Vertex of the quadratic equation

For a given quadratic y = ax^2 + bx + c, the vertex (h, k) is found by computing

h = –b/2a, 

k = (4ac – b^2)/4a

y=2(x - 3)^2 - 2

If we expand:

y = 2x^2 - 12x + 16

a = 2

b = -12

c = 16

h = -(-12)/(2*2)

h = 3

k = (4*2*16 - (-12)^2)/(4*2)

k = -2

The vertex is (3, -2).

Axis of symmetry is x = 3

The domain is the entire set of real numbers R.

The range is [-2, +~).