What is the vertex of # y=2(x-2)^2-11 #?

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Answer 1 · 2016-02-06T09:08:08

Vertex is at #(2, -11)#

This is a parabola which opens upward

of the form #(x-h)^2=4p(y-k)# where vertex is #(h, k)#

from the given #y=2(x-2)^2-11#

transform first to the form

#y=2(x-2)^2-11#

#y+11=2(x-2)^2#

#(y+11)/2=(2(x-2)^2)/2#

#(y+11)/2=(cancel2(x-2)^2)/cancel2#

#1/2*(y+11)=(x-2)^2#

#(x-2)^2=1/2*(y+11)#

#(x-2)^2=1/2*(y--11)#

so that #h=2# and #k=-11#

vertex is at #(2, -11)#

Kindly see the graph
graph{y=2(x-2)^2-11[-5,40,-15,10]}

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