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

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Answer 1 · 2018-03-30T08:31:43

Vertex is #(-5/6,-71/12)#

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

= #-(x^2+4x+4)-2x^2-x-4#

= #-3x^2-5x-8#

= #-3(x^2+5/3x+(5/6)^2)-(-3)(5/6)^2-8#

= #-3(x+5/6)^2+25/12-8#

= #-3(x+5/6)^2-71/12#

Now it is in vertex form #y=a(x-h)^2+k#

and vertex is #(-5/6,-71/12)#

graph{-(x+2)^2-2x^2-x-4 [-6.876, 3.124, -8.7, -3.7]}