# Calculus

the curve: (x)(y^2)-(x^3)(y)=6

(dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3))

a) find all points on the curve whose x-coordinate is 1 and write an equation for the tangent line of each of these points

b)find the x-coordinate of each point on the curve where the tangent line is vertical

1. a) If x = 1, y^2 -y -6 = 0
(y-3)(y+2) - 0
y = 3 or -2

b) Find the points where dy/dx = infinity

Those would be the places where the denominator of dy/dx = 0. x=0 is one such point. Any point where x^2 = 2y would be another, if there is such a point on the curve.

posted by drwls

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