y=x^2-2x-8. Zeroes when (x-4)(x+2)=0, so x=4 and -2 are zeroes. Axis of symmetry lies midway between zeroes at x=1/2(4+(-2))=1/2(2)=1. Also found by completing the square: y=x^2-2x+1-9=(x-1)^2-9 or (y+9)=(x-1)^2. The parabola y=x^2 is symmetrical about the y axis. The parabola (y+9)=(x-1)^2 is symmetrical about the line x=1. The vertex lies on the axis of symmetry. The vertex of y=x^2 is at the origin (0,0) so the vertex of (y+9)=(x-1)^2 is at the "displaced origin" (1,-9). The vertex is the minimum found by differentiating the function then setting it to zero: 2x-2=0, so 2x=2 and x=1 and y=1-2-8=-9. So there are different ways of establishing the properties.