How do you graph #y=2^(2x)# using a table of values?

1 Answer
Gió
Feb 2, 2016

I tried few values:

Explanation:

I would use "friendly" values for #x# to avoid difficult and complicated evaluations (basically, values that I can easily evaluate).
I'll try:
#x=-1# so that #y=2^(-2)=1/4=0.25#
#x=-1/2# so that #y=2^(-2*1/2)=2^-1=1/2=0.5#
#x=0# so that #y=2^0=1#
#x=1/2# so that #y=2^(2*1/2)=2^1=2#
#x=1# so that #y=2^(2*1)=4#
#x=3/2# so that #y=2^(2*3/2)=2^3=8#

Graphically:
enter image source here

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