What is the vertex of #f(x)=x^2-6x+8#?

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Answer 1 · 2014-09-16T00:06:46

We have 3 options to solve this problem:

1) Graph It

2) Complete the Square

3) Use the expressions: #(-b/(2a),f(-b/(2a)))#

I will use Option #2: Completing the Square

#x^2-6x+8=0#

#x^2-6x=-8#

#(-6/2)^2=(-3)^2=9#, Add #9# to both sides of the equation

#x^2-6x+9=-8+9#

#x^2-6x+9=1#

#(x-3)^2=1#

#(x-3)^2-1=0#

#(x-3)^2-1=f(x)#

#vertex -> (3,-1)#