posted by Anonymous .
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?
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
would it by (2,sqrt(10)/2)?