## Task B-2.2.

The function $f(x)=x^{2}+p x+q$ takes negative values only for $x \in\langle-3,14\rangle$. How many integer values from the set $[-100,-10]$ can the function $f$ take?
## Solution.

If the function $f$ takes negative values only on the interval $\langle-3,14\rangle$, the numbers -3 and 14 are its roots.

Then, according to Vieta's formulas,

$$
-3+14=-p, -3 \cdot 14=q, \quad \text{or} \quad p=-11, q=-42
$$

which means $f(x)=x^{2}-11 x-42$.

The minimum value of this function is

$$
y_{\min }=f\left(\frac{-3+14}{2}\right)=-\frac{289}{4}=-72.25
$$

so $f(x) \geqslant-72.25$ for all real numbers $x$.

Therefore, $f(x)$ can only be an integer that is greater than or equal to -72,

and on the interval $[-100,-10]$ the function $f$ can take integer values from the set $\{-72,-71,-70, \ldots,-11,-10\}$, which contains 63 values.

1 point