Question 215000


{{{-18x^2-177x-190}}} Start with the given expression.



{{{-18(x^2+(59/6)x+95/9)}}} Factor out the {{{x^2}}} coefficient {{{-18}}}. 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 {{{59/6}}} to get {{{59/12}}}. In other words, {{{(1/2)(59/6)=59/12}}}.



Now square {{{59/12}}} to get {{{3481/144}}}. In other words, {{{(59/12)^2=(59/12)(59/12)=3481/144}}}



{{{-18(x^2+(59/6)x+highlight(3481/144-3481/144)+95/9)}}} Now add <font size=4><b>and</b></font> subtract {{{3481/144}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{3481/144-3481/144=0}}}. So the expression is not changed.



{{{-18((x^2+(59/6)x+3481/144)-3481/144+95/9)}}} Group the first three terms.



{{{-18((x+59/12)^2-3481/144+95/9)}}} Factor {{{x^2+(59/6)x+3481/144}}} to get {{{(x+59/12)^2}}}.



{{{-18((x+59/12)^2-1961/144)}}} Combine like terms.



{{{-18(x+59/12)^2-18(-1961/144)}}} Distribute.



{{{-18(x+59/12)^2+1961/8}}} Multiply.



So after completing the square, {{{-18x^2-177x-190}}} transforms to {{{-18(x+59/12)^2+1961/8}}}. So {{{-18x^2-177x-190=-18(x+59/12)^2+1961/8}}}.



So {{{-18x^2-177x-190=0}}} is equivalent to {{{-18(x+59/12)^2+1961/8=0}}}.



So {{{y=-18x^2-177x-190}}} is equivalent to {{{y=-18(x+59/12)^2+1961/8}}}.



So the equation {{{y=-18(x+59/12)^2+1961/8}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=-18}}}, {{{h=-59/12}}}, and {{{k=1961/8}}}



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



So the vertex of {{{y=-18(x+59/12)^2+1961/8}}} is *[Tex \LARGE \left(-\frac{59}{12},\frac{1961}{8}\right)] since {{{h=-59/12}}} and {{{k=1961/8}}}