# What is the slope of the tangent line of #(x-y)^3-y^2/x= C #, where C is an arbitrary constant, at #(1,-2)#?

##### 1 Answer

Dec 27, 2016

#### Explanation:

The point

From

(x-y)^3-y^2/x=23,

y' at P =31/29.

The equation to the tangent at P is

graph{((x-y)^3-y^2/x-23)(y-31/29x+89/29)=0x^2 [-5, 5, -10, 10]}