SOLUTION: If f(x)=ax^2+bx+c Show that (f(x-h)-f(x))/h = 2ax+ah+b

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Question 353398: If f(x)=ax^2+bx+c
Show that (f(x-h)-f(x))/h = 2ax+ah+b
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = ax^2 + bx + c
f(x-h) = a(x-h)^2 + b(x-h) + c
Simplifying. We start, of course, with the exponent. TO square (x-h) either use FOIL or the pattern %28a-b%29%5E2+=+a%5E2+-2ab+%2B+b%5E2:
f(x-h) = a(x^2 -2(x)(h) + h^2) + b(x-h) + c
Now we multiply (using the Distributive Property):
f(x-h) = ax^2 - 2ahx + ah^2 + bx - bh + c

f(x-h) - f(x) = (ax^2 - 2ahx + ah^2 + bx - bh + c) - (ax^2 + bx + c)
Subtracting we get:
f(x-h) - f(x) = -2ahx + ah^2 - bh
%28f%28x-h%29+-+f%28x%29%29%2Fh+=+%28-2ahx+%2B+ah%5E2+-+bh%29%2Fh
Factoring out h in the numerator on the right side we get:
%28f%28x-h%29+-+f%28x%29%29%2Fh+=+%28h%28-2ax+%2B+ah+-+b%29%29%2Fh
Now we can cancel the h's:
%28f%28x-h%29+-+f%28x%29%29%2Fh+=+%28cross%28h%29%28-2ax+%2B+ah+-+b%29%29%2Fcross%28h%29
leaving:
%28f%28x-h%29+-+f%28x%29%29%2Fh+=+-2ax+%2B+ah+-+b
Since this did not work out as the problem says it should, it suggests that an error has been made. I do not think my work has an error. I believe the error is in the way you posted your problem. I believe you were supposed to find
(f(x+h)-f(x))/h
and not
(f(x-h)-f(x))/h
If you rework the problem with (x+h), taking the same steps as above, you will end up with the correct answer.