Rewrite the quadratic equation in standard form by completing the square.

`f(x)=-(1/3)x^2+6x+6`

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`f(x)=-1/3x^2+6x+6`

`=-1/3[x^2-18x-18]` We need a leading coefficient of 1

`=-1/3[x^2-18x+81-81-18]` Add/subtract 1/2 "b" squared

`=-1/3[(x-9)^2-99]` Rewrite trinomial as binomial squared

`=-1/3(x-9)^2+33` Distribute the -1/3

Thus the solution is `f(x)=-1/3(x-9)^2+33`

`-1/3 x^2 +6x+6` `=-3( 1/9x^2-2x-2)=` `-3(1/9 x^2- 2x xx1/3 xx 3+9-11)` `=-3[(1/3 x-3)^2-11]`

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