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How do you write #y=3x^2-6x+1# into vertex form?

1 Answer

May 3, 2015

Vertex form for a quadratic is
#y = m(x-a)+b#
(where #(a,b)# is the vertex)

#y=3x^2-6x+1#

#y=3(x^2-2x)+1 " extracting the "m" constant"#

#y=3(x^2-2x+1) -3 +1 " [completing the square](http://socratic.org/algebra/quadratic-equations-and-functions/completing-the-square-1)"#

#y=3(x-1)^2 -2 " simplification"#

#y=3(x-1)^2+(-2)" into completely vertex form"#

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