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

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Answer 1 · 2017-03-24T03:34:47

The vertex is #(4/3,-47/3)#

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

This is not in vertex form yet, so we need to expand and organize the quadratic, complete the square, then determine the vertex.

Expand:
#y=-x^2-4x-3-2(x^2-6x+9)#

#y=-x^2-4x-3-2x^2+12x-18#

Organize:
#y=-3x^2+8x-21#

Complete the square:
#y=-3[x^2-(8x)/3+7]#

#y=-3[(x-4/3)^2-16/9+7]#

#y=-3[(x-4/3)^2+47/9]#

#y=-3(x-4/3)^2-3(47/9)#

#y=-3(x-4/3)^2-47/3#

Determine vertex:
Vertex form is #y=a(x-color(red)(h))^2+color(blue)(k)# where #(color(red)(h),color(blue)(k))# is the vertex of the parabola.

The vertex is therefore at #(color(red)(4/3),color(blue)(-47/3))#.

Double check with graph:
graph{y=-x^2-4x-3-2(x-3)^2 [-30, 30, -30, 5]}