find the equation of the tangent line to x^2y-x=y^3-8 at x=0

Answer 1

derivative:

x^2 y'' + 2xy - 1 = 3y^2 y'

when x=0 then y=2, so (0,2)

plugging these in:

-1 = 3*4 * y'

so the slope is M = y' = -1/12

B = y - mx = 3 - (-1/12)(0) = 3

y = (-1/12)x + 3