Question 335938

{{{x^2+10x+7}}} Start with the given expression.



Take half of the {{{x}}} coefficient {{{10}}} to get {{{5}}}. In other words, {{{(1/2)(10)=5}}}.



Now square {{{5}}} to get {{{25}}}. In other words, {{{(5)^2=(5)(5)=25}}}



{{{x^2+10x+highlight(25-25)+7}}} Now add <font size=4><b>and</b></font> subtract {{{25}}}. Make sure to place this after the "x" term. Notice how {{{25-25=0}}}. So the expression is not changed.



{{{(x^2+10x+25)-25+7}}} Group the first three terms.



{{{(x+5)^2-25+7}}} Factor {{{x^2+10x+25}}} to get {{{(x+5)^2}}}.



{{{(x+5)^2-18}}} Combine like terms.



So after completing the square, {{{x^2+10x+7}}} transforms to {{{(x+5)^2-18}}}. So {{{x^2+10x+7=(x+5)^2-18}}}.



This means that {{{f(x)=x^2+10x+7}}} is equivalent to {{{f(x)=(x+5)^2-18}}}.



So the equation {{{f(x)=(x+5)^2-18}}} is now in vertex form {{{f(x)=a(x-h)^2+k}}} where {{{a=1}}}, {{{h=-5}}}, and {{{k=-18}}}



Remember, the vertex of {{{f(x)=a(x-h)^2+k}}} is (h,k).



So the vertex of {{{f(x)=(x+5)^2-18}}} is (-5,-18) since h=-5 and k=-18 



This means that the vertex of {{{f(x)=x^2+10x+7}}} is (-5,-18)



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Jim