= Solution
Factorization gives $y=(x-2)^2(x+1)$, so the zeros are $x=-1$ and $x=2$, with a double root at two. The <derivatives> are $y'=3x(x-2)$ and $y''=6(x-1)$. Thus
$$
\boxed{\text{local maximum }(0,4),\quad\text{local minimum }(2,0),\quad\text{inflection point }(1,2).}
$$
The second <derivative> changes sign at one, confirming the <inflection point>. The curve rises from $-\infty$, crosses at $-1$, reaches its local maximum, falls to touch the axis at two, then rises to $+\infty$. Neither local extremum is a global extremum on the whole real line.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-2-cubic-extrema.png]
{title=Zeros, two stationary points and an inflection point of the cubic}
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