the 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
Sketch a possible graph for a function that has the following characteristics. f(0)=2.5
-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.
1 answer
1 answer