Question

Найдите промежутки знакопостоянства функции a)y=x^2+2x-8 b)y=-2x^2+2x-7

Answer 1

$f(x)=ax^2+bx+c\\\\x^2+2x-8\\D=4+4\cdot8=36\\\left.\begin{array}{cc}x_1=\frac{-2+\sqrt{36}}2=2&x_2=\frac{-2-\sqrt{36}}2=-4\end{array}\right.\\f(x)>0,x\in(-\infty;-4)\cup(2;\infty)\\f(x)<0,x\in(-4;2)\\\\-2x^2+2x-7\\D=4-56<0\\x\in\emptyset\\f(x)<0,x\in R$