Question 241771


{{{-16x^2+88x+3}}} Start with the given right side of the equation.



{{{-16(x^2-(11/2)x-3/16)}}} Factor out the {{{x^2}}} coefficient {{{-16}}}. 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 {{{-11/2}}} to get {{{-11/4}}}. In other words, {{{(1/2)(-11/2)=-11/4}}}.



Now square {{{-11/4}}} to get {{{121/16}}}. In other words, {{{(-11/4)^2=(-11/4)(-11/4)=121/16}}}



{{{-16(x^2-(11/2)x+highlight(121/16-121/16)-3/16)}}} Now add <font size=4><b>and</b></font> subtract {{{121/16}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{121/16-121/16=0}}}. So the expression is not changed.



{{{-16((x^2-(11/2)x+121/16)-121/16-3/16)}}} Group the first three terms.



{{{-16((x-11/4)^2-121/16-3/16)}}} Factor {{{x^2-(11/2)x+121/16}}} to get {{{(x-11/4)^2}}}.



{{{-16((x-11/4)^2-31/4)}}} Combine like terms.



{{{-16(x-11/4)^2-16(-31/4)}}} Distribute.



{{{-16(x-11/4)^2+124}}} Multiply.



So after completing the square, {{{-16x^2+88x+3}}} transforms to {{{-16(x-11/4)^2+124}}}. So {{{-16x^2+88x+3=-16(x-11/4)^2+124}}}.



So {{{y=-16x^2+88x+3}}} is equivalent to {{{y=-16(x-11/4)^2+124}}}.



So the equation {{{y=-16(x-11/4)^2+124}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=-16}}}, {{{h=11/4}}}, and {{{k=124}}}



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



So the vertex of {{{y=-16(x-11/4)^2+124}}} is *[Tex \LARGE \left(\frac{11}{4},124\right)].