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

1 Answer
Aug 3, 2018

Answer:

#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 !