Question 90019
f(x)={{{(x^2-2x+1)}}}
a) Let x=0 and solve for the y-term
f(0)={{{((0)^2-2(0)+1)}}}
f(0)=1
.
b) let y=0 and solve for the x-term
0={{{(x^2-2x+1)}}}
0={{{(x^2-2x+1)}}}  [factor]
0=(x-1)(x-1)
.
x-1=0 [set each factor equal to zero and solve for the x=term]
x=1  [same for the other factor]
So, x = 1 and x = 1