Task B-2.4. For the function $f(x)=x^{2}-b x+c$, it holds that $f(5-x)=f(5+x)$ for every $x \in \mathbb{R}$. For which real numbers $b$ and $c$ does the product $f(1) \cdot f(2)$ have the minimum value?
## Solution.

From the equality $f(5-x)=f(5+x)$, we conclude that $x_{0}=5$ if $\left(x_{0}, f\left(x_{0}\right)\right)$ is the vertex of the graph of the given quadratic function.

(2 points)

It follows that

$$
\frac{-b}{2}=5 \Rightarrow b=10
$$

Then the product is

$$
\begin{aligned}
f(1) \cdot f(2) & =(1-b+c)(4-2 b+c)= \\
& =(-9+c)(-16+c)= \\
& =c^{2}-25 c+144
\end{aligned}
$$

a quadratic function which attains its minimum value for $c=\frac{25}{2}$.