Question 170146


{{{x^2-4x+6}}} Start with the right side of the equation.



Take half of the {{{x}}} coefficient {{{-4}}} to get {{{-2}}}. In other words, {{{(1/2)(-4)=-2}}}.



Now square {{{-2}}} to get {{{4}}}. In other words, {{{(-2)^2=(-2)(-2)=4}}}



{{{x^2-4x+highlight(4-4)+6}}} Now add <font size=4><b>and</b></font> subtract {{{4}}}. Make sure to place this after the "x" term. Notice how {{{4-4=0}}}. So the expression is not changed.



{{{(x^2-4x+4)-4+6}}} Group the first three terms.



{{{(x-2)^2-4+6}}} Factor {{{x^2-4x+4}}} to get {{{(x-2)^2}}}.



{{{(x-2)^2+2}}} Combine like terms.



So after completing the square, {{{x^2-4x+6}}} transforms to {{{(x-2)^2+2}}}. So {{{x^2-4x+6=(x-2)^2+2}}}.



So {{{y=x^2-4x+6}}} is equivalent to {{{y=(x-2)^2+2}}}.



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Answer:



So the equation {{{y=x^2-4x+6}}} in vertex form is {{{y=(x-2)^2+2}}}