Question

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

Answer 1

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$`.