Answer
$8$
Work Step by Step
Consider the tangent line that contains the point $(x, c)$. The slope the tangent line to the graph of $f(x)$ at $(x, c)$ can be written as
$f'(x) =\lim\limits_{x \to c} \dfrac{f(x)-f(c)}{x-c}$
Consider $c=-1$ and $f(x)=x^3-2x^2+x$. Thus, we have:
$f'(-1) =\lim\limits_{x \to 1} \dfrac{f(x)-f(-1)}{x-(-1)} \\=\lim\limits_{x \to -1} \dfrac{(x^3-2x^2+x)-(-4)}{x+1}\\=\lim\limits_{x \to -1} \dfrac{(x^2-3x+4)(x+1)}{x+1} \\=\lim\limits_{x \to -1} x^2-3x+4 \\=8 $
