Answer: \(2\left(x+\dfrac34\right)^2-\dfrac{41}{8}\); stationary point \(\left(-\dfrac34,-\dfrac{41}{8}\right)\); \(k=\dfrac{41}{8}\).

(a) Complete the square:

\(2x^2+3x-4 =2\left(x^2+\frac32x\right)-4.\)

Now

\(x^2+\frac32x =\left(x+\frac34\right)^2-\frac9{16}.\)

Therefore

\(2x^2+3x-4 =2\left[\left(x+\frac34\right)^2-\frac9{16}\right]-4.\)

So

\(2x^2+3x-4 =2\left(x+\frac34\right)^2-\frac98-4.\)

Hence

\(2x^2+3x-4 =2\left(x+\frac34\right)^2-\frac{41}{8}.\)

(b) From the completed-square form, the stationary point of \(y=2x^2+3x-4\) is

\(\left(-\frac34,-\frac{41}{8}\right).\)

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

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

So the \(y\)-intercept is \((0,4)\).

The \(x\)-intercepts occur when

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

Using the quadratic formula,

\(x=\frac{-3\pm\sqrt{3^2-4(2)(-4)}}{2(2)} =\frac{-3\pm\sqrt{41}}{4}.\)

So the graph meets the \(x\)-axis at

\(\left(\frac{-3-\sqrt{41}}4,0\right) \quad\text{and}\quad \left(\frac{-3+\sqrt{41}}4,0\right).\)

The graph of \(y=|2x^2+3x-4|\) is obtained by reflecting the part of \(y=2x^2+3x-4\) below the \(x\)-axis above the \(x\)-axis. So it has cusps at the two \(x\)-intercepts and a maximum between them at

\(\left(-\frac34,\frac{41}{8}\right).\)

(d) The equation

\(|2x^2+3x-4|=k\)

has exactly 3 solutions when the horizontal line \(y=k\) just touches the reflected middle part at its maximum, while still meeting the two outer branches.

Therefore

\(k=\frac{41}{8}.\)