What is the vertex form of #y=11x^2 - 4x + 31 #?

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Answer 1 · 2017-10-22T11:41:28

The vertex form of equation is #y= 11(x-2/11)^2 +30 7/11#
of which vertex is at #(2/11 , 30 7/11#)

# y= 11x^2-4x+31 or y= 11(x^2-4/11x) +31# or

#y= 11(x^2-4/11x +(2/11)^2)- 11*4/11^2 +31# or

#y= 11(x-2/11)^2- 4/11 +31# or

#y= 11(x-2/11)^2 +337/11# or

#y= 11(x-2/11)^2 +30 7/11#

The vertex form of equation is #y= 11(x-2/11)^2 +30 7/11#

of which vertex is at #(2/11 , 30 7/11#) [Ans]