Answer: vertex \(\left(\dfrac32,0\right)\), intercepts \(\left(\dfrac32,0\right)\) and \((0,6)\); \(x=1\) or \(x=3\).

The graph of \(y=|4x-6|\) is a V-shaped graph.

The vertex occurs where \(4x-6=0\), so \(x=\frac32\). Hence the vertex is \(\left(\frac32,0\right)\).

The \(y\)-intercept is found by putting \(x=0\):

\(y=|4(0)-6|=6.\)

So the graph meets the axes at \(\left(\frac32,0\right)\) and \((0,6)\).

For part (b), solve the absolute value equation by using the two possible signs:

\(4x-6=2x\quad\text{or}\quad 4x-6=-2x.\)

From \(4x-6=2x\),

\(2x=6\), so \(x=3\).

From \(4x-6=-2x\),

\(6x=6\), so \(x=1\).

Therefore the solutions are

\(x=1\quad\text{or}\quad x=3.\)

This completes the solution and gives the required result.