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NEEKALIAIN
Answered

not sure how difficult this would be to solve

Not Sure How Difficult This Would Be To Solve class=

Answer :

AUGUSTINETHAWNPIIN
f (x) = x^2 - 2x + 1
f (x) = (-2)^2 - 2(-2) + 1
= 4 + 4 + 1
= 9
f (x) = ( 0 )^2 - 2 (0) + 1
= 0 - 0 + 1
= 1
0 = x^2 - 2x + 1
x^2 = 2x - 1 = 1
f (2) = 2^2 - 2(2) + 1
= 4 - 4 + 1
= 1
f (3) = 3^2 - 2(3) + 1
= 9 - 6 + 1
= 3 + 1
= 4

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