1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 14, and 14, respectively. Find the greatest possible value of $f(x)$.
Answer: 15.

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 $f(n)=6, f(n+1)=14$, we get $\frac{9}{4} a+c=6, \frac{a}{4}+c=14$, from which $a=-4, c=15$. But $c=f\left(x_{\text{v}}\right)$ and is the maximum value of the function.