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March 1, 2015

March 1, 2015

Posted by **Jonathan** on Friday, March 2, 2012 at 12:43am.

-horizontal asymptote is y=3

-Vertical x=4

-f'(x)<0 and f''(x)>0 for x>4.

-f'(x)<0 and f''(x) <0 for x<4

I got the horizontal and vertical but I don't know what to do for the last two.

- Calculus -
**Steve**, Friday, March 2, 2012 at 2:51pmthe asymptotes make

y = 3x/(x-4)

a good starting place. We want f to be falling on both sides of the asymptote, kind of like y=1/x.

y' = -12/(x-4)^2

y' < 0 on both sides of x=4

y'' = 24/(x-4)^3

y'' < 0 for x<4

y'' > 0 for x>4

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