Answer: intercepts \((-1,0)\), \((2,0)\), \((4,0)\), \((0,-24)\); \(x\lt -1\) or \(2\lt x\lt 4\).

(a) The \(x\)-intercepts occur when

\(-3(x-2)(x-4)(x+1)=0.\)

So

\(x=2,\qquad x=4,\qquad x=-1.\)

Therefore the curve meets the \(x\)-axis at

\((-1,0),\quad(2,0),\quad(4,0).\)

The \(y\)-intercept occurs when \(x=0\):

\(y=-3(-2)(-4)(1)=-24.\)

So the curve meets the \(y\)-axis at

\((0,-24).\)

The leading term is \(-3x^3\), so the curve rises to the left and falls to the right. It crosses the \(x\)-axis at \(-1\), \(2\) and \(4\), with a minimum in the fourth quadrant and a maximum in the first quadrant.

(b) From the sketch, the curve is above the \(x\)-axis when

\(x\lt -1\)

or when

\(2\lt x\lt 4.\)

Therefore

\(-3(x-2)(x-4)(x+1)\gt 0\)

for

\(x\lt -1\quad\text{or}\quad2\lt x\lt 4.\)