Show that x^2 - 6x + 11> 0 for all values of x

Using completing the square:x^2 - 6x + 11 = (x-3)^2 - (3)^2 + 11= (x - 3)^2 + 2So the turning point is (3,2). Since this curve is a positive quadratic, the turning point is a minimum the y value is 2 at it's lowest point meaning that it is >0 for all values of x.