Answer: \(a=2,\ b=1,\ c=-1\), and \(x\leqslant -1.45\) or \(-0.40\leqslant x\leqslant0.85\) approximately.

(a) From the graph, the curve touches the \(x\)-axis at \(x=-1\), so \(x=-1\) is a repeated root.

It crosses the \(x\)-axis at \(x=1\), so \(x=1\) is a simple root.

Therefore

\(f(x)=a(x+1)^2(x-1).\)

Comparing with \(a(x+b)^2(x+c)\),

\(b=1,\qquad c=-1.\)

The graph also shows the \(y\)-intercept is \(-2\). Since

\(f(0)=a(1)^2(-1)=-a,\)

we have

\(-a=-2,\)

so

\(a=2.\)

(b) To solve \(f(x)\leqslant -1\), draw or imagine the horizontal line \(y=-1\) on the graph.

The intersections occur approximately at

\(x=-1.45,\quad x=-0.40,\quad x=0.85.\)

The graph is at or below \(y=-1\) to the left of the first intersection, and between the second and third intersections. Hence