Решебник по алгебре 9 класс Мерзляк Задание 343

 

\[\boxed{\text{343\ (343).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[\text{M\ }(0;\ - 8):\]

\[- 6 \cdot 0 + 0 + c = - 8\]

\[c = - 8\]

\[Ответ:\ c = - 8.\]

\[\boxed{\text{343.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ f(x) = \frac{x^{2} - 16}{x + 4}\]

\[x + 4 \neq 0\]

\[x \neq - 4\]

\[D(f) = ( - \infty; - 4) \cup ( - 4; + \infty).\]

\[\frac{x^{2} - 16}{x + 4} = \frac{(x + 4)(x - 4)}{(x + 4)} =\]

\[= x - 4\]

\[y = x - 4\]

\[2)\ f(x) = \frac{12x - 72}{x^{2} - 6x}\]

\[x^{2} - 6x \neq 0\]

\[x \cdot (x - 6) \neq 0\]

\[x \neq 0\]

\[x \neq 6.\]

\[D(f) =\]

\[= ( - \infty;0) \cup (0;6) \cup (6; + \infty).\]

\[\frac{12x - 72}{x(x - 6)} = \frac{12 \cdot (x - 6)}{x \cdot (x - 6)} = \frac{12}{x}\]

\[y = \frac{12}{x}\]

\[3)\ f(x) = \frac{x^{2} - 9}{x^{2} - 9}\]

\[x^{2} - 9 \neq 0\]

\[x \neq 3\]

\[x \neq - 3.\]

\[D(f) =\]

\[= ( - \infty; - 3) \cup ( - 3;3) \cup (3; + \infty).\]

\[\frac{x^{2} - 9}{x^{2} - 9} = 1\]

\[y = 1\]

\(\ \)

\[4)\ \ f(x) = \frac{x^{3} - 4x}{x^{2} - 2x} = \frac{x\left( x^{2} - 4 \right)}{x(x - 2)} =\]

\[= \frac{(x - 2)(x + 2)}{x - 2} = x + 2;\ \]

\[\ x \neq 0;\ \ x \neq 2\]