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

##### 1 Answer

Jan 12, 2017

#### Explanation:

As# P(-3, 1) is on the graph,

C = ((-3)(1)-1^2)(1-3)=0#.

Differentiating,

At P,

I have used a parallel line, in proximity of the tangent, to keep off the

gap, at the point of contact. In this graphics method, P appears as a

gap, for the exact equation of the tangent. For the interested reader,

this graph is also included. In this graph, the pixels at P do not glow.

graph{((xy-y^2)(1+x)-8)(3x-5y+13.7)=0 [-9.79, 9.785, -4.895, 4.895]}

graph{((xy-y^2)(1+x)-8)(3x-5y+14)=0 [-9.79, 9.785, -4.895, 4.895]}