Odpowiedzi

2009-04-14T22:16:25+02:00
F(x)=-3x²+2x+5

f(x)=a(x-b/2a)²-Δ/4a
Δ=4+60=64

Postac kononiczna
f(x)=-3(x+1/3)²+64/12
f(x)=-3(x+1/3)²+16/3

x1=(-2-8)/(-6)=5/3
x2=(-2+8)/(-6)=-1

postac iloczynowa
f(x)=-3(x-5/3)(x+1)