Question 586052


{{{x^2+12x+19}}} Start with the given expression.



Take half of the {{{x}}} coefficient {{{12}}} to get {{{6}}}. In other words, {{{(1/2)(12)=6}}}.



Now square {{{6}}} to get {{{36}}}. In other words, {{{(6)^2=(6)(6)=36}}}



{{{x^2+12x+highlight(36-36)+19}}} Now add <font size=4><b>and</b></font> subtract {{{36}}}. Make sure to place this after the "x" term. Notice how {{{36-36=0}}}. So the expression is not changed.



{{{(x^2+12x+36)-36+19}}} Group the first three terms.



{{{(x+6)^2-36+19}}} Factor {{{x^2+12x+36}}} to get {{{(x+6)^2}}}.



{{{(x+6)^2-17}}} Combine like terms.



So after completing the square, {{{x^2+12x+19}}} transforms to {{{(x+6)^2-17}}}. So {{{x^2+12x+19=(x+6)^2-17}}}.



So {{{y=x^2+12x+19}}} is equivalent to {{{y=(x+6)^2-17}}}.



So the equation {{{y=(x+6)^2-17}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=1}}}, {{{h=-6}}}, and {{{k=-17}}}



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



So the vertex of {{{y=(x+6)^2-17}}} is (-6,-17) since h = -6 and k = -17



A graph will confirm this 



{{{ drawing(500, 500, -10, 10, -18, 2,
 grid(1),
 graph( 500, 500, -10, 10, -18, 2,x^2+12x+19)

)}}}
Graph of {{{y=x^2+12x+19}}} with vertex (-6,-17)

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