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Rewrite this standard form quadratic equation into vertex form by completing the square. Upload your work here to show your thinking.

y=x2 - 6x +3

1 Answer

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Solution :

Given the equation :


$y = x^2 - 6x+3 $


$y=1(x^2 - 6x +n)+3$


$n = \left((b)/(2)\right)^2$


$n = \left((6)/(2)\right)^2$

n = 9


$y=1(x^2 - 6x +9-9)+3$


$y=1(x^2 - 6x +9)-9+3$


$y=1(x-3)^2- 6$

Therefore the vertex form of
$y = x^2 - 6x+3 $ is
$y=1(x-3)^2- 6$.

User Daxon
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