The minimum value of the slope of the tangent to curve $\mathrm{y}=x^3-3 x^2+2 x+93$ is

The slope of the tangent to the curve $y = x^3 - 3x^2 + 2x + 93$ is given by the derivative $m(x) = \frac{dy}{dx} = 3x^2 - 6x + 2$.

Since $m(x)$ is a quadratic with a positive leading coefficient, it attains a minimum at its vertex. The vertex occurs at $x = -\frac{b}{2a} = -\frac{-6}{2 \times 3} = 1$.

Evaluating the slope at this point gives $m(1) = 3(1)^2 - 6(1) + 2 = -1$.

The minimum slope is $-1$.