What is the vertex form of #y= x^2-3x+108 #?

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Answer 1 · 2015-12-31T15:52:32

Complete The square to find the vertex

y = #x^2 # - 3x + 108

y = 1(#x^2# - 3x + - ) + 108

___= #(b / 2)^2#

___= #(3/2)^2#

___= #9/4#

y = 1(#x^2# - 3x + 9/4 - 9/4) + 108

y = 1#(x - #3/2#)^2# - #9/4# + 108

y = 1#(x - #3/2#)^2# + #423/4#

The vertex is at (#3/2#, #423/4#)