What is the vertex form of #y= 3x^2+x-55#?

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Answer 1 · 2017-03-02T09:14:37

#y = 3 x^2 + x - 55# has a minimum #-661/12# at #(-1/6, -661/12)#

#y = 3 x^2 + x - 55#

#y = [3(x^2 + x/3)] - 55#

solve using completing a square,
#y = [3(x + 1/6)^2 - 3*(1/6)^2] - 55#

#y = 3(x + 1/6)^2 - 3*(1/36) - 55#

#y = 3(x + 1/6)^2 - 1/12 - 55#

#y = 3(x + 1/6)^2 - 661/12#

Therefore,
#y = 3 x^2 + x - 55# has a minimum #-661/12# at #(-1/6, -661/12)#