Question 586056
{{{x^2+3x-4}}} Start with the given expression.



Take half of the {{{x}}} coefficient {{{3}}} to get {{{3/2}}}. In other words, {{{(1/2)(3)=3/2}}}.



Now square {{{3/2}}} to get {{{9/4}}}. In other words, {{{(3/2)^2=(3/2)(3/2)=9/4}}}



{{{x^2+3x+highlight(9/4-9/4)-4}}} Now add <font size=4><b>and</b></font> subtract {{{9/4}}}. Make sure to place this after the "x" term. Notice how {{{9/4-9/4=0}}}. So the expression is not changed.



{{{(x^2+3x+9/4)-9/4-4}}} Group the first three terms.



{{{(x+3/2)^2-9/4-4}}} Factor {{{x^2+3x+9/4}}} to get {{{(x+3/2)^2}}}.



{{{(x+3/2)^2-25/4}}} Combine like terms.



So after completing the square, {{{x^2+3x-4}}} transforms to {{{(x+3/2)^2-25/4}}}. So {{{x^2+3x-4=(x+3/2)^2-25/4}}}.



So {{{y=x^2+3x-4}}} is equivalent to {{{y=(x+3/2)^2-25/4}}}.



So the equation {{{y=(x+3/2)^2-25/4}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=1}}}, {{{h=-3/2}}}, and {{{k=-25/4}}}



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



So the vertex of {{{y=(x+3/2)^2-25/4}}} is (-3/2,-25/4). 

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