Respostas

2014-05-16T19:24:39-03:00
As coordenadas do vértice são o Xv e Yv

f(x)=x^2-2x-3\\
\\\Delta=(-2)^2-4*1*(-3)\\\Delta=4+12\\\Delta=16

\boxed{Xv= \frac{-b}{2a} ~\to~Xv= \frac{-(-2)}{2*1} ~\to~Xv=1}

\boxed{Yv= \frac{-\Delta}{4a} ~\to~ Yv=\frac{-16}{4} ~\to~-4}

\boxed{V(1;-4)}
1 5 1