1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 5, and 5, respectively. Find the smallest possible value of $f(x)$.
Answer: $\frac{39}{8}$.

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 $x_{\text {v }}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented as $f(x)=a(x-n-1.5)^{2}+c$. Since $f(n)=6, f(n+1)=5$, we get $\frac{9}{4} a+c=6, \frac{a}{4}+c=5$, from which $a=\frac{1}{2}, c=\frac{39}{8}$. But $c=f\left(x_{\text {v }}\right)$ and is the minimum value of the function.