What is the vertex of #y=-2(x+3)^2+12x+4 #?

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Answer 1 · 2017-04-05T07:09:47

vertex #(0, -14)#

Given -

#y=-2(x+3)^2+12x+4#

#y=-2(x^2+6x+9)+12x+4#

#y=-2x^2-12x-18+12x+4#

#y=-2x^2-14#

#x# term is missing in the expression #-2x^2-14#
Let us supply it.

#y=-2x^2+0x-14#

#x=(-b)/(2xxa)=0/(2xx (-2))=0#

At #x=0#

#y=-2(0)^2-14=-14#

vertex #(0, -14)#