Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 139

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\[1)\ y = x^{2} + 2x - 4\]

\[\textbf{а)}\ x_{0} = - \frac{2}{2} = - 1;\ \ \]

\[y_{0} = 1 - 2 - 4 = - 5.\]

\[\textbf{б)}\ x^{2} + 2x - 4 = 0\]

\[D_{1} = 1 + 4 = 5\]

\[x_{1} = - 1 + \sqrt{5};\ \ x_{2} = - 1 - \sqrt{5}.\]

\[\textbf{в)}\ y > 0\ при\ x < - 1 - \sqrt{5};\ \]

\[\ x > - 1 + \sqrt{5};\]

\[y < 0\ при\ \ \]

\[- 1 - \sqrt{5} < x < - 1 + \sqrt{5}.\]

\[\textbf{г)}\ y_{наим} = - 5.\ \]

\[2)\ y = - x^{2} + 4x - 5\]

\[\textbf{а)}\ x_{0} = \frac{4}{2} = 2;\ \]

\[\ y_{0} = - 4 + 8 - 5 = - 1.\]

\[\textbf{б)}\ нулей\ нет.\ \]

\[\textbf{в)}\ y > 0 - нет;\]

\[y < 0\ при\ любом\ \text{x.}\]

\[\textbf{г)}\ y_{наиб} = - 1.\]

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

\[\textbf{а)}\ x_{0} = - \frac{1}{2} = - 0,5;\ \ \]

\[y_{0} = - 0,25 + 0,5 + 2 = 2,25.\]

\[\textbf{б)}\ x = - 2;\ \ x = 1.\ \]

\[\textbf{в)}\ y > 0\ при\ \ - 2 < x < 1;\]

\[y < 0\ при\ x < - 2;\ \ x > 1.\]

\[\textbf{г)}\ y_{наиб} = 2,25.\]

\[4)\ y = x^{2} + 4x - 5\]

\[\textbf{а)}\ x_{0} = - \frac{4}{2} = - 2;\ \]

\[\ y_{0} = 4 - 8 - 5 = - 9.\]

\[\textbf{б)}\ x_{1} = - 5;x_{2} = 1.\]

\[\textbf{в)}\ y > 0\ при\ x < - 5;\ \ x > 1.\]

\[y < 0\ при\ \ - 5 < x < 1.\]

\[\textbf{г)}\ y_{наим} = - 9.\ \]