Answer: (a) \(x=-3,\ -\frac23\); (b) intercepts \((-3,0)\), \((\frac12,0)\), \((0,3)\); (c) \(k\lt0\) or \(k\gt\frac{343}{81}\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

(a) Differentiate

\(\mathrm q(x)=-\frac13(2x-1)(x+3)^2.\)

Using the product rule,

\(\mathrm q'(x)=-\frac13\left(2(x+3)^2+2(2x-1)(x+3)\right).\)

Factorising,

\(\mathrm q'(x)=-\frac23(x+3)(3x+2).\)

Stationary points occur when \(\mathrm q'(x)=0\), so

\(x=-3\quad\text{or}\quad x=-\frac23.\)

(b) The \(x\)-intercepts are found from

\(-\frac13(2x-1)(x+3)^2=0.\)

So

\(x=\frac12\quad\text{or}\quad x=-3.\)

The root \(x=-3\) is repeated, so the graph touches the \(x\)-axis there.

The \(y\)-intercept is

\(\mathrm q(0)=-\frac13(-1)(3)^2=3.\)

The leading term is negative, so the cubic rises to the left and falls to the right.

The graph has a local minimum at \((-3,0)\) and a local maximum at

\(\left(-\frac23,\frac{343}{81}\right).\)

(c) A horizontal line \(y=k\) has exactly one intersection with the curve when it lies below the local minimum or above the local maximum.

Therefore

\(k\lt0\quad\text{or}\quad k\gt\frac{343}{81}.\)