Answer: stationary points \(\left(\frac23,\frac{343}{27}\right)\) and \((3,0)\); \(k=\frac{343}{27}\).

(a) Let

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

Differentiate using the product rule:

Factorising,

Stationary points occur when

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

So

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

When \(x=3\), \(y=0\). When \(x=\frac23\),

Hence the stationary points are

\(\left(\frac23,\frac{343}{27}\right)\) and \((3,0)\).

(b) The original curve has roots at

\(2x+1=0\quad\text{and}\quad x-3=0,\)

so \(x=-\frac12\) and \(x=3\). The \(y\)-intercept is

\(y=(1)(-3)^2=9.\)

For \(y=\left|(2x+1)(x-3)^2\right|\), the part below the \(x\)-axis is reflected above the \(x\)-axis. The graph therefore meets the axes at

\(\left(-\frac12,0\right)\), \((3,0)\), and \((0,9)\).

(c) A horizontal line gives exactly 3 distinct intersections when it passes through the maximum turning value. Therefore

\(k=\frac{343}{27}.\)