What is the vertex form of #y=-3x^2-6x-3#?

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Answer 1 · 2017-06-15T05:40:44

#y=-3(x+1)^2#

Given -

#y=-3x^2-6x-3#

Vertex -

x-coordinate of the vertex

#x=(-b)/(2a)=(-(-6))/(2 xx(-3))=6/(-6)=-1#

Y-coordinate of the vertex

At #x=-1#

#y=-3(-1)^2-6(-1)-3=-3+6-3=0#

Vertex #(-1,0)#

Vertex form of the equation is -

#y=a(x-h)^2+k#

#a=-3# - coefficient of #x^2#
#h=-1#
#k=2#

#y=-3(x+1)^2+0#

#y=-3(x+1)^2#