1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values $-9, -9$, and $-15$ respectively. Find the greatest possible value of $f(x)$.
Answer: $-\frac{33}{4}$.

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 $x_{\text {v }}$, 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 $f(n)=-9, f(n+2)=-15$, we get $\frac{a}{4}+c=-9, \frac{9 a}{4}+c=-15$, from which $a=-3, c=-\frac{33}{4}$. But $c=f\left(x_{\text {v }}\right)$ and is the maximum value of the function.