Which of the following equations does not define y as a function of x?
A. xy^2 = 8
Solve for y by dividing both sides by x, (where x is not 0)
That does not define a function because there is NOT just
ONE y for each x, since there are TWO y's for each x.
For example when x = 2,
This is not a function because, for example, when x = 2, there are
TWO values for y, +2 and -2. So that's why it doesn't define a function.
B. x^2 + y^2 = 0
There is only one ordered pair that satisfies that.
That is (x,y) = (0,0)
It does define a function whose domain is {0} and whose range is {0}
It defines a function because there is just one y, 0, for each x, 0,
since the only x and y are both 0.
So, it's a function because there is just one y for every x.
C. 3y = 7
y = 7/x
That defines a function because there is just one y for each x.
D. y = square root of x - 3
That defines a function because there is just one y for each x.
Edwin