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http://www.purplemath.com/modules/fcnnot2.htm

this means whereever there is a x in the original equasion that you substituate in x+h instead

|dw:1362544377090:dw|

were the directions to expand it?

No just to find the value

|dw:1362545462751:dw|

oh then pluggign in x+h is all ya had to do

oh.... so just 2(x+h)^2+5(x+h)+2 would be my answer?

And as you'll find out soon enough, \[\frac{d}{dx}[2x^2+5x+2] = 4x+5\]

So confusing...so is it 4x+5 ? @whpalmer4

Gotcha! Thank you so much! What about for -f(x)? @whpalmer4

No no -f(x) of 2x^2+5x+2 @whpalmer4

if \(f(x) = 2x^2+5x+2\) then \(-f(x) = -(2x^2+5x+2) = -2x^2-5x-2\), no?

Not to be confused with \(f(-x) = 2(-x)^2+5(-x)+2 = 2x^2-5x+2\)