Answer: The graph meets the \(x\)-axis at \((-2,0)\), \((-1,0)\) and \((2,0)\), and the \(y\)-axis at \((0,4)\).

The graph is

\(y=\left|(x-2)(x+1)(x+2)\right|.\)

The graph meets the \(x\)-axis when

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

So

\(x=-2,\qquad x=-1,\qquad x=2.\)

Therefore the \(x\)-intercepts are

\((-2,0),\quad(-1,0),\quad(2,0).\)

The graph meets the \(y\)-axis when \(x=0\):

\(y=\left|(-2)(1)(2)\right|=4.\)

So the \(y\)-intercept is

\((0,4).\)

Since the expression is inside a modulus, the graph is never below the \(x\)-axis. The parts of the cubic which would be below the \(x\)-axis are reflected above it, giving sharp turns at the three roots and a maximum in the first quadrant.