Answer: \(x^2-x-6=\left(x-\frac12\right)^2-\frac{25}{4}\), the stationary point is \(\left(\frac12,-\frac{25}{4}\right)\), and \(\left|x^2-x-6\right|\lt4\) for \(\dfrac{1-\sqrt{41}}{2}\lt x\lt-1\) or \(2\lt x\lt\dfrac{1+\sqrt{41}}{2}\).
(a)(i) Complete the square:
\(x^2-x-6=x^2-x+\frac14-\frac14-6\).
Hence
\(x^2-x-6=\left(x-\frac12\right)^2-\frac{25}{4}\).
(a)(ii) From the completed-square form, the stationary point of
\(y=x^2-x-6\)
is
\(\left(\frac12,-\frac{25}{4}\right)\).
(b) First consider the graph of
\(y=x^2-x-6\).
Its roots are found from
\(x^2-x-6=(x-3)(x+2)=0\),
so the roots are \(x=-2\) and \(x=3\).
Therefore the graph of \(y=\left|x^2-x-6\right|\) touches the \(x\)-axis at
\((-2,0)\quad\text{and}\quad(3,0)\).
The part of the quadratic below the \(x\)-axis is reflected above the \(x\)-axis. Thus the stationary point \(\left(\frac12,-\frac{25}{4}\right)\) becomes
\(\left(\frac12,\frac{25}{4}\right)\)
on the modulus graph.
Useful points for the sketch include
\((-4,14),\quad(-2,0),\quad\left(\frac12,\frac{25}{4}\right),\quad(3,0),\quad(4,6)\).
(c) To solve
\(\left|x^2-x-6\right|\lt4\),
use the equivalent double inequality
\(-4\lt x^2-x-6\lt4\).
First solve
\(x^2-x-6\lt4\).
This gives
\(x^2-x-10\lt0\).
The roots of \(x^2-x-10=0\) are
\(x=\dfrac{1\pm\sqrt{41}}{2}\).
Since the quadratic opens upwards,
\(\dfrac{1-\sqrt{41}}{2}\lt x\lt\dfrac{1+\sqrt{41}}{2}\).
Now solve
\(x^2-x-6\gt-4\).
This gives
\(x^2-x-2\gt0\),
so
\((x-2)(x+1)\gt0\).
Thus
\(x\lt-1\quad\text{or}\quad x\gt2\).
Combining both conditions gives
\(\dfrac{1-\sqrt{41}}{2}\lt x\lt-1\quad\text{or}\quad2\lt x\lt\dfrac{1+\sqrt{41}}{2}\).
