What is the answer to [f(2)-f(1)]-[f(1)-f(0)] ? It says "Consider the function f(s)=s(to the power of 2)-6s+9"

Answer 1

f(s) = s^2 - 6s + 9

f(2) = 2^2 - 6(2) + 9
f(2) = 4 - 12 + 9
f(2) = -8 + 9
f(2) = 1

f(1) = 1^2 - 6(1) + 9
f(1) = 1 - 6 + 9
f(1) = -5 + 9
f(1) = 4

f(0) = 0^2 - 6(0) + 9
f(0) = 9

(f(2) - f(1)) - (f(1) - f(0)) = (1 - 4) - (4 - 9) = (-3) - (-5) = 2

Answer 2

[f(2)-f(1)]-[f(1)-f(0)]
=f(2)-2f(1)+f(0)
=2^2 -6(2)+9 -2( 1-6+9) + 9
=4 -12 +9 -8+9
=4-12+10
=2

Answer 3

F(s) = s^2 -6*s + 9; Just plug in the numbers you are given for s (2, 1,1,and 0) and evalutate each F(s) then add them.