ГДЗ по алгебре 9 класс Макарычев Задание 90

\[\boxed{\text{90\ (90).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = \frac{1}{4}x^{2}\]

\[\textbf{а)}\ y( - 2,5) = \frac{1}{4} \cdot ( - 2,5)^{2} =\]

\[= \frac{{2,5}^{2}}{4} = 1,5625;\]

\[y( - 1,5) = \frac{1}{4} \cdot ( - 1,5)^{2} = 0,5625;\]

\[y(3,5) = \frac{1}{4} \cdot {3,5}^{2} = 3,0625.\]

\[\textbf{б)}\ y = 5:\]

\[5 = \frac{1}{4}x^{2}\ \]

\[x^{2} = 20\ \ \]

\[x = \pm \sqrt{20}\text{\ \ }\]

\[x = \pm 2\sqrt{5}.\]

\[y = 3:\]

\[3 = \frac{1}{4}x^{2}\text{\ \ }\]

\[x^{2} = 12\]

\[x = \pm \sqrt{12}\text{\ \ }\]

\[x = \pm 2\sqrt{3}.\]

\[y = 2:\]

\[2 = \frac{1}{4}x^{2}\ \]

\[x^{2} = 8\ \ \]

\[x = \pm 2\sqrt{2}.\]

\[\textbf{в)}\ Функция\ убывает\ на\ \]

\[промежутке\ ( - \infty;0\rbrack\ и\ \]

\[возрастает\]

\[на\ промежутке\ \lbrack 0;\ + \infty).\]