Question #c3ce8

1 Answer
Dec 19, 2017

It is a function.

Explanation:

Well, first remember that #f(x)# is a function when only one #y# value can match to a specific #x# value.

Ok. Let us now solve #y# this in terms of #x#, assuming #y=f(x)#.

#x=2y^2#
#x/2=y^2#
#sqrt (x/2)=y#

We can see that there can only be one value of #y# for a specific value of #x#. Therefore, this is a function. (Limited, of course, since there can't be negative #x# values defined as a real number.