Asked by Joseph
I have no idea what to do here.
[f(2)-f(1)]-[f(1)-f(0)]
f(s)=s^2-6s-9
[f(2)-f(1)]-[f(1)-f(0)]
f(s)=s^2-6s-9
Answers
Answered by
Steve
The "s" in f(s) is just a place holder. To evaluate f(2), just replace every "s" with a "2" and then simplify it all.
To save yourself some work, note that
[f(2)-f(1)]-[f(1)-f(0)]
= f(2)-f(1)-f(1)+f(0)
= f(2)-2f(1)+f(0)
so, [f(2)-f(1)]-[f(1)-f(0)]
= (2^2-6*2-9) - 2(1^2-6*1-9) + (0^2-6*0-9)
= (4-12-9) - 2(1-6-9) + (-9)
= -17 - 2(-14) - 9
= -17+28-9
= 2
To save yourself some work, note that
[f(2)-f(1)]-[f(1)-f(0)]
= f(2)-f(1)-f(1)+f(0)
= f(2)-2f(1)+f(0)
so, [f(2)-f(1)]-[f(1)-f(0)]
= (2^2-6*2-9) - 2(1^2-6*1-9) + (0^2-6*0-9)
= (4-12-9) - 2(1-6-9) + (-9)
= -17 - 2(-14) - 9
= -17+28-9
= 2
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