1. The slope of the curve is given by the derivative:

$y=\frac{1}{2}x^4-5x^3+18x^2-19x$

So,

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

2. We need the maximum slope, i.e. the maximum value of

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

To find where this slope is extremum, differentiate again:

$m'(x)=\frac{d^2y}{dx^2}=6x^2-30x+36$

$=6(x^2-5x+6)=6(x-2)(x-3)$

Thus critical points for the slope occur at

$x=2,\;3$

3. To determine where the slope is maximum, use the third derivative:

$\frac{d^3y}{dx^3}=12x-30$

• At $x=2$: $\frac{d^3y}{dx^3}=24-30=-6<0$ So $\dfrac{dy}{dx}$ has a local maximum at $x=2$.

• At $x=3$: $\frac{d^3y}{dx^3}=36-30=6>0$ So $\dfrac{dy}{dx}$ has a local minimum at $x=3$.

Hence the maximum slope occurs at $x=2$.

4. Now find the corresponding point on the curve:

$y(2)=\frac{1}{2}(2^4)-5(2^3)+18(2^2)-19(2)$

$=\frac{1}{2}(16)-5(8)+18(4)-38$

$=8-40+72-38=2$

So the point is

$(2,2)$

5. Checking options:

• A: $\left(3,\frac{21}{2}\right)$ — no
• B: $(0,0)$ — no
• C: $(2,9)$ — no
• D: $(2,2)$ — yes

Therefore, the maximum slope occurs at

$\boxed{(2,2)}$