How do you write the Vertex form equation of the parabola #y=x^2-6x+5#?

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Answer 1 · 2016-12-05T23:55:49

Please see the explanation.

Given #y = x^2 - 6x + 5" [1]"#

The vertex form of a parabola of this type is:

#y = a(x - h)^2 + k" [2]"#

where #(h,k) # is the vertex and "a" is the coefficient of the #x^2# term in standard form.

The first step is to add 0 to equation [1] in the form #ah^2 - ah^2#. Because #a = 1#, we add #h^2 - h^2# to equation [1]:

#y = x^2 - 6x + h^2 - h^2 + 5" [3]"#

If we expand the square in equation [2], #(x - h)^2 = x^2 - 2hx + h^2#, we can see that #-2hx# must equal #-6x# in equation [3]. Write this equation:

#-2hx = -6x#

Solve for h:

#h = 3#

Substitute #(x - h)^2# for #x^2 - 6x + h^2# in equation [3]:

#y = (x - h)^2 - h^2 + 5" [4]"#

Substitute 3 for every h:

#y = (x - 3)^2 - 3^2 + 5#

Simplify the constant terms:

#y = (x - 3)^2 - 4#

This is the vertex form.