Answer: \(2\left(x+\dfrac54\right)^2-\dfrac{49}{8}\); stationary point \(\left(-\dfrac54,-\dfrac{49}{8}\right)\); intercepts \((-3,0)\), \(\left(\dfrac12,0\right)\), \((0,3)\); \(k=\dfrac{49}{8}\).

(a) Complete the square:

\(2x^2+5x-3=2\left(x^2+\frac52x\right)-3.\)

Now

Therefore

(b) The stationary point of \(y=2x^2+5x-3\) is

\(\left(-\frac54,-\frac{49}{8}\right).\)

(c) The \(x\)-intercepts of \(y=|2x^2+5x-3|\) occur when

\(2x^2+5x-3=0.\)

Factorising,

\((2x-1)(x+3)=0.\)

So the \(x\)-intercepts are

The \(y\)-intercept is

\(y=|-3|=3,\)

so it is \((0,3)\).

The part of the original quadratic below the \(x\)-axis is reflected above the \(x\)-axis. Hence the reflected central maximum is

\(\left(-\frac54,\frac{49}{8}\right).\)

(d) A horizontal line \(y=k\) cuts the modulus graph in exactly 3 distinct points when it passes through this central maximum. Therefore

\(k=\frac{49}{8}.\)