How do you graph #f(x)=-(2/3)^x+3# and state the domain and range?

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Answer 1 · 2017-11-11T12:06:45

See below.

#y=-(2/3)^x+3#

First find any #y# axis intercepts. These will occur where #x=0#:

#y=-(2/3)^0+3=2#

#y# axis intercept at #( 0 , 2 )#

#x# axis intercept when #y=0#

#-(2/3)^x+3=0#

#(2/3)^x=3#

#xln(2/3)=ln(3)=>x=(ln(3)/(ln(2/3)))~~-2.71#

#(-2.71 , 0 )#

as #x -> oo# , #color(white)(88)-(2/3)^x+3->3#

as #x -> -oo# , #color(white)(88)-(2/3)^x+3->-oo#

The line #y=3# is a horizontal asymptote:

Graph:

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