Question 355007
First, we need to complete the square for the expression {{{5x^2+10x+7}}}





{{{5x^2+10x+7}}} Start with the given expression.



{{{5(x^2+2x+7/5)}}} Factor out the {{{x^2}}} coefficient {{{5}}}. This step is very important: the {{{x^2}}} coefficient <font size=4><b>must</b></font> be equal to 1.



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



Now square {{{1}}} to get {{{1}}}. In other words, {{{(1)^2=(1)(1)=1}}}



{{{5(x^2+2x+highlight(1-1)+7/5)}}} Now add <font size=4><b>and</b></font> subtract {{{1}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{1-1=0}}}. So the expression is not changed.



{{{5((x^2+2x+1)-1+7/5)}}} Group the first three terms.



{{{5((x+1)^2-1+7/5)}}} Factor {{{x^2+2x+1}}} to get {{{(x+1)^2}}}.



{{{5((x+1)^2+2/5)}}} Combine like terms.



{{{5(x+1)^2+5(2/5)}}} Distribute.



{{{5(x+1)^2+2}}} Multiply.



So after completing the square, {{{5x^2+10x+7}}} transforms to {{{5(x+1)^2+2}}}. So {{{5x^2+10x+7=5(x+1)^2+2}}}.



So {{{y=5x^2+10x+7}}} is equivalent to {{{y=5(x+1)^2+2}}}.



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



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



So the vertex of {{{y=5(x+1)^2+2}}} is (-1,2). 



A graph will confirm this


{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,5x^2+10x+7)

)}}}


Graph of {{{y=5x^2+10x+7}}} with vertex (-1,2)



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Jim