Answer
$$
f(x)=-x^{2}-10x-25
$$
Critical number: -5.
A relative maximum occurs at $x=-5$
Work Step by Step
$$
f(x)=-x^{2}-10x-25
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=-(2)x-10-0\\
&=-2x-10
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =-2x-10&=0\\
-2x & =10\\
x & =\frac{10}{-2}=-5
\end{aligned}
$$
Critical number: -5.
Now use the second derivative test. The second derivative is
$$
f^{\prime\prime}(x)=-2.
$$
Evaluate $f^{\prime\prime}(x)$ at $-5$ , getting
$$
f^{\prime\prime}(-5)=-2 \lt 0.
$$
so that by Part 2 of the second derivative test, leads to a relative maximum of
$$
f(-5)=-(-5)^{2}-10(-5)-25=0
$$
