# How do you find the vertex of the parabola by completing the square x^2-6x+8=y?

$y = {x}^{2} - 6 x + 8 = {x}^{2} - 6 x + 9 - 1 = {\left(x - 3\right)}^{2} - 1$
This has its vertex where $x - 3 = 0$, that is $x = 3$.
When $x = 3$ we have $y = {0}^{2} - 1 = - 1$
So the vertex is at $\left(3 , - 1\right)$