Страница 59 номер 237, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

Пересечение с осью X при \(y = 0\), пересечение с осью Y при \(x = 0\).

1) \(f(x) = \frac{1}{6}x - 7\)
\(x = 0 \Rightarrow y = f(0) = \dfrac{1}{6} \cdot 0 - 7 = - 7\)
\(y = 0 \Rightarrow \dfrac{1}{6}x - 7 = 0 \Rightarrow \dfrac{1}{6}x = 7 \Rightarrow x = 42\)

2) \(f(x) = \frac{20 + 4x}{3x - 5}\)
\(x = 0 \Rightarrow f(0) = \dfrac{20 + 4 \cdot 0}{3 \cdot 0 - 5} = \dfrac{20}{- 5} = - 4\)
\(y = 0 \Rightarrow \dfrac{20 + 4x}{3x - 5} = 0 \Rightarrow \begin{cases} 20 + 4x = 0 \\ 3x - 5 \neq 0 \end{cases}\ \Rightarrow \begin{cases} x = - 5 \\ x \neq \frac{5}{3} \end{cases}\ \Rightarrow x = - 5\)

3) \(g(x) = 9 - x^{2}\)
\(x = 0 \Rightarrow g(0) = 9 - 0 = 9 \Rightarrow g(x) = 9\)
\(y = 0 \Rightarrow 9 - x^{2} = 0 \Rightarrow x^{2} = 9 \Rightarrow x = \pm 3\)

4) \(\varphi(x) = x^{2} + 2x - 3\)
\(x = 0 \Rightarrow \varphi(0) = 0^{2} + 2 \cdot 0 - 3 = - 3\)
\(y = 0 \Rightarrow x^{2} + 2x - 3 = 0 \Rightarrow (x - 1)(x + 3) = 0 \Rightarrow\)
\(\Rightarrow x = 1;\ x = - 3\)

Ответ: 1) \(x = 0 \Rightarrow f(0) = - 7;\ y = 0 \Rightarrow x = 42\); 2) \(x = 0 \Rightarrow f(0) = - 4;\ y = 0 \Rightarrow x = - 5\); 3) \(x = 0 \Rightarrow g(0) = 9;\ y = 0 \Rightarrow x = \pm 3\); 4) \(x = 0 \Rightarrow \varphi(0) = - 3;\ y = 0 \Rightarrow x = 1;\ x = - 3\).