Question

Com relação a função quadrática definida por y=-2x²-4x+6 quais são as coordenadas do vértice do gráfico dessa função

Answer 1

Δ=b²-4ac
Δ=-4²-4*(-2)*6
Δ= 16+48
Δ= 64

VERTICES

Xv= $\frac{-b}{2a} = \frac{-(-4)}{2*(-2)} = \frac{4}{-4}$=-1
Yv=$\frac{ delta }{4a} = \frac{-64}{4*(-2)} = \frac{-64}{-8}$=8

Answer 2

Primeiramente iremos descobrir as raízes:

$Primeiramente\ iremos\ descobrir\ as\ raizes:\\\\\Delta=b^{2} -4ac\\\\\Delta=4^{2} -4(-2)6\\\\\Delta=16+48\\\\\Delta=64$

$x'\ e\ x"=-\frac{-b\ +-\ \sqrt{\Delta} }{2.a} \\\\x'\ e\ x"= \frac{-(-4)+ou -\ 8}{2.(-2)} \\\\x'\ = \frac{-4+8}{-4} = -1\\\\x"\ =\frac{-4\ -\ 8}{-4} = 3\\\\x'=-1\\x"=3$

$agora\ iremos\ descobrir\ o\ x^{vertice}\ e\ o\ y^{vertice} \\\\x^{v}=\frac{-b}{2a} \\\\x^{v}=\frac{4}{2(-2)} \\\\x^{v}=-1\\\\y^{v}=\frac{-\Delta}{4a} \\\\y^{v}=\frac{-64}{4(-2)} \\\\y^{v}=\frac{-64}{-8} \\\\y^{2v}=8$

$logo\ abaixo\ esta\ o\ grafico:$