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f(x)=2x^2+5x+2 ... find f(x+h)

Mathematics
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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
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Other answers:

@Goten77 so I got to 2x^2+4xh+2h^2+5x+5h+2 ... is that as far as I can get?
were the directions to expand it?
No just to find the value
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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?
That's the best you can do. You need to expand it, though. Usually this particular construct is in the context of \[\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}\]and the subtraction of f(x) removes many of the terms and the h in the denominator cancels one or more, hopefully leaving you with a limit you can evaluate without too much pain and suffering. In the case of this problem, you would have \[\lim_{h\rightarrow0}\frac{2x^2+4xh+2h^2+5x+5h+2-(2x^2+5x+2)}{h} = \]\[\lim_{h\rightarrow0}\frac{4xh+2h^2+5h}{h} =\lim_{h\rightarrow0}4x+2h+5 = 4x+5 \]
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
Sorry, that was a preview of coming attractions :-) No your answer is \[2x^2+4xh+2h^2+5x+5h+2\]just like you asked about.
Gotcha! Thank you so much! What about for -f(x)? @whpalmer4
Unless there's a part of your problem statement that you didn't include or I didn't see, -f(x) isn't part of your problem, only my digression about computing derivatives...
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 \(f(-x) = 2(-x)^2+5(-x)+2 = 2x^2-5x+2\)

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