1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 13, 13, and 35, respectively. Find the smallest possible value of $f(x)$.
Answer: $\frac{41}{4}$.

Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\mathrm{B}}$, then $x_{\mathrm{B}}=n+0.5$, and thus $f(x)$ can be represented as $f(x)=a(x-n-0.5)^{2}+c$. Since $f(n)=13, f(n+2)=35$, we get $\frac{a}{4}+c=13, \frac{9 a}{4}+c=35$, from which $a=11, c=\frac{41}{4}$. But $c=f\left(x_{\text {v }}\right)$ and is the minimum value of the function.