Solution :
Given the equation :
[tex]$y = x^2 - 6x+3 $[/tex]
[tex]$y=1(x^2 - 6x +n)+3$[/tex]
[tex]$n = \left(\frac{b}{2}\right)^2$[/tex]
[tex]$n = \left(\frac{6}{2}\right)^2$[/tex]
n = 9
[tex]$y=1(x^2 - 6x +9-9)+3$[/tex]
[tex]$y=1(x^2 - 6x +9)-9+3$[/tex]
[tex]$y=1(x-3)^2- 6$[/tex]
Therefore the vertex form of [tex]$y = x^2 - 6x+3 $[/tex] is [tex]$y=1(x-3)^2- 6$[/tex].
