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 ( $3/2$ , $423/4$ )

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