(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = ax^2 + bx - 4\).

The integral is \(y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x + c\).

Since the curve passes through (0, 11), substitute \(x = 0\) and \(y = 11\) to find \(c\):

\(11 = \frac{1}{3}a(0)^3 + \frac{1}{2}b(0)^2 - 4(0) + c\)

\(c = 11\)

Thus, the equation of the curve is \(y = \frac{1}{3}ax^3 + \frac{1}{2}bx^2 - 4x + 11\).

(ii) At the stationary point (2, 3), \(\frac{dy}{dx} = 0\).

Substitute \(x = 2\) into \(\frac{dy}{dx} = ax^2 + bx - 4\):

\(4a + 2b - 4 = 0\)

Substitute \(x = 2\) and \(y = 3\) into the integrated equation:

\(3 = \frac{1}{3}a(2)^3 + \frac{1}{2}b(2)^2 - 4(2) + 11\)

\(\frac{8}{3}a + 2b - 8 + 11 = 3\)

Solve the simultaneous equations:

1. \(4a + 2b - 4 = 0\)

2. \(\frac{8}{3}a + 2b + 3 = 3\)

Solving these gives \(a = 3\) and \(b = -4\).