How do you write #y=3x^2-12x+1# in vertex form?

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Answer 1 · 2018-04-21T05:07:46

Given -

#y=3x^2-12x+1#

#x=(-b)/(2a)=(-(-12))/(2xx3)=12/6=2#

At #(x=2)#

#y=3(2^2)#-12(2)+1=12-24+1=-11

Vertex #(2,-11)#

The vertex form is -

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

Where -

#a=3# is the coefficient of #x^2#
#h=2# is the x - co-ordinate of vertex
#k=-11# is the y - co-ordinate of vertex

#y=3(x-2)^2-11#