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# converting a quadratic written in standard form, to vertex.

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hello everybody,
hope you've had an awesome holiday and happy new year!
i do not understand how this problem turns into\(f(x)=9(x-1/3)+2\)

here is the original thing:
\(f(x)=9x^2-6x+3\)

i got an answer that is totally off:
\(f(x)=9(x-1/3)^2+6\)

i don't understand where i went wrong, and every time i attempt to do the problem, i keep getting different results each time.

can someone please explain how and why this equals to \(f(x)=9(x-1/3)^2+2\)

thank you! <3

Jan 3, 2020

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f(x) =9x^2-6x+3      'complete the square'    to do this , leading coeff has to be 1

divide by 9  (we'll put it back in shortly)

f(x)/9   =  x^2 - 2/3 x +1/3     now complete the square

x^2 - 2/3 x + 1/9       -1/9  +1/3

f(x) / 9 =  (x-1/3)^2     - 1/9 + 3/9

f(x) / 9   =  (x- 1/3)^2   + 2/9                   Now remember that division by 9?   Put the 9 back in by multiplying by 9

f(x) = 9 (x-1/3)^2  + 2          END !

Jan 3, 2020
edited by ElectricPavlov  Jan 3, 2020
edited by ElectricPavlov  Jan 3, 2020
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thanks so much EP !!

Nirvana  Jan 4, 2020