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.