Question 341749
To find the vertex and axis of symmetry, we need to complete the square for {{{ (2/3)x^2-3x+6}}}



{{{(2/3)x^2-3x+6}}} Start with the given expression.



{{{(2/3)(x^2-(9/2)x+9)}}} Factor out the {{{x^2}}} coefficient {{{2/3}}}. 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 {{{-9/2}}} to get {{{-9/4}}}. In other words, {{{(1/2)(-9/2)=-9/4}}}.



Now square {{{-9/4}}} to get {{{81/16}}}. In other words, {{{(-9/4)^2=(-9/4)(-9/4)=81/16}}}



{{{(2/3)(x^2-(9/2)x+highlight(81/16-81/16)+9)}}} Now add <font size=4><b>and</b></font> subtract {{{81/16}}} inside the parenthesis. Make sure to place this after the "x" term. Notice how {{{81/16-81/16=0}}}. So the expression is not changed.



{{{(2/3)((x^2-(9/2)x+81/16)-81/16+9)}}} Group the first three terms.



{{{(2/3)((x-9/4)^2-81/16+9)}}} Factor {{{x^2-(9/2)x+81/16}}} to get {{{(x-9/4)^2}}}.



{{{(2/3)((x-9/4)^2+63/16)}}} Combine like terms.



{{{(2/3)(x-9/4)^2+(2/3)(63/16)}}} Distribute.



{{{(2/3)(x-9/4)^2+21/8}}} Multiply.



So after completing the square, {{{(2/3)x^2-3x+6}}} transforms to {{{(2/3)(x-9/4)^2+21/8}}}. So {{{(2/3)x^2-3x+6=(2/3)(x-9/4)^2+21/8}}}.



So {{{y=(2/3)x^2-3x+6}}} is equivalent to {{{y=(2/3)(x-9/4)^2+21/8}}}.



So the equation {{{y=(2/3)(x-9/4)^2+21/8}}} is now in vertex form {{{y=a(x-h)^2+k}}} where {{{a=2/3}}}, {{{h=9/4}}}, and {{{k=21/8}}}



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



So the vertex of {{{y=(2/3)(x-9/4)^2+21/8}}} is *[Tex \LARGE \left(\frac{9}{4},\frac{21}{8}\right)] since {{{h=9/4}}} and {{{k=21/8}}}



Also, remember that the axis of symmetry is simply {{{x=h}}}. So the axis of symmetry is {{{x=9/4}}}



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