Answer: Stationary points at \(x=-2\) and \(x=\frac13\). Intercepts: \((-2,0)\), \((\frac32,0)\), \((0,6)\). Also \(0\lt k\lt \frac{343}{54}\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

Expand the function:

\(y=\frac12(3-2x)(x+2)^2=-x^3-\frac52x^2+2x+6.\)

Differentiate:

\(\frac{\mathrm dy}{\mathrm dx}=-3x^2-5x+2.\)

Set the derivative equal to zero:

\(-3x^2-5x+2=0.\)

This is equivalent to

\(3x^2+5x-2=0,\)

so

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

Hence the stationary points have \(x\)-coordinates

\(x=-2,\quad x=\frac13.\)

The \(x\)-intercepts are found from \(\frac12(3-2x)(x+2)^2=0\), giving \(x=-2\) and \(x=\frac32\). The \(y\)-intercept is

\(y=\frac12(3)(2)^2=6.\)

At \(x=-2\), \(y=0\), and at \(x=\frac13\),

\(y=\frac12\left(3-\frac23\right)\left(\frac73\right)^2=\frac{343}{54}.\)

A horizontal line \(y=k\) cuts the cubic in three distinct real roots when it lies strictly between the local minimum \(0\) and the local maximum \(\frac{343}{54}\). Therefore

\(0\lt k\lt \frac{343}{54}.\)

This gives the required answer: Stationary points at \(x=-2\) and \(x=\frac13\). Intercepts: \((-2,0)\), \((\frac32,0)\), \((0,6)\). Also \(0\lt k\lt \frac{343}{54}\).