Question 166637
In order to write {{{x^2+4x}}} in vertex form, we need to complete the square for {{{x^2+4x}}}



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Completing the Square:



{{{x^2+4x}}} Start with the given expression.



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)}}} Now add <font size=4><b>and</b></font> subtract {{{4}}}. Notice how {{{4-4=0}}}. So the expression is not changed.



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



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



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




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



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{{{y=(x+2)^2-4}}} Start with the previous equation



{{{y=(x-(-2))^2-4}}} Rewrite {{{x+2}}} as {{{x-(-2)}}}



{{{y=1(x-(-2))^2-4}}} Place a "1" outside the parenthesis.



Now {{{y=1(x-(-2))^2-4}}} is in vertex form {{{y=a(x-h)^2+k}}} where {{{a=1}}}, {{{h=-2}}} and {{{k=-4}}}