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Qual as coordenadas do vertice da parabola das função.

a) y=x²-4x-12

b)y=x²-6x+9

c)y-2x²-5x+2

d)y=4x²-9x+2



RESOLVENDO

A) y=x²-4x-12
Xv= - b/2a ==> xv= -(-4)/2.1 ==> Xv=4/2 ==> Xv= 2
delta= (-4)^2 -4.1.(-12)=16+48=64
Yv= - delta = - 64 ==> Yv = - 16
4a 4.1
b)y=x²-6x+9
Xv= - b/2a ==> xv= -(-6)/2.1 ==> Xv=6/2 ==> Xv= 3
delta= (-6)^2 -4.1.9=36-36=0
Yv= - delta = - 0 ==> Yv = 0
4a 4.1
c)y= -2x²-5x+2
Xv= - b/2a ==> xv= -(-5)/2.(-2) ==> Xv= - 5/4
delta= (-5)^2 -4.(-2).2=25+16=41
Yv= - delta = - 41 ==> Yv = 41
4a 4.(-2) 8
d)y=4x²-9x+2
Xv= - b/2a ==> xv= -(-9)/2.4 ==> Xv=9/8
delta= (-9)^2-4.4.2=81-32=49
Yv= - delta = - 49 ==> Yv = - 49
4a 4.1 4



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