2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to -1. Initially, the fly moved exactly along the parabola up to the point with an abscissa of 2, but then it started moving along a straight line until it hit the parabola again at the point with an abscissa of 4. Find $f(3)$, given that the line $y=2023 x$ intersects the fly's path along the straight line segment at its midpoint.
Answer: 6070.

Solution. Let the quadratic function be of the form $y=-x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, and on the other hand, $(3 ; 6069)$. Since $f(2)=-4+2 b+c, f(4)=-16+4 b+c$, then $-20+6 b+2 c=12138$ or $3 b+c=6079$. Therefore, $f(3)=-9+3 b+c=-9+6079=6070$.

Grading criteria. Full points for a complete and justified solution - 12 points. For a correct approach with arithmetic errors, minus 3 points.