Answer: stationary points \(\left(-\dfrac12,0\right)\), \(\left(\dfrac{11}{6},-\dfrac{686}{27}\right)\); intercepts \(x=-\dfrac12,3\), \(y=-3\); exactly one solution when \(k\gt 0\) or \(k\lt -\dfrac{686}{27}\).
We start with the main method. Use the roots, intercepts and sign of the leading coefficient to determine the shape of the cubic and its modulus.
The curve is
\(y=(2x+1)^2(x-3).\)
Differentiate using the product rule:
\(\frac{dy}{dx}=4(2x+1)(x-3)+(2x+1)^2.\)
Factorise:
\(\frac{dy}{dx}=(2x+1)\{4(x-3)+(2x+1)\}.\)
So
\(\frac{dy}{dx}=(2x+1)(6x-11).\)
For stationary points, set \(\frac{dy}{dx}=0\):
\(2x+1=0\quad\text{or}\quad6x-11=0.\)
Hence
\(x=-\frac12\quad\text{or}\quad x=\frac{11}{6}.\)
When \(x=-\frac12\),
\(y=0.\)
When \(x=\frac{11}{6}\),
\(2x+1=\frac{14}{3}\quad\text{and}\quad x-3=-\frac76.\)
Therefore
\(y=\left(\frac{14}{3}\right)^2\left(-\frac76\right)=-\frac{686}{27}.\)
The stationary points are
\(\left(-\frac12,0\right)\quad\text{and}\quad\left(\frac{11}{6},-\frac{686}{27}\right).\)
For the sketch, the \(x\)-intercepts are found from
\((2x+1)^2(x-3)=0.\)
So
\(x=-\frac12\) and \(x=3\).
The \(y\)-intercept is found by putting \(x=0\):
\(y=(1)^2(-3)=-3.\)
The root \(x=-\frac12\) is a repeated root, so the graph touches the \(x\)-axis there.
The graph crosses the \(x\)-axis at \(x=3\).
For \((2x+1)^2(x-3)=k\), the number of solutions is the number of intersections between the cubic and the horizontal line \(y=k\).
From the shape of the graph, there is exactly one solution when the horizontal line lies above the local maximum or below the local minimum.
The local maximum is \(0\) and the local minimum is \(-\frac{686}{27}\).
Therefore exactly one solution occurs when
\(k\gt 0\quad\text{or}\quad k\lt -\frac{686}{27}.\)
This completes the solution and gives the required result.
