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x^2−(y^2)x = (3x − 3)y

Using this graph and the equation I need to find two things.

(a) Formula which gives the slope dy/dx at every point (x, y) on the graph.

(b) As you can see in the picture, there are two points on the graph which have x-coordinate equal to 1. What are the exact slopes of the tangent lines at those two points?

I believe I correctly figured out (a) to be y'(x)= (2x-y^2-3y)/(2xy+3x-3)

However, I am not sure how to use this formula to find two different slopes for one value of x.

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# Find dy/dx of x^2 − (y^2) x = (3x − 3)y

**Physics Forums | Science Articles, Homework Help, Discussion**