The maximum slope of the curve $y=\frac{1}{2}{x}^4-5x^3+18x^2-19x$ occurs at the point:

Let $m$ be the slope:
$$m=\frac{dy}{dx}=2x^3-15x^2+36x-19$$
To maximize $m$, find $\frac{dm}{dx}=0$:
$$\frac{dm}{dx}=6x^2-30x+36=0$$
$$x^2-5x+6=0\Rightarrow (x-2)(x-3)=0$$
Possible points: $x=2,3$.Check second derivative $\frac{d^{2}m}{d{x}^2}=12x-30$:At $x=2$, $12(2)-30=-6<0$ (Maxima).At $x=3$, $12(3)-30=6>0$ (Minima).So max slope is at $x=2$.Find $y$ at $x=2$:
$$y=\frac{1}{2}(2{)}^4-5(2{)}^3+18(2{)}^2-19(2)$$
$$y=8-40+72-38=2$$
Point is $(2,2)$.