Question

Найдите координаты вершины параболы и точек ее пересечения с осями Ох и Оу
У= - 2х^2+х - 1/9.
У=8х^2-х- 1/4.

Answer 1

$1)\; \; y=-2x^2+x-\frac{1}{9}\\\\x(versh.)=-\frac{b}{2a}=-\frac{1}{-4}=\frac{1}{4}\\\\y(versh.)=y(\frac{1}{4})=-2\cdot \frac{1}{16}+\frac{1}{4}-\frac{1}{9}=-\frac{1}{8}+\frac{1}{4}-\frac{1}{9}=\frac{-9+18-8}{72}=\frac{1}{72}\\\\V(\frac{1}{4}\, ;\, \frac{1}{72})\\\\OY:\; \; x=0\; \; \Rightarrow \; \; \; y=-\frac{1}{9}\; \; ,\; \; \underline {A(0,-\frac{1}{9})}\\\\OX:\; \; y=0\; \; \Rightarrow \; \; -2x^2+x-\frac{1}{9}=0\\\\D=1-\frac{8}{9}=\frac{1}{9}\; \; ,\; \; x_{1,2}=\frac{-1\pm \frac{1}{3}}{-4}$

$x_1=\frac{-1-\frac{1}{3}}{-4}=\frac{-4}{-3\cdot 4}=\frac{1}{3}\; \; ,\; \; x_2=\frac{-1+\frac{1}{3}}{-4}=\frac{-2}{-3\cdot 4}=\frac{1}{6}\\\\\underline {B(\frac{1}{3}\, ;\, 0)\; \; ,\; \; C(\frac{1}{6}\, ;\, 0)}$

$2)\; \; y=8x^2-x-\frac{1}{4}\\\\x(versh.)=\frac{1}{2\cdot 8}=\frac{1}{16}\\\\y(versh.)=8\cdot \frac{1}{256}-\frac{1}{16}-\frac{1}{4}=\frac{1-2-8}{32}=-\frac{9}{32}\\\\\underline {V(\frac{1}{16}\, ;\, -\frac{9}{32})}\\\\OY:\; \; x=0\; \; \Rightarrow \; \; \; y=-\frac{1}{4}\; \; ,\; \; \underline {A(0\, ,\, -\frac{1}{4})}$

$OX:\; \; y=0\; \; \Rightarrow \; \; \; 8x^2-x-\frac{1}{4}=0\\\\D=1+8=9\; \; ,\; \; x_{1,2}=\frac{1\pm 3}{16}\\\\x_1=\frac{1-3}{16}=-\frac{1}{8}\; \; ,\; \; x_2=\frac{1+3}{16}=\frac{1}{4}\\\\\underline {B(-\frac{1}{8}\, ,\, 0)\; \; ,\; \; C(\, \frac{1}{4}\, ,\, 0)}$