ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 1108

\[y = - x^{3} + ax^{2} + bx + c;\text{\ c} > 0:\]

\[1)\ x_{a}\ и\ x_{b} - абсциссы\ точек\ A\ и\ B:\]

\[y = - \left( x - x_{b} \right)\left( x - x_{a} \right)^{2} =\]

\[= \left( x_{b} - x \right)\left( x - x_{a} \right)^{2};\]

\[= - \left( x_{b} - x_{a} \right)^{2}.\]

\[2)\ y^{'}\left( x_{b} \right) = - \left( x_{b} - x_{a} \right)^{2};\]

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

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

\[3)\ Точка\ D:\]

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

\[4)\ Прямая\ проходит\ через\ D:\]

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

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

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

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

\[c = x_{b}x_{a}^{2}.\]

\[6)\ Получаем:\]

\[x_{b}x_{a}^{2} = x_{b}\left( x_{b} - x_{a} \right)^{2}\]

\[x_{a}^{2} = x_{b}^{2} - 2x_{b}x_{a} + x_{a}^{2}\]

\[2x_{b}x_{a} = x_{b}^{2}\]

\[x_{b} = 2x_{a};\]

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

\[7)\ Площадь\ \mathrm{\Delta}ABD:\]

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

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

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

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

\[x_{a} = 1;\text{\ \ \ }x_{b} = 2.\]

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

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

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

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

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

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