Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1619

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\[f(x) = x^{2} + px + q;\ y = 2x - 3.\]

\[x = \ 1:\]

\[1^{2} + p \bullet 1 + q = 2 \bullet 1 - 3\]

\[1 + p + q = 2 - 3\]

\[1 + p + q = - 1\]

\[p + q = - 2\]

\[q = - 2 - p.\]

\[Вершина\ параболы:\]

\[x = - \frac{b}{2a} = - \frac{p}{2};\]

\[y = f\left( - \frac{p}{2} \right) = \frac{p^{2}}{4} - \frac{p^{2}}{2} + q =\]

\[= \frac{p^{2} - 2p^{2} + 4q}{4} = \frac{4q - p^{2}}{4}.\]

\[От\ вершины\ до\ оси\ Ox:\]

\[d(p) = \left| \frac{4q - p^{2}}{4} \right| =\]

\[= \left| \frac{4( - 2 - p) - p^{2}}{4} \right| =\]

\[= \left| - \frac{8 + 4p + p^{2}}{4} \right| =\]

\[= \frac{p^{2} + 4p + 8}{4}.\ \]

\[d^{'}(p) = \frac{1}{4} \bullet (\left( p^{2} \right)^{'} + (4p + 8)^{'} =\]

\[= \frac{1}{4} \bullet (2p + 4) = \frac{1}{2} \bullet (p + 2).\]

\[Промежуток\ возрастания:\]

\[p + 2 > 0\]

\[\ p > - 2.\]

\[p = - 2 - точка\ минимума;\]

\[q = - 2 - ( - 2) = - 2 + 2 = 0;\]

\[d( - 2) = \frac{4 + 4 \bullet ( - 2) + 8}{4} =\]

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

\[Ответ:\ \ p = - 2;\ \ q = 0;\ \ d = 1.\]

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