How do you graph #y=(1/2)*3^x#?

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Answer 1 · 2017-10-08T18:42:21

See below.

First you need to find critical points:

For: #y=(1/2)*3^x#

#y# axis intercept occurs when #x=0#. So:

#y= (1/2)*3^0=> y=(1/2)*1=> color(red)(y=1/2)#

As:

#lim_(x->oo^+)(1/2)*3^x= oo#

For #x<0#

#3^x->1/(3^x)#

So:

#lim_(x->oo^-)(1/2)*3^x=0#

The #color(red)(x)# axis is a horizontal asymptote:

Hope this helps.

Graph of #y=(1/2)*3^x#

graph{y=(1/2)*3^x [-16.02, 16.02, -8.01, 8.01]}