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ineedcalchelp · Sep 27, 2007, 06:02 PM

I found the dy/dx. But how do you find a vertical tangent line? And how do you find a vertical tangent when x = (?)?

galactus · Sep 29, 2007, 03:33 PM

$\frac{dy}{dx}=\frac{2x-y}{x-3y^{2}}$

What gives division by 0? When $x=3y^{2}$ ... [1]

Sub this back into the original function:

$(3y^{2})^{2}-(3y^{2})y+y^{3}=10$

$9y^{4}-2y^{3}-10=0$

Solving this for y gives:

y=-0.97533487561 and y=1.08709662767

Sub these into [1] gives:

x=2.85383435874 and x=3.5433723367

These are the coordinates of the points of vertical tangency.