Y=x3 is that a function?
1 answer:
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Actually Welcome to the concept of expo functions.
f(x) = -8(2)^x - 12 ,
for f(0) ,here substitute x = 0
so we get as ,
==> f(0) = -8(2)^0 -12
==> f(0) = -8-12
==> f(0) = -20
hence, f(0) = -20