1. The problem statement, all variables and given/known data Write an equation of each horizontal tangent line to the curve. 2. Relevant equations y = 2y^3 + 6x^2y - 12x + 6y = 1 y' = (4x - 2xy) / x^2 + y^2 + 1) 3. The attempt at a solution Well, horizontal tangent line means the derivative equals zero. Thus, 4x - 2xy = 0 2x(2-y) = 0 x = 0, y = 2 Since I need an equation, I plug x = 0 back into the original function and end up with 2y^3 + 6y = 1 Then I need to solve for y. I guess this is more of an algebra question than a calculus question? But nevertheless, I don't know how to solve for y.. I could do 2y(y^2 + 3) = 1 but that doesn't help me.