How do I find the equation of the tangent line to the graph of f(x)=(x^2+n)(x-n) at a given point (-x,-y)
I'm not sure if i'm right, but
expand:
(x^2+n)(x-n)
=x^3-x^2*n+nx-n^2
take the first derivative
so f'(x)=3x^2-2xn+n
-f'(x)=3x^2-2xn+n
m=-(3*(-x)^2-2*-x*n+n)
m=-3x^2-2xn-n
so the tangent line is
y=mx+b
y=(-3x^2-2xn-n)*x+b
b depends on your n, i don't feel like working it out though sorry
I think that's right but i've never done a problem like this so...