Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1107

\[y = x^{3} + ax^{2} + bx + c;\ c < 0.\]

\[1)\ x_{m}\ и\ x_{n} - абсциссы\text{\ M\ }и\ N:\]

\[y = \left( x - x_{m} \right)\left( x - x_{n} \right)^{2};\]

\[= \left( x_{m} - x_{n} \right)^{2}.\]

\[2)\ y^{'}\left( x_{m} \right) = \left( x_{m} - x_{n} \right)^{2};\text{\ \ \ }\]

\[y\left( x_{m} \right) = 0;\]

\[y = \left( x_{m} - x_{n} \right)^{2} \bullet \left( x - x_{m} \right).\]

\[3)\ Точка\ A:\]

\[y(0) = 0^{3} + a \bullet 0^{2} + b \bullet 0 + c = c.\]

\[4)\ Прямая\ проходит\ через\ A:\]

\[\left( x_{m} - x_{n} \right)^{2} \bullet \left( 0 - x_{m} \right) = c\]

\[c = - x_{m}\left( x_{m} - x_{n} \right)^{2}.\]

\[5)\ Функция\ проходит\ через\ A:\ \]

\[y(0) = \left( 0 - x_{m} \right)\left( 0 - x_{n} \right)^{2}\]

\[c = - x_{m}x_{n}^{2}.\]

\[6)\ Получаем:\]

\[- x_{m}x_{n}^{2} = - x_{m}\left( x_{m} - x_{n} \right)^{2}\]

\[x_{n}^{2} = x_{m}^{2} - 2x_{m}x_{n} + x_{n}^{2}\]

\[2x_{m}x_{n} = x_{m}^{2}\]

\[x_{m} = 2x_{n};\]

\[c = - 2x_{n} \bullet x_{n}^{2} = - 2x_{n}^{3}.\]

\[7)\ Площадь\ \mathrm{\Delta}AMN:\]

\[S = \frac{1}{2} \bullet \left| - 2x_{n}^{3} \right| \bullet \left( x_{m} - x_{n} \right) = 1\]

\[2x_{n}^{3} \bullet \left( 2x_{n} - x_{n} \right) = 2\]

\[x_{n}^{3} \bullet x_{n} = 1\]

\[x_{n}^{4} = 1\]

\[x_{n} = 1;\text{\ \ \ }x_{m} = 2.\]

\[8)\ y = (x - 2)(x - 1)^{2} =\]

\[= (x - 2)\left( x^{2} - 2x + 1 \right);\]

\[y = x^{3} - 2x^{2} + x - 2x^{2} + 4x - 2 =\]

\[= x^{3} - 4x^{2} + 5x - 2;\]

\[a = - 4;\ \ \ b = 5;\ \ \ c = - 2.\]

\[Ответ:\ \ a = - 4;\ b = 5;\ c = - 2.\]