Y=-x^2+2x+1 how do you get the vertex, and x intercept and the symmetry point to the y intercept

In y=ax²+bx+c, the vertex is when x=-b/(2a), 
the x intercept is the value of x when y=0, and the y intercept is the value of y when x=0
in this case, a=-1, b=2, so the vertex is when x=-2/[2*(-1)]=1
when x=1, y=-(1)²+2*1+1=2, so the vertex is (1,2)

when x=0, y=1, the x intercept is (0,1)
when y=0, -x²+2x+1=0, use the quadratic formula to find x: x=-1+√2, x=-1-√2
so the symmetry point the the y intercept is (-1+√2, 0) (-1-√2, 0)

what is the question? I dont understand..