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

The given equation of the curve is $y=\frac{1}{2}{x}^4-5x^3+18x^2-19x$

            Differentiate both sides

$m=\frac{dy}{dx}=2x^3-15x^2+36x-19$

$\begin{array}{c}\frac{dm}{dx}=6x^2-30x+36\\ =6\left(x^2-5x+6\right)\\ =6\left(x-2\right)\left(x-3\right)\end{array}$

If $\frac{dm}{dx}=0$ $\Rightarrow$

$x-2=0,x-3=0$

$\frac{d^{2}m}{d{x}^2}=6\left(2x-5\right)<0$ for $x=2$

so the slope has maximum value at $x=2$, when $x=2$ then $y=8-40+72-38=\boxed{2}$

Therefore, the point is $\left(2,2\right)$