# Calculus

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

1. 0
2. 3
1. y^4 = y^2 - x^2
4y^3 y' = 2y y' - 2x
y' = -2x/(4y^3-2y)

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?

1. 0
posted by Steve
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?

1. 0
posted by Ashley
3. if x=0,

y^4 = y^2
y^2(y^2-1) = 0
y = 0 or ±1
but y=0 is not a possibility.

1. 0
posted by Steve
4. 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?

1. 0
posted by Ashley
5. Hello? This is calculus. Have you forgotten you algebra I?

y^4 = y^2
y^4 - y^2 = 0
y^2(y^2-1) = 0
...

1. 0
posted by Steve

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