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

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Feb 21, 2018

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Hence, the vertex is
I have approached by the method of calculus (maxima and minima)
#V-=(x,y)=V-=(-1/4,-34/16)#

Explanation:

I have approached by the method of calculus (maxima and minima)
The curve is symmetrical about an axis parallel to y axis.
The vertex is the point where #dy/dx=0#

Given:

#y=-x^2-2x-3(x/3-2/3)^2#

Differentiating wrt x

#dy/dx=-2x-2-3xx2(x/3-2/3)xx1/3#

#dy/dx=0#

#-2x-2-3xx2(x/3-2/3)xx1/3=0#

#-2x-2-2/3x+4/3=0#

#-2x-2/3x=2-4/3#
#-6/3x-2/3x=6/3-4/3#

#-6x-2x=6-4#

#-8x=2#

#8/8x=-2/8#

#x=-1/4#

#y=-x^2-2x-3(x/3-2/3)^2#

#y=-(-1/4)^2-2(-1/4)-3((-1/4)/3-2/3)^2#

#=-1/16+1/2-3(-1/12-2/3)^2#

#=-1/16+8/16-3(-1/12-8/12)^2#

#=(-1+8)/16-3((-1-8)/12)^2#

#=7/16-3(-9/12)^2#

#=7/16-3(-3/4)^2#

#=7/16-3xx9/16#

#7/16-27/16#

#y=-34/16#

Hence, the vertex is

#V-=(x,y)=V-=(-1/4,-34/16)#

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