Change this equation to vertex form and find the vertex and axis of symmetry: y=x^2-8x+2

The vertex  form of parabola is y = a(x-h)^2+c, where (h, k) are
the coordinates of the vertex.  (1/a)/4 = 1/4a is focal distance from the vertex (h,k), and x = h
is the axis of of symmetry

The given  parabola is y = x^2-8x+2. To
convert this to vertex form we have to complete the x^2-8x into a perfect square by adding 4^2 
so that x^2-8x+4^2 = (x-4)^2.

Therefore we add and subtract
4^2:

y = (x^2-8x+4^2) - 4^2+2.

y =
1(x-8)^2 -18 is in the required form.

The coordinates of the vertex
= (8, -18)

The focal length = 1/4*1 = 1/4 . So the ocus is 1/4 units
above the vertex (8,-18).

The axis of symmetry is  x= 8, a || line
to y axis.