Answer: intercepts \((0,4)\) and \(\left(\frac43,0\right)\); \(x\leq -1\) or \(x\geq \frac{11}{3}\).

The key point is to split the absolute-value expression into the correct cases before solving.

For the \(y\)-intercept, put \(x=0\):

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

So the graph cuts the \(y\)-axis at \((0,4)\).

For the \(x\)-intercept, put \(y=0\):

\(|4-3x|=0.\)

Hence

\(4-3x=0,\qquad x=\frac43.\)

So the graph cuts the \(x\)-axis at \(\left(\frac43,0\right)\).

To solve the inequality, split the modulus into two cases:

\(|4-3x|\geq 7\)

means

\(4-3x\geq 7 \quad\text{or}\quad 4-3x\leq -7.\)

From the first inequality,

\(-3x\geq 3,\qquad x\leq -1.\)

From the second inequality,

\(-3x\leq -11,\qquad x\geq \frac{11}{3}.\)

Therefore

\(x\leq -1\quad\text{or}\quad x\geq \frac{11}{3}.\)