How do you graph, find the intercepts and state the domain and range of #f(x)=1/2(2^x-8)#?

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Answer 1 · 2018-07-29T06:45:29

The x-intercept is #(3,0)#. The y-intercept is #(0,-7/2)#.
The domain is #x in RR#. The range is #y in (-4,+oo)#

The function is

#y=1/2(2^x-8)#

The intercepts are

when #x=0#

#=>#, #y=1/2(2^0-8)=1/2(1-8)=-7/2#

The y-intercept is #(0,-7/2)#

when #y=0#

#=>#, #0=1/2(2^x-8)#

#=>#, #2^x-8=0#

#=>#, #2^x=2^3#

#=>#, #x=3#

The x-intercept is #(3,0)#

The domain is #x in RR#

To find the range, proceed as follows

#y=1/2(2^x-8)#

#2y=2^x-8#

#2^x=2y+8#

Taking logs

#ln(2^x)=ln(2y+8)#

#xln2=ln(2y+8)#

Therefore,

#2y+8>0#

#2y>-8#

#y>-4#

The range is #y in (-4,+oo)#

See the graph belox

graph{(y-1/2(2^x-8))(y+4)=0 [-14.24, 14.24, -7.12, 7.12]}