Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1587

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\[f(x) = - \frac{3}{5}x^{2};\ A\left( 2;\ - \frac{12}{5} \right):\]

\[f(2) = - \frac{3}{5} \bullet 2^{2} = - \frac{3}{5} \bullet 4 = - \frac{12}{5};\]

\[f^{'}(x) = - \frac{3}{5}\left( x^{2} \right)^{'} =\]

\[= - \frac{3}{5} \bullet 2x = - \frac{6}{5}x;\]

\[f^{'}(2) = - \frac{6}{5} \bullet 2 = - \frac{12}{5};\]

\[y = - \frac{12}{5} - \frac{12}{5}(x - 2) =\]

\[= - \frac{12}{5} - \frac{12}{5}x + \frac{24}{5} = \frac{12}{5}(1 - x).\]

\[С\ осью\ Oy\ (x = 0):\]

\[y(0) = \frac{12}{5} \bullet (1 - 0) = \frac{12}{5};\]

\[C\left( 0;\ \frac{12}{5} \right).\]

\[С\ осью\ Ox\ (y = 0):\]

\[\frac{12}{5}(1 - x) = 0\]

\[1 - x = 0\]

\[x = 1;\]

\[B(1;\ 0).\]

\[S_{\mathrm{\Delta}} = \frac{1}{2} \bullet OC \bullet OB = \frac{1}{2} \bullet \frac{12}{5} \bullet 1 = \frac{6}{5}.\]

\[p = \frac{OC + OB + BC}{2} =\]

\[= \frac{\frac{12}{5} + 1 + \sqrt{\left( \frac{12}{5} \right)^{2} + 1^{2}}}{2} =\]

\[= \frac{\frac{12}{5} + \frac{5}{5} + \sqrt{\frac{144}{25} + \frac{25}{25}}}{2} =\]

\[= \frac{\frac{17}{5} + \sqrt{\frac{169}{25}}}{2} = \left( \frac{17}{5} + \frac{13}{5} \right) \bullet \frac{1}{2} =\]

\[= \frac{30}{5} \bullet \frac{1}{2} = \frac{15}{5} = 3.\]

\[r = \frac{S}{p} = \frac{6}{5}\ :3 = \frac{2}{5} = 0,4.\]

\[Ответ:\ \ r = 0,4.\]