Решебник по алгебре 9 класс Мерзляк Задание 347

\[\boxed{\text{347\ (347).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[f(x) = x^{2} - 6x + 8\]

\[x_{0} = \frac{6}{2} = 3;\]

\[y_{0} = 9 - 18 + 8 = - 1.\]

\[Ox:\ \ \ \ \ \]

\[x^{2} - 6x + 8 = 0\]

\[x_{1} + x_{2} = 6,\ \ x_{1}x_{2} = 8,\ \ \]

\[x_{1} = 4,\ \ x_{2} = 2\]

\[(4;0),\ \ (2;0).\]

\[\text{Oy}:\ \text{\ \ }\]

\[y = 8,\ \ \ \ (0;8).\]

\[f(1) = 1 - 6 + 8 = 3\]

\[1)f(6) = 8,\ \ \]

\[f(1) = 3.\]

\[2)\ f(x) = 8:\ \]

\[x = 0,\ \ x = 6.\]

\[f(x) = - 1:\ \ \]

\[x = 3.\]

\[f(x) \neq - 2.\]

\[2)\ y_{наим.} = - 1;\ y_{наиб.} - нет.\]

\[4)\ E(y) = \lbrack - 1; + \infty).\]

\[5)\ убывает\ ( - \infty;3\rbrack;\ \ \]

\[возрастает\ \lbrack 3;\ + \infty).\]

\[6)\ y > 0\ при\ \]

\[x \in ( - \infty;2) \cup (4; + \infty);\]

\[y < 0\ при\ x \in (2;4).\]