Answer: \(\mathrm{f}(x)=\frac{1}{3}(x+2)(2x+1)(x-3)\) or \(\mathrm{f}(x)=-\frac{1}{3}(x+2)(2x+1)(x-3)\).

The graph shown is \(y=|\mathrm{f}(x)|\), so the graph gives the zeros of \(\mathrm{f}(x)\), but it does not determine whether \(\mathrm{f}(x)\) or \(-\mathrm{f}(x)\) was the original cubic.

From the graph, the \(x\)-intercepts are

\(x=-2,\quad x=-\frac12,\quad x=3.\)

Therefore \(\mathrm{f}(x)\) has factors

\((x+2),\quad (2x+1),\quad (x-3).\)

So

\(\mathrm{f}(x)=k(x+2)(2x+1)(x-3)\)

for some constant \(k\). Using the scale shown on the graph gives \(|k|=\frac13\).

Since the modulus graph has removed the sign of the original cubic, both signs are possible:

\(\mathrm{f}(x)=\frac13(x+2)(2x+1)(x-3)\)

or

\(\mathrm{f}(x)=-\frac13(x+2)(2x+1)(x-3).\)