Posted by **Ashley** on Monday, October 1, 2012 at 3:00pm.

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

- Calculus -
**Steve**, Monday, October 1, 2012 at 3:12pm
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?

- Calculus -
**Ashley**, Monday, October 1, 2012 at 3:27pm
how did you know it was + or - l? I got what x equals but I was confused on how to find the y values?

- Calculus -
**Steve**, Monday, October 1, 2012 at 3:29pm
if x=0,

y^4 = y^2

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

y = 0 or ±1

but y=0 is not a possibility.

- Calculus -
**Ashley**, Monday, October 1, 2012 at 3:43pm
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?

- Calculus -
**Steve**, Monday, October 1, 2012 at 4:16pm
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

...

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