Answer
$$
f(x)=x(x+5)^{2}
$$
$f(x)$ concave upward on
$$(\frac{ -10}{3},\infty )$$.
$f(x)$ concave downward on
$$(-\infty , \frac{ -10}{3}).$$
Inflection point at
$$(\frac{ -10}{3} , \frac{ -250}{27})$$
Work Step by Step
$$
f(x)=x(x+5)^{2}
$$
Rewrite the function $f(x)$ as
$$
f(x)=x(x^{2}+10x+25)=x^{3}+10x^{2}+25x
$$
The first derivative is
$$
\begin{aligned}
f^{\prime}(x) &=(3)x^{2}+10(2)x+25 \\
&=3x^{2}+20x+25 ,
\end{aligned}
$$
and the second derivative is
$$
\begin{aligned}
f^{\prime\prime}(x) &=3(2)x+20+0\\
&=6x+20
\end{aligned}
$$
A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) =6x+20 & \gt 0 \\
6x & \gt -20\\
x & \gt \frac{ -20}{6}= \frac{ -10}{3}
\end{aligned}
$$
so $f(x)$ is concave upward on $(\frac{ -10}{3},\infty )$.
Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so
$$
\begin{aligned}
f^{\prime\prime}(x) =6x+20 & \lt 0 \\
6x & \lt -20\\
x &\lt \frac{ -20}{6}= \frac{ -10}{3}
\end{aligned}
$$
so $f(x)$ is concave downward on $(-\infty , \frac{ -10}{3})$.
Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $\frac{ -10}{3} $, the graph changes from concave upward to concave downward at that point, so $(\frac{ -10}{3} , f(\frac{ -10}{3}))$ is inflection point.
Now ,we find $ f(\frac{ -10}{3})$as follows:
$$
\begin{aligned}
f(\frac{ -10}{3}) &=(\frac{ -10}{3})((\frac{ -10}{3})+5)^{2}\\
&=(\frac{ -10}{3})(\frac{ -10+15}{3})^{2}\\
&=(\frac{ -10}{3})(\frac{ 25}{9})\\
&=\frac{ -250}{27}
\end{aligned}
$$
Finally, we have inflection point at $(\frac{ -10}{3}, f(-4))=(\frac{ -10}{3} , \frac{ -250}{27})$.
