What is the vertex of the function x2 - 6x + 8 = 0

you can also use this formula to get the vertex.
a(x+b/2a)^2+4ac-b^2/4a

if you plug the numbers

1(x-6/2(1))^2 +4(1)(8)-(-6)^2/4(1)

you simplify
you get
1(x-3)^2-1
vertex is (3,-1)

Solve quadratic by formula

a.x^2 +b.x + c = o

x = [ -b + or - (Square root(b^2 - 4.a.c)]/2a

In your case a = 1 b = -6 and c = 8

x = [ 6 + or - (square root ( 36 -4.1.8)]/2.1 = 6 + or - 1

so x is 5 or 7

Vertex will be at x max i.e. at 7

Value at vertex = 49 -6.7 + 8 = =1

use completing the square to get it in the form (x-h)^2 +k=
so
(x-3)^2-1 = 0
and the vertex is (h,k) (note that you subtract h) so the vertex is (3, -1)