eventual asymptotes, you have to find all critical points.
Also, `y=e^ln(x)<=>3*2^(x-1)=e^ln(3*2^(x-1))=3e^((x-1)ln(2))=3/2e^(xln2)`

`y^'=(3ln2)/2e^(xln2)`

Now, critical points exist if at any points, `y^'=0`, but `e^x` never decrease to 0 for any `x in RR`, so there is no asymptotes in `+-oo` for function `y`, however, when `x to 0`, you can use a limited development to find an oblic asymptote :

`y=3/2(xln(2)+1)`

\0/ Here's our answer !