Roman C. answered • 04/22/16

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y

^{3}+ x^{3}- 2xy = 83y

^{2}y' + 3x^{2}- 2y - 2xy' = 0(3y

^{2}- 2x)y' = 2y - 3x^{2}y' = (2y - 3x

^{2}) / (3y^{2}- 2x)We verify that (2,2) is on the curve.

2

^{3}+ 2^{3}- 2·2·2 = 8 + 8 - 8 = 8 Yes.The slope at (2,2) is

y'(2,2) = (2·2 - 3·2

^{2}) / (3·2^{2}- 2·2) = -1Tangent line:

Point-slope form: y = -1(x - 2) + 2

Slope-intercept form: y = -x + 4