What is the vertex form of #y= x^2 -6x+8 #?

1 Answer
Alan P.
Feb 26, 2016

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

Explanation:

The general vertex form is
#color(white)("XXX")y=m(x-a)^2+b# for a parabola with vertex at #(a,b)#

To convert #y=x^2-6x+8# into vertex form, perform the process called "completing the square":

For a squared binomial #(x+k)^2 = color(blue)(x^2+2kx)+k^2#
So if #color(blue)(x^2-6x)# are the first two terms of an expanded squared binomial, then #k=-3# and the third term must be #k^2=9#

We can add #9# to the given expression to "complete the square", but we we also need to subtract #9# so that the value of the expression stays the same.

#y=x^2-6x color(red)(+9)+8 color(red)(-9)#

#y=(x-3)^2-1#
or, in explicit vertex form:
#y=1(x-3)^2+(-1)#

Typically I leave the value #m# off when it is #1# (the default anyway) but find that writing the constant term as #+(-1)# helps me remember that the #y# coordinate of the vertex is #(-1)#

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