What is the vertex of #y=3x^2+2x+5#?

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Answer 1 · 2017-10-04T16:31:56

Vertex #->(x,y)=(-1/3,14/3)#

Given: #y=3x^2+2x+5#

This is part of the process of completing the square.

Write as #y=3(x^2color(red)(+2/3)x)+5#

To complete the square you would 'do other things' to this. I am not going to do that!

#x_("vertex")=(-1/2)xx( color(red)(+2/3))= -1/3#

Substitute for #x# to determine #y_("vertex")#

#y_("vertex")= 3(-1/3)^2+2(-1/3)+5 #

#y_("vertex")= +1/3-2/3+5 = 4 2/3 -> 14/3#

Vertex #->(x,y)=(-1/3,14/3)#