Posted by **Anonymous** on Sunday, November 18, 2007 at 12:46am.

At what point(s) does the curve defined by x^2-6x=6-6y^2 have horizontal tangents?

I took the derivative and solved for y'. I got y'=-1/6x/y+1/2y. hmm...what am I suppose to do?

- Calculus -
**Reiny**, Sunday, November 18, 2007 at 1:24am
leave your derivative as

dy/dx = (2x-6)/(-12y) after you differentiated implicitly.

then (2x-6)/(-12y) = 0 can only be true if

2x-6 = 0

x = 3 now put that back into the original equation to get y = ...

your equation will be y = that value

- Calculus -
**Anonymous**, Sunday, November 18, 2007 at 11:51am
would it by (2,sqrt(10)/2)?

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