(i) Differentiate the curve: \(\frac{dy}{dx} = 3x^2 - 6x - 9\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 6x - 9 = 0\).

Solving gives \(x = 3\) or \(x = -1\).

Since the minimum point is on the \(x\)-axis, \(y = 0\) at \(x = 3\).

Substitute \(x = 3\) into the curve equation: \(0 = 3^3 - 3(3)^2 - 9(3) + k\).

Simplify: \(0 = 27 - 27 - 27 + k\).

Thus, \(k = 27\).

(iv) Integrate \(y\) to find the area: \(\int (x^3 - 3x^2 - 9x + 27) \, dx\).

The integral is \(\frac{x^4}{4} - x^3 - \frac{9x^2}{2} + 27x\).

Evaluate from \(x = 0\) to \(x = 3\):

\(\left[ \frac{3^4}{4} - 3^3 - \frac{9(3)^2}{2} + 27(3) \right] - \left[ \frac{0^4}{4} - 0^3 - \frac{9(0)^2}{2} + 27(0) \right]\).

Calculate: \(\frac{81}{4} - 27 - 40.5 + 81 = 33.75\).

Thus, the area of the shaded region is 33.75.