(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: Write an equation of each horizontal tangent line to the curve.

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