The sign table and the shape of the graph

A sign table lays out the sign of the derivative interval by interval. We build one for $f(x) = x^3 - 6x^2 + 9x$.

The sign of the derivative

Since $f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)$, we have $f' = 0$ at $x = 1$ and $x = 3$. Those two points cut the number line into three parts, and we examine the sign on each.

The sign table

At $x = 1$, where increase turns to decrease, there is a local maximum; at $x = 3$, where decrease turns to increase, a local minimum. The values are $f(1) = 4$ and $f(3) = 0$.

Crossings of the $x$-axis

Factoring $f(x) = x(x - 3)^2$ shows that the graph meets the $x$-axis at $x = 0$ and $x = 3$. Since $x = 3$ is a double root, the graph touches the axis there without crossing. That is the same statement as the local minimum being exactly $0$.

The shape of the graph

The curve climbs from the left, turns at $(1, 4)$, touches the $x$-axis at $(3, 0)$ and climbs again. As $x \to -\infty$ we have $f \to -\infty$ and as $x \to \infty$ we have $f \to \infty$: the typical shape of a cubic.

The graph of the derivative

$f'$ is an upward parabola crossing the $x$-axis at $x = 1$ and $x = 3$. Only between those two points is $f'$ negative, and that is the interval on which $f$ decreases. The vertex of $f'$ is at $x = 2$, and $f'(2) = -3$ is the least value of the slope.

The center of symmetry

The graph of a cubic has point symmetry about its inflection point, the midpoint of the local maximum and the local minimum. Here that center is $\left( \dfrac{1 + 3}{2}, \dfrac{4 + 0}{2} \right) = (2, 2)$.

Whether a point is an extremum is decided by the change of sign of $f'$. If $f' = 0$ but the sign does not change, it is not an extremum. The trouble of writing out a sign table is what keeps that change of sign from being missed.

The cubic on the graph is $y = x^3 - 6x^2 + 9x$, the parabola is the derivative, and the large dots are the local maximum $(1, 4)$ and the local minimum $(3, 0)$.