*Originally posted by webby *

**I made it that far. By the way your dy/dx should end in a 1 instead of a 6.**

I set it equal to zero and found horizontal tangent points at x=0 and y=2. I am not used to finding y values when doing that. Normally when it is not implicit you find just the x values and plug it in to the original equation to find the y values. If I plug 0 in for x, I end up with my first post on this topic: 2y^3 + 6y - 1 = 0

And I'm still stuck.

By the way, this problem came from an old AP Calculus exam.

You are correct on the 1..I was scribbling in pen and missed canceling a 6. I looked at it a bit more; here are my thoughts, not guaranteed to be correct.

The roots for dy/dx=0 should be x=0 and y=1/2, I believe. Now, substituting these roots back into the original expression gives:

2y^3+6y-1=0

I believe the equation for the tangent line would then be its' derivative: 6y^2+6.

Substituting y=1/2 into the original expression gives x^2-1/4=0

The equation for its' tangent line would be 2x.

Now, on to part 2: we are looking for points on the original curve that satisfy two conditions: y=-x and dy/dx=-1 (in other words, there may be multiple points that satisfy the first condition but the curve isn't tangent to the line y=-x). Substituting y=-x into the original equation, we get:

8y^3-12y^2+6y-1=0.

I cheated and used a cubic root calculator at this point...

All three roots are real and equal: y=1/2. Therefore x=-1/2. Checking this by plugging this value into dy/dx; it evaluates to -1, satisfying our second condition.