(iii) To express \(x^2 - 6x\) in the form \((x-p)^2 - q\), we complete the square:

\(x^2 - 6x = (x-3)^2 - 9\)

Thus, \(p = 3\) and \(q = 9\).

(iv) To find \(h^{-1}(x)\), start with \(y = (x-3)^2 - 9\).

Solve for \(x\):

\(y + 9 = (x-3)^2\)

\(x - 3 = \pm \sqrt{y + 9}\)

Since \(x \geq 3\), we take the positive root:

\(x = \sqrt{y + 9} + 3\)

Thus, \(y = h^{-1}(x) = \sqrt{x + 9} + 3\).

The domain of \(h^{-1}\) is \(x \geq -9\) because the expression under the square root must be non-negative.