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

Question

Theodore Amidon · Accepted Answer

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

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 !

Sadie Alexander · Answer

undefined To find the asymptotes of the function $ y = 3 \cdot 2^{x-1} $, we need to consider horizontal, vertical, and slant asymptotes.

1. Horizontal asymptote: There is no horizontal asymptote because as $ x $ approaches positive or negative infinity, $ 2^{x-1} $ grows without bound, causing the entire function to grow without bound as well.

2. Vertical asymptote: There are no vertical asymptotes because the function is defined for all real numbers $ x $.

3. Slant asymptote: There is no slant asymptote because the degree of the numerator (which is a constant, $3$) is less than the degree of the denominator (which grows without bound as $ x $ increases or decreases). Therefore, there is no slant asymptote.

In conclusion, the function $ y = 3 \cdot 2^{x-1} $ has no asymptotes.