calculus (derivativs)
posted by Eve .
Find the points on the graph of f where the tangent line is horizontal.
original: x/x^2+25)
After I did the quotient rule I got:
(x^2+25)(1)(x)(2x)/((x^2+25))^2
Then of course I know you set that equal to 0, but I do not know how to set it up or "finalize" the problem because of the denominator. Please help!

calculus (derivativs) 
Graham
((x^2+25)(1)(x)(2x))/(x^2+25)^2 = 0
Simplify
(25x^2)/(x^2+25)^2 = 0
As long as x^2+25 <> 0 then you can multiply both sides by the denominator
(25x^2) = 0
Thus
x^2 = 25
x = ±5
( Check: 25+25<>0 .. okay)
Find y when x=5 and x=+5. These are your points 
calculus (derivativs) 
Eve
does that go into the derivative equation or the original?
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