Question 219646

{{{3x^2-36x-117}}} Start with the right side of the given equation.



{{{3(x^2-12x-39)}}} Factor out the {{{x^2}}} coefficient {{{3}}}. 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 {{{-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}}}



{{{3(x^2-12x+highlight(36-36)-39)}}} Now add <font size=4><b>and</b></font> subtract {{{36}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{36-36=0}}}. So the expression is not changed.



{{{3((x^2-12x+36)-36-39)}}} Group the first three terms.



{{{3((x-6)^2-36-39)}}} Factor {{{x^2-12x+36}}} to get {{{(x-6)^2}}}.



{{{3((x-6)^2-75)}}} Combine like terms.



{{{3(x-6)^2+3(-75)}}} Distribute.



{{{3(x-6)^2-225}}} Multiply.



So after completing the square, {{{3x^2-36x-117}}} transforms to {{{3(x-6)^2-225}}}. So {{{3x^2-36x-117=3(x-6)^2-225}}}.



So {{{y=3x^2-36x-117}}} is equivalent to {{{y=3(x-6)^2-225}}}.



Now {{{y=3(x-6)^2-225}}} is in vertex form {{{y=a(x-h)^2+k}}} where (h,k) is the vertex. In this case, {{{h=6}}} and {{{k=-225}}}. So the vertex is (6, -225).



Note: to verify the answer, you could either graph the two equations (and you'll see that they are really the same) or you can expand out {{{3(x-6)^2-225}}} to get the original expression again.