Dada a funcao y = x2 – 4x + 3, podemos que o vértice da parábola tem coordenadas:
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Oi!

V(Xv, Yv)


Resposta: V(- 2, - 1)
Bons estudos.
V(Xv, Yv)
Resposta: V(- 2, - 1)
Bons estudos.
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