Could someone Pleas check my work. I don't know what I'm doing wrong:
1. f(x)=x^2-2x+1
a=1, b=-2, c=1
Axis of symmetry: -b/2(a)
-(-2)/2(1)= 2/2= 1
Vertex: f(1)=(1)^2-2(1)+1=0
(1,0)
y-intercept: x=0 f(0)=(0)^2-2(0)+1
(0,1)

Question

Could someone Pleas check my work. I don't know what I'm doing wrong:
1. f(x)=x^2-2x+1
        a=1, b=-2, c=1
Axis of symmetry: -b/2(a) 
-(-2)/2(1)= 2/2= 1
Vertex: f(1)=(1)^2-2(1)+1=0
(1,0)
y-intercept: x=0 f(0)=(0)^2-2(0)+1 
(0,1)

cj49 · Answer

u need to find the roots ryt?

anonymous · Answer

Could you explain? I'm not sure what that is, that wasn't in my notes

anonymous · Answer

Oh you were asking. The directions say "use the axis of symmetry, y-intercept, and vertex to graph each function

anonymous · Answer

I graphed it. I just don't know if I did the work right