ГДЗ по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 73

\[\boxed{\mathbf{73.}}\]

\[Уравнение\ полуокружности:\]

\[y = \sqrt{R^{2} - x^{2}};\]

\[x = \sqrt{R^{2} - y^{2}}.\]

\[\textbf{а)}\ R = 2;\ \ O(0;0):\]

\[y = f(x) = \sqrt{2^{2} - x^{2}} = \sqrt{4 - x^{2}}.\]

\[\textbf{б)}\ R = 2;\ \ O(0;0):\]

\[y = f(x) = - \sqrt{2^{2} - x^{2}} =\]

\[= - \sqrt{4 - x^{2}}.\]

\[\textbf{в)}\ R = 2;\ \ O(0;0):\]

\[x = \varphi(y) = \sqrt{2^{2} - y^{2}} = \sqrt{4 - y^{2}}.\]

\[\textbf{г)}\ R = 2;\ \ O(0;0):\]

\[x = \varphi(y) = - \sqrt{2^{2} - y^{2}} =\]

\[= - \sqrt{4 - y^{2}}.\]

\[\textbf{д)}\ R = 2;O(1;1):\]

\[y = f(x) = 1 + \sqrt{2^{2} - (x - 1)^{2}} =\]

\[= 1 + \sqrt{4 - (x - 1)^{2}}.\]

\[\textbf{е)}\ R = 2;\ \ O( - 1;1):\]

\[y = f(x) = - \sqrt{2^{2} - (x + 1)^{2}} +\]

\[+ 1 = 1 - \sqrt{4 - (x + 1)^{2}}\text{.\ }\]