
Steve S. answered 01/09/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
x^2=y^2
sqrt(y^2) = sqrt(x^2)
abs(y) = abs(x)
y = ±x
Since there are two values of y for each value of x,
y is not a function of x.
y is a relation of x. Here are four points that satisfy the original relation:
(-2, -2), (-2, +2), (+2, +2), (+2, -2).