# How do you find the slope of the tangent line of f(x) = 3-2x^3 at (-1, 5)?

Aug 5, 2015

I found: slope$= - 6$

#### Explanation:

The slope of the tangent will be the derivative of your function evaluated at $x = - 1$ (the $x$ coordinate of the tangence point).
So:
$f ' \left(x\right) = 0 - 6 {x}^{2} = - 6 {x}^{2}$
evaluated at $x = - 1$:
$f ' \left(- 1\right) = - 6$
So the slope of the tangent will be $- 6$.