Matemática, perguntado por PABLOAMORIM, 1 ano atrás

PFV É PRA DE NOITE ME AJUDEM
VERTISE DE UMA PARABOLA
X²-5X+6=0


PABLOAMORIM: POR FAVORRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR!!!!!!!!!!!!!!!!

Soluções para a tarefa

Respondido por emicosonia
1
PFV É PRA DE NOITE ME AJUDEMVÉRTICE DE UMA PARÁBOLA
Vértice da parábola

Xv = (Xis do vértice)
Yv = ( ipsylon do vértice)
ax² + bx + c = 0

X²-5X+6=0     -------------a > 0 pois a = 1 (concavidade voltada para CIMA)
a = 1
b = - 5
c = 6
Δ = b² - 4ac
Δ = (-5)² - 4(1)(6)
Δ = + 25 - 24
Δ = 1

Xv = -b/2a
Xv = -(-5)/2(1)
Xv = + 5/2     ou        Xv = 2,5
e
Yv = - 
Δ/4a
Yv = -1/4(1)
Yv = - 1/4        ou     Yv = 0,25

VÉRTICE { 5/2; -1/4}
ou
VÉRTICE { 2,5; -0,25} 

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