f(x)=4x2-8x+2=4(x2-2x+½),
f(x)=4(x2-2x+1-½),
f(x)=4((x-1)2-½).
Vertex when x=1, f(1)=-2, that is, the point (1,-2). Axis of symmetry is x=1.
To find two symmetrical points we need f(1-p) and f(1+p) where p is an arbitrary x-displacement. If p=1 then we have:
f(0)=f(2)=2 (y-intercept is (0,2)). So the symmetrical points are (0,2) and (2,2).
More generally: f(1-p)=f(1+p)=4(p2-½). When p=1/√2 the symmetrical points are on the x-axis (x=1±1/√2) and these are the x-intercepts.