Posted by **stephanie** on Wednesday, November 12, 2008 at 8:01pm.

original curve: 2y^3+6(x^2)y-12x^2+6y=1

dy/dx=(4x-2xy)/(x^2+y^2+1)

a) write an equation of each horizontal tangent line to the curve

b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P.

- please help; calc -
**bobpursley**, Wednesday, November 12, 2008 at 8:05pm
if dy/dx is zero, that must mean 4x-2xy is zero, or y=2

When y=2, solve for x in the original equation.

- please help; calc -
**stephanie**, Wednesday, November 12, 2008 at 10:11pm
when you plug it in the original eqn. it doesn't work

- it doesn't work -
**stephanie**, Wednesday, November 12, 2008 at 10:44pm
28=1 is the value of x and that doesn't work

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