y=x^2-6x+13, y=x^2-6x+9+4=(x-3)^2+4; y-4=(x-3)^2. Since a square is always positive y-4>0, so y>4, and y=4 represents the minimum value of y. There is no definable maximum, since y increases as x increases positively or negatively. y increases to infinity as x approaches infinity or minus infinity.
(Also, y'=2x-6. When y'=0 there is a turning point, so 2x=6, x=3, y=9-18+13=4.)