Since [tex]x^2[/tex] is the square of x and 6x is twice the product between x and 3, the second square must be 3 squared, i.e. 9.
So, if we think of 15 as 9+6, we have
[tex]x^2-6x+9+6 = (x-3)^2+6[/tex]
Which is the required vertex form. This form tells us imediately that the vertex is the point (3,6).
Since the leading coefficient is 1, the parabola is facing upwards (it's U shaped), so the vertex is a minimum.
