Постройте график функции y=x^2-2x-8. Найдите с помощью графика: промежуток, в котором функция убывает.

Ответ:

\[\ f\ убывает\ \ при\ x < 1.\]

\[y = - 5x^{2} + 6x\]

\[y = - 5 \cdot \left( x - \frac{3}{5} \right)^{2} + \frac{9}{5}.\]

\[y_{\max} = y\left( \frac{3}{5} \right) = \frac{9}{5} = 1,8.\]

\[y = - 3x^{2} + 7x + 1\ \ \ \ и\ \ \ y = - 5\]

\[- 5 = - 3x^{2} + 7x + 1\]

\[3x^{2} - 7x - 6 = 0\]

\[D = b^{2} - 4ac = 49 - 4 \cdot 3 \cdot ( - 6) =\]

\[= 49 + 72 = 121\]

\[x_{1} = \frac{7 + 11}{6} = \frac{18}{6} = 3\]

\[x_{2} = \frac{7 - 11}{6} = - \frac{4}{6} = - \frac{2}{3}\]

\[Ответ:x = 3\ \ и\ \ x = - \frac{2}{3}.\]

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

\[1)\ x_{0} = - \frac{b}{2a} = - \frac{4}{2} = - 2.\]

\[y_{0}( - 2) = 4 - 8 - 2 = - 6.\]

\[2)\ x² + 4x - 2 = 0\]

\[D = b^{2} - 4ac = 16 - 4 \cdot 1 \cdot ( - 2) =\]

\[= 16 + 8 = 24\]

\[x_{1} = \frac{- 4 + 2\sqrt{6}}{2} = - 2 + \sqrt{6}\]

\[x_{2} = \frac{- 4 - 2\sqrt{6}}{2} = - 2 - \sqrt{6}\]

\[3)\ x = 1 \Longrightarrow y = 1 + 4 - 3 = 3\]

\[x = 2 \Longrightarrow y = 4 + 8 - 2 = 10\]

\[\ x = 15 \Longrightarrow y = 6,25\]