#### Answer

$$
f(x)=f(x)=-x(x-3)^{2}
$$
$f(x)$ concave upward on
$$(-\infty, 2 )$$.
$f(x)$ concave downward on
$$(2, \infty )$$
Inflection point at
$$(2, -2)$$

#### Work Step by Step

$$
f(x)=-x(x-3)^{2}
$$
Rewrite the function $f(x)$ as
$$
f(x)=-x(x^{2}-6x+9)=-x^{3}+6x^{2}-9x
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=-(3)x^{2}+6(2)x-9 \\
&=-3x^{2}+12x-9 ,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=-3(2)x+12-0 ,\\
&=-6x+12
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) =-6x+12 & \gt 0 \\
-6x & \gt -12\\
x & \lt \frac{ -12}{-6}= 2
\end{aligned}
$$
so $f(x)$ is concave upward on $(-\infty, 2 )$.
Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) =-6x+12 & \lt 0 \\
-6x & \lt -12\\
x & \gt \frac{ -12}{-6}= 2
\end{aligned}
$$
so $f(x)$ is concave downward on $(2, \infty )$.
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $2 $, the graph changes from concave upward to concave downward at that point, so $(2 , f(2))$ is inflection point.
Now ,we find $ f(2)$as follows:
$$
\begin{aligned}
f(2) &=-(2)((2)-3)^{2}\\
&=-(2)(-1)^{2}\\
&=-2
\end{aligned}
$$
Finally, we have inflection point at $(2, f(2))=(2, -2)$.