Question 215091


{{{-4x^2-12x+6}}} Start with the given expression.



{{{-4(x^2+3x-3/2)}}} Factor out the {{{x^2}}} coefficient {{{-4}}}. 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 {{{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}}}



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



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



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



{{{-4((x+3/2)^2-15/4)}}} Combine like terms.



{{{-4(x+3/2)^2-4(-15/4)}}} Distribute.



{{{-4(x+3/2)^2+15}}} Multiply.



So after completing the square, {{{-4x^2-12x+6}}} transforms to {{{-4(x+3/2)^2+15}}}. So {{{-4x^2-12x+6=-4(x+3/2)^2+15}}}.



So {{{y=-4x^2-12x+6}}} is equivalent to {{{y=-4(x+3/2)^2+15}}}.



So the equation {{{y=-4(x+3/2)^2+15}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=-4}}}, {{{h=-3/2}}}, and {{{k=15}}}



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



So the vertex of {{{y=-4(x+3/2)^2+15}}} is *[Tex \LARGE \left(-\frac{3}{2},15\right)] because {{{h=-3/2}}} and {{{k=15}}}. 



You can graph this to confirm the answer.