Find the tangent line to the curve at the point

y^3 = x^2 + x, at (0,0) and (1,2)

please tell me what i should be trying to do in this problem and what steps should be taken

(dy/dx)3y^2 = 2x + 1
dy/dx = (2x+1)/(3y^2)

Plug in your x and y variables to get your answer.
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explanation:

Think of an equation where you only have one variable. Lets say Y = 2x^2
Since Y is just "Y", the derivative of that is 1. The derivative of the whole equation is
1(dy/dx) = 4x
dy/dx is isolated at this point, so dy/dx equals 4x.

Now with y^3 = x^2 + x
Since Y has an exponent, dy/dx isn't isolated after you take the derivative initially. You must isolate for it first.

Get the derivative: (dy/dx)(3y^2) = 2x + 1
Then isolate for dy/dx
dy/dx = (2x+1)/(3y^2)

y^3 = x^2 + x
y' = (2x + 1)/(3y^2)
y'(0 , 0) = undefined
so the tangent is a vertical line at x = 0
http://www.wolframalpha.com/input/?i=y^3…

y'(1 , 2) = 3/12 = 1/4
y - 2 = 1/4(x - 1)
y = 1/4*x + 7/4