Answer: Intercepts: \((1,0)\), \((0,4)\), \((-\frac23,0)\), \((0,2)\). Also \(\frac27\leqslant x\leqslant6\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

For \(y=4|x-1|\), the vertex is at \((1,0)\). The \(y\)-intercept is found by putting \(x=0\):

\(y=4|-1|=4.\)

So this graph has intercepts \((1,0)\) and \((0,4)\).

For \(y=|3x+2|\), the vertex is where \(3x+2=0\), so \(x=-\frac23\). The \(y\)-intercept is \(2\).

So this graph has intercepts \((-\frac23,0)\) and \((0,2)\).

To solve the inequality, square both sides. This is valid because both sides are non-negative:

\(16(x-1)^2\leq (3x+2)^2.\)

Expand and simplify:

\(16x^2-32x+16\leq9x^2+12x+4,\)

so

\(7x^2-44x+12\leq0.\)

The roots are

\(x=\frac{44\pm\sqrt{44^2-4(7)(12)}}{14}=\frac{44\pm40}{14}.\)

Hence the critical values are \(\frac27\) and \(6\). Since the quadratic opens upwards, it is non-positive between the roots:

\(\frac27\leqslant x\leqslant6.\)

This gives the required answer: Intercepts: \((1,0)\), \((0,4)\), \((-\frac23,0)\), \((0,2)\). Also \(\frac27\leqslant x\leqslant6\).