1. Find the first derivative of the function to get the slope:

\(y = -x^3 + 3x^2 + 8x - 20\)

\(\frac{dy}{dx} = -3x^2 + 6x + 8\)

2. To find the maximum slope, find the critical points of the slope function by taking the derivative of the slope function (second derivative of the original function) and setting it to zero:

\(\frac{d^2y}{dx^2} = -6x + 6\)

Set \(\frac{d^2y}{dx^2} = 0\):

\(-6x + 6 = 0\)

\(6x = 6\)

\(x = 1\)

3. To confirm that this is a maximum, we can check the third derivative, but in this case, since we are given options, we can simply substitute \(x = 1\) into the original function and the first derivative to find the point and the slope at that point.

Substitute \(x = 1\) into the original function:

\(y = -(1)^3 + 3(1)^2 + 8(1) - 20\)

\(y = -1 + 3 + 8 - 20\)

\(y = -10\)

So, the point is \((1, -10)\).

4. The question asks where the slope is maximum, which occurs at x=1. The point on the curve is (1, -10).