Answer
The curve, observing from left to right,
- rises on $(-\infty,1)$, is concave down, and crosses the x-axis at the origin,
- at $(1,16)$ reaches a local maximum, turns down,
- starts decreasing on the interval $(1,3),$
- at $(2,8)$ changes concavity to "up" (inflection point),
and contiues falling to
- the point $(3,0)$, a local minimum, from where
- it rises on the interval $(3,\infty)$
Work Step by Step
$y=f(x)=x(6-2x)^{2}=x(36-24x+4x^{2})=36x-24x^{2}+4x^{3}$
Intercepts are
$\left\{\begin{array}{l}
f(0)=0\\
f(3)=0
\end{array}\right.\qquad (0,0), (3,0).$
(some points on the graph to begin with.)
$f'(x)=36-48x+12x^{2}=12(x^{2}-4x+3)=12( x-1)(x-3)$.
$f''(x)=48-24x=24(2-x)$
Analyzing $f'(x)$, we see that it is defined everywhere,
and has zeros $ x=1,3\qquad$... (critical points)
Applying the 2nd derivative test,
$\left\{\begin{array}{lllll}
x=1, & f''(1)=-24 \lt 0 & \Rightarrow(\max), & f(1)=16 & (1,16)\\
x=3, & f''(1)=24 \gt 0 & \Rightarrow(\min), & f(3)=0 & (3,10)
\end{array}\right.$
$f''(x)=0$ for $x=2,\qquad f(2)=8,\quad(2,8)$ is a point of inflection.
For graphing purposes:
The graph of $f'(x)$ is a parabola that opens up. The graph is below the x-axis between the zeros, and above for the other two intervals.
$\left[\begin{array}{cccccccccc}
f'(x): & -\infty & ++ & 1 & -- & 3 & ++ & \infty\\
f(x): & & \nearrow & & \searrow & & \nearrow &
\end{array}\right]$
Concavity and points of inflection:
$\left[\begin{array}{ccccccccc}
f''(x): & -\infty & -- & 2 & ++ & \infty & \\
f(x): & & \cap & inf. & \cup & &
\end{array}\right]$
