- #1
lude1
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Homework Statement
Write an equation of each horizontal tangent line to the curve.
Homework Equations
y = 2y^3 + 6x^2y - 12x + 6y = 1
y' = (4x - 2xy) / x^2 + y^2 + 1)
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
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.