Answer: Stationary points \((1,0)\), \((2,-1)\); intercepts \((1,0)\), \(\left(\frac52,0\right)\), \((0,-5)\); exactly one solution when \(k\lt -1\) or \(k\gt 0\).
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.
\(\mathrm f(x)=(2x-5)(x-1)^2\).
Differentiate using the product rule:
\(\mathrm f'(x)=2(x-1)^2+2(2x-5)(x-1)\).
Factorising gives
\(\mathrm f'(x)=2(x-1)\big((x-1)+(2x-5)\big)=6(x-1)(x-2)\).
So the stationary points occur at \(x=1\) and \(x=2\).
\(\mathrm f(1)=0\), and \(\mathrm f(2)=-1\), so the stationary points are \((1,0)\) and \((2,-1)\).
The \(x\)-intercepts are found from \((2x-5)(x-1)^2=0\), giving \(x=1\) and \(x=\frac52\). The \(y\)-intercept is \(\mathrm f(0)=-5\).
The curve touches the \(x\)-axis at \(x=1\), then has a minimum at \((2,-1)\), and crosses the \(x\)-axis at \(x=\frac52\).
A horizontal line \(y=k\) cuts the curve exactly once only when it lies above the local maximum \(0\), or below the local minimum \(-1\).
Therefore \(k\gt 0\) or \(k\lt -1\).
The result has been obtained using the exact condition from the question, so no additional solutions are introduced.
