Originally Posted by

**moonman** Find all points where the tangent line is horizontal.

The equation is x^(2) + xy + y^(2) = 1

Okay so from a previous thread I learned that the horizontal tangent line is when a single point touches the equation and in this case the cirlce/oval.

I took the derivative and i got:

2x + xy' + y + 2y^(2) y' = 0 Mr F says: Wrong. It's 2x + xy' + y + 2y y' = 0. Now solve this equation for y'.

I'm looking at the graph of this and I think there are only two horizontal tangents but I dont know how to compute them. Please show me how. Thanks.