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 equal to 2, but then it started moving along a straight line until it hit the parabola again at the point with an abscissa equal to 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: 6068.

Solution. Let the quadratic function be of the form $y=x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, on the other hand, $\left(3 ; 6069\right)$. Since $f(2)=4+2 b+c, f(4)=16+4 b+c$, then $20+6 b+2 c=12138$ or $3 b+c=6059$. Therefore, $f(3)=9+3 b+c=9+6059=6068$.

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