What is the vertex of # y= 2|x|^2 – 1 #?

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Answer 1 · 2016-01-30T09:51:25

Vertex : #(0,-1)#

#y=2abs(x)^2-1#

This should give us a parabola and this equation is same as

#y=2x^2-1# as #abs(x)^2 # and #x^2# would give the same value as on squaring we would get only the positive value.

The vertex of #y=2x^2-1# can be found by comparing it with the vertex form
#y=a(x-h)^2+k# where #(h,k)# is the vertex

#y=2(x-0)^2-1#
#y=a(x-h)^2 +k#

We can see #h=0# and #k=-1#

Vertex is #(0,-1)#