Answer: (a) \(5\left(x-\frac75\right)^2-\frac95\); (b) \(\left(\frac75,-\frac95\right)\); (d) \(0\lt k\lt\frac95\).

Work through the problem step by step, using exact values where possible before giving any final numerical approximation.

(a) Complete the square:

\(5x^2-14x+8=5\left(x^2-\frac{14}{5}x\right)+8.\)

Half of \(-\frac{14}{5}\) is \(-\frac75\), so

\(5x^2-14x+8=5\left[\left(x-\frac75\right)^2-\frac{49}{25}\right]+8.\)

Hence

(b) The stationary point of \(y=5\left(x-\frac75\right)^2-\frac95\) is

\(\left(\frac75,-\frac95\right).\)

(c) First find where the original quadratic is zero:

\(5x^2-14x+8=0.\)

Factorising gives

\((5x-4)(x-2)=0,\)

so the \(x\)-intercepts are

\(\left(\frac45,0\right)\quad\text{and}\quad(2,0).\)

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

\(y=\lvert8\rvert=8.\)

So the curve passes through \((0,8)\). The part of \(5x^2-14x+8\) below the \(x\)-axis is reflected above the \(x\)-axis, giving a maximum point on the reflected section at

\(\left(\frac75,\frac95\right).\)

(d) The horizontal line \(y=k\) cuts the absolute value graph in four distinct points only when it lies strictly between the \(x\)-axis and the top of the reflected middle section. Therefore

\(0\lt k\lt \frac95.\)