Respostas

2014-04-29T13:25:50-03:00
2x^2-4x+5\\\Delta=(-4)^2-4.2.5\\\Delta=16-40\\\Delta=-24

\boxed{Xv= \frac{-b}{2a}~\to~Xv= \frac{4}{4}~\to~Xv= 1}

\boxed{Yv= \frac{-\Delta}{4a} ~\to~Yv= \frac{24}{8} ~\to~Yv=3}


V(1~:~3)

 Ou deriva e iguala a zero

y=2x^2-4x+5\\y'=4x-4

4x-4=0\\4x=4\\ \\x=1

y=2(1)^2-4.1+5\\y=2-4+5\\y=3
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