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

The slope will be the derivative of your function evaluated in $x = - 1$.
$f ' \left(x\right) = 3 {x}^{2} - 2$
$f ' \left(- 1\right) = 3 - 2 = 1$