Calculus
posted by Ashley .
determine the point(s)at which the graph of y^4 = y^2 x^2 has a horizontal tangent.

y^4 = y^2  x^2
4y^3 y' = 2y y'  2x
y' = 2x/(4y^32y)
Now, assuming that y ≠ 0 or ±1/√2,
y'=0 when x=0
so, y^4 = y^2, and y = ±1, so
(0,1) and (0,1)
extra credit: what happens at y = 0 or ±1/√2? 
how did you know it was + or  l? I got what x equals but I was confused on how to find the y values?

if x=0,
y^4 = y^2
y^2(y^21) = 0
y = 0 or ±1
but y=0 is not a possibility. 
where did you get y^2 1 from? Did you subtract y^2 from the right side? if so where did the 1 come from?

Hello? This is calculus. Have you forgotten you algebra I?
y^4 = y^2
y^4  y^2 = 0
y^2(y^21) = 0
...