Question

1. Знайдіть координати вершини параболи
1)y=x²+2x-8
2)y=-2x²-4x+1

Answer 1

$\displaystyle\bf\\1)\\\\y=x^{2} +2x-8\\\\\\X_{v} =-\frac{b}{2a} =-\frac{2}{2\cdot 1} =-1\\\\\\Y_{v}=(-1)^{2} +2\cdot(-1)-8=1-2-8=-9\\\\\\Otvet:(-1 \ ; \ -9)\\\\\\2)\\\\y=-2x^{2} -4x+1\\\\\\X_{v} =-\frac{b}{2a} =-\frac{-4}{2\cdot (-2)} =-1\\\\\\Y_{v}=-2\cdot(-1)^{2} -4\cdot(-1)+1=-2+4+1=3\\\\\ Otvet: (-1 \ ; \ 3)$