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

1. May 2, 2010

### lude1

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.

2. May 2, 2010

### tiny-tim

Hi lude1!

(try using the X2 tag just above the Reply box )
Where does the 4x come from?

(Or should the 12x be 12x2 ?)

3. May 2, 2010

### lude1

Ah, yes, you are correct! It should be 12x2

Thus:

y = 2y3 + 6x2y - 12x2 + 6y = 1

and

y' = (4x - 2xy) / (x2 + y2 + 1)

4. May 2, 2010

### tiny-tim

(why do you keep starting with "y =" ? )

hmm … then, sorry, I don't think your 2y3 + 6y = 1 has a straightforward answer

(unless the = 1 should be = 8)

oh, and don't forget you still have to deal with the other case, y = 2.