ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 302

\[y = a(x - b)^{2};\ \ a < 0.\]

\[1)\ Касается\ y = \frac{x}{2} - 3:\]

\[y^{'}(x) = a \bullet 2(x - b) = 2ax - 2ab;\]

\[2ax - 2ab = \frac{1}{2}\]

\[x - b = \frac{1}{4a}\]

\[b = x - \frac{1}{4a}.\]

\[2)\ Точка\ касания:\]

\[a(x - b)^{2} = \frac{x}{2} - 3\]

\[a\left( x - x + \frac{1}{4a} \right)^{2} = \frac{x}{2} - 3\]

\[a \bullet \frac{1}{16a^{2}} = \frac{x}{2} - 3\]

\[\frac{1}{16a} = \frac{x}{2} - 3\]

\[\frac{x}{2} = 3 + \frac{1}{16a}\]

\[x = 6 + \frac{1}{8a};\]

\[b = 6 + \frac{1}{8a} - \frac{1}{4a} = 6 - \frac{1}{8a}.\]

\[3)\ x = b = 6 - \frac{1}{8a}.\]

\[4)\ y = - a(0 - b)^{2} = - ab^{2} =\]

\[= - a \bullet \left( 6 - \frac{1}{8a} \right)^{2} =\]

\[= - a\left( 36 - \frac{3}{2a} + \frac{1}{64a^{2}} \right) =\]

\[= \frac{3}{2} - 36a - \frac{1}{64a}.\]

\[5)\ Сумма\ искомых\ отрезков:\]

\[S(a) = x + y =\]

\[= 6 - \frac{1}{8a} + \frac{3}{2} - 36a - \frac{1}{64a} =\]

\[= \frac{15}{2} - 36a - \frac{9}{64a};\]

\[S^{'}(a) = 0 - 36 - \frac{9}{64} \bullet \left( - \frac{1}{a^{2}} \right) =\]

\[= \frac{9\left( 1 - 256a^{2} \right)}{64a^{2}}.\]

\[6)\ 1 - 256a^{2} \geq 0\]

\[256a^{2} - 1 \leq 0\]

\[(16a + 1)(16a - 1) \leq 0\]

\[- \frac{1}{16} \leq a \leq \frac{1}{16}.\]

\[7)\ Точка\ минимума:\]

\[a = - \frac{1}{16};\]

\[b = 6 + 2 = 8.\]

\[Ответ:\ \ y = - \frac{1}{16}(x - 8)^{2}.\]