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$\textbf{Составим систему уравнений}$ по точкам (-1; 2), (-2; -3) и (-3; -4):

$\begin{cases} a - b + c = 2 \\ 4a - 2b + c = -3 \\ 9a - 3b + c = -4 \end{cases}$

$\textbf{Исключим переменную $c$}$ , вычитая уравнения друг из друга:

$\begin{cases} 3a - b = -5 \\ 5a - b = -1 \end{cases}$

$\textbf{Найдем коэффициенты:}$

$2a = 4 \implies a = 2$

$3(2) - b = -5 \implies b = 11$

$2 - 11 + c = 2 \implies c = 11$

$\textbf{Вычислим $f(2)$:}$

$f(x) = 2x^2 + 11x + 11$

$f(2) = 2 \cdot 2^2 + 11 \cdot 2 + 11 = 8 + 22 + 11 = 41</script</article></body></html>$

Задание №41815 ЕГЭ по Математике