Answer
$$
f(x)=x^{2}-12x+36
$$
Critical number: 6.
by Part 2 of the second derivative test, 6 leads to a relative minimum .
Work Step by Step
$$
f(x)=x^{2}-12x+36
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=(2)x-12+0\\
&=2x-12
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =2x-12 &=0\\
2x & =12\\
x & =\frac{12}{2}=6
\end{aligned}
$$
Critical number: 6.
Now use the second derivative test. The second derivative is
$$
f^{\prime\prime}(x)=2.
$$
Evaluate $f^{\prime\prime}(x)$ at $6$ , getting
$$
f^{\prime\prime}(6)=2 \gt 0.
$$
so that by Part 2 of the second derivative test, $6$ leads to a relative minimum of
$$
f(6)=(6)^{2}-12(6)+36=0
$$
