What is the vertex form of #y= 3x^2-11x+6 #?

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Answer 1 · 2017-08-11T07:08:36

#(11/6, -49/12)#

The #x# value of the axis of symmetry is the same as the #x# value of the vertex.

Use the axis of symmetry formula #x=-b/(2a)# to find the #x# value of the vertex.

#x = (-(-11))/(2(3))#

#x = 11/6#

Substitute #x = 11/6# into the original equation for the #y# value of the vertex.

#y = 3(11/6)^2 - 11(11/6) + 6#

#y = -49/12#

Therefore, the vertex is at #(11/6, -49/12)#.