Question 353398
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 {{{(a-b)^2 = a^2 -2ab + b^2}}}:
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<br>
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
{{{(f(x-h) - f(x))/h = (-2ahx + ah^2 - bh)/h}}}
Factoring out h in the numerator on the right side we get:
{{{(f(x-h) - f(x))/h = (h(-2ax + ah - b))/h}}}
Now we can cancel the h's:
{{{(f(x-h) - f(x))/h = (cross(h)(-2ax + ah - b))/cross(h)}}}
leaving:
{{{(f(x-h) - f(x))/h = -2ax + 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.