1. The problem statement, all variables and given/known data Given 2y^3 +(6x^2)y-12x^2+6y = 1 A. Show that dy/dx = 4x -2xy/x^2 + y^2 + 1 B. Write an equation of each horizontal tangent line to the curve C. The line through the origin with slope -1 is tangent to the curve at point P. Find the x-y coordinates of point P. 3. The attempt at a solution A. 6y^2(dy/dx) + 12xy + 6x^2(dy/dx) - 24x + 6(dy/dx) = 0 6y^2(dy/dx) + 6x^2(dy/dx) + 6(dy/dx) = 24x - 12xy (dy/dx) (6y^2 + 6x^2 +6 ) = 24x - 12xy (dy/dx) = 24x - 12xy/ (6y^2 + 6x^2 +6 ) (dy/dx) = 4x - 2xy/ x^2 + y^2 + 1 <--- (factoring out a 6) B. 4x - 2xy = 0 2x( 2 - y )= 0 x=0 , y=2 This part confuses me a bit. Plugging x=0 back in the original equation gives me a nonsensical answer ( 28=1) and plugging y=2 in the original nets me a cubic equation. Is y=2 the equation for the tangent or what? C. 4x -2xy/x^2 + y^2 + 1 = -1 4x -2xy = -x^2 - y^2 - 1 Since the line goes through the origin with slope of -1, y= -x 4x - 2x(-x) = -x^2 - (-x)^2 -1 Rearranging gives 4x^2 + 4x +1 = 0 , Factoring that gives x= -1/2 And again, since y= -x, then the coords are (-1/2, 1/2) **I may be doing this totally wrong. I haven't been in a calculus class for 2 years and I'm trying to brush up on it a bit.** Sorry for any confusion on my format. This is my first time posting here. Thanks!!!