It is a parabola because it has a form y = a*x^2 + bx + c. Probably you know that. Such a parabola has a vertex at a minimum (if a>0 and branches go up) or minimum if a<0 and branches go down.
In both cases it is at x where y'(x) = 0 (probably also known). Then I differentiate and found the root.
There is another method: y(x) = x^2 - 6x + 11 = (x-3)^2 + 2. Because (x-3)^2 >= 0 and =0 only for x=3, then y(x) has a minimum of 2 at the point x=3. Also, if we change x to 6-x then y don't change, y(6-x) = y(x). So if a point (x; y) belongs to the parabola, then the point (6-x; y) also belongs. This is the same that "parabola is symmetric with respect to the line x=3".