What is the vertex of # y= 2|x|^2 – 4x+1 #?

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Answer 1 · 2018-02-16T09:04:32

#y_"vertex" = (1,-1)#

#y= 2abs(x)^2-4x+1#

First note that #absx^2 =x^2#

Hence, #y= 2x^2-4x+1#

#y# is a parabolic function of the form #y=ax^2+bx+c# which has a vertex at #x=-b/(2a)#

#x= - (-4)/(2*2) =1#

#y(1) = 2-4+1 =-1#

Hence, #y_"vertex" = (1,-1)#

We can see this result from the graph of #y# below:

graph{2abs(x)^2-4x+1 [-5.55, 6.936, -2.45, 3.796]}