Answer: Stationary point \(\left(\frac12,-\frac{49}{4}\right)\); intercepts \((-3,0)\), \((4,0)\), \((0,12)\); exactly 2 roots when \(k\gt\frac{49}{4}\).
Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.
Expand the quadratic:
\(y=(x+3)(x-4)=x^2-x-12.\)
The stationary point is found from
\(\frac{\mathrm dy}{\mathrm dx}=2x-1=0\), so \(x=\frac12\).
Then
\(y=\left(\frac12+3\right)\left(\frac12-4\right)=\frac72\cdot\left(-\frac72\right)=-\frac{49}{4}.\)
So the stationary point is \(\left(\frac12,-\frac{49}{4}\right)\).
The graph of \(y=|(x+3)(x-4)|\) is obtained by reflecting the part of the quadratic below the \(x\)-axis above the \(x\)-axis.
The \(x\)-intercepts are \(x=-3\) and \(x=4\), and the \(y\)-intercept is \(y=12\).
The reflected central part has maximum value \(\frac{49}{4}\). Therefore a horizontal line \(y=k\), with \(k\gt0\), cuts the graph in exactly 2 distinct points only when it is above this maximum.
Hence \(k\gt\frac{49}{4}\).
The result has been obtained using the exact condition from the question, so no additional solutions are introduced.
