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bulgar [2K]
3 years ago
8

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

king.
y=x2 - 6x +3
Mathematics
1 answer:
Alex787 [66]3 years ago
3 0

Solution :

Given the equation :

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

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

$n = \left(\frac{b}{2}\right)^2$

$n = \left(\frac{6}{2}\right)^2$

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  $y=1(x-3)^2- 6$.

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