Question 998886
The function should be C(t)=5t/(0.01t^2+3.3) ={{{5t/(0.01t^2+3.3)}}} ,
because 5t/0.01t^2+3.3 ={{{5t/0.01t^2}}}+{{{3.3}}} ,
and that would mean infinite concentration at {{{t=0}}} ,
which does not make sense.


A) The graph for {{{C(t)=5t/(0.01t^2+3.3)}}} is
{{{drawing(300,300,-50,450,-2,18,
graph(300,300,-50,450,-2,18,5x/(0.01x^2+3.3)),
locate(10,18,"C ,mcg/mL"),locate(340,-1,"t , minutes")
)}}}
The graph above is not a sketch, but a more precise graph, such as we would get with a graphing calculator, otr some software with f-graphing capabilities.

B) Calculus tells us that the function will have a maximum, when the derivative {{{dC/dt}}} is zero.
{{{dC/dt=(5(0.01t^2+3.3)-5t(0.01*2t))/(0.01t^2+3.3)^2=(0.05t^2+16.5-0.1t^2)/(0.01t^2+3.3)^2=(-0.05t^2+16.5)}}}
{{{dC/dt=0}}} when {{{-0.05t[MAX]^2+16.5=0}}}-->{{{t[MAX]=sqrt(16.5/0.05)=sqrt(330)=about18.2minutes}}}
For {{{t<t[MAX]}}} , {{dC/dt>0, meaning that {{{c(t)}}} is increasing.
For {{{t>t[MAX]}}} , {{dC/dt<0, meaning that {{{c(t)}}} is decreasing.
At {{{t=t[MAX]}}} , {{{C(t[MAX])=C[MAX]=5sqrt(330)/(0.01*330+3.3)=5sqrt(330)/6.6=about13.8}}}{{{"mcg/mL"}}}
{{{drawing(300,300,-5,45,-2,18,
graph(300,300,-5,45,-2,18,5x/(0.01x^2+3.3)),
red(circle(18.2,13.8,0.5)), locate(18,16,red(C[MAX])),
locate(1,18,"C ,mcg/mL"),locate(34,-1,"t , minutes")
)}}}


C) Between taking the drug orally (t=0) and the maximum point the blood concentration increases rapidly at first, and the absorption rate decreases towards the maximum. (We see the changes in absorption rate as changes of the slope on the graph of {{{C(t)}}} ).
After the early {{{t[MAX]=18.2minutes}}} , the concentration of the drug in the blood decreases, slowly at first, followed by a faster rate of decrease, and the s slower rate of decrease towards {{{4.1}}}{{{"mcg/mL"}}} at two hours.
{{{C(20minutes)=5*20/(0.01*400+3.3)=100/7.3=about 13.7}}}{{{"mcg/mL"}}}
{{{C(30minutes)=5*30/(0.01*900+3.3)=150/12.3=about 12.2}}}{{{"mcg/mL"}}}
{{{C(60minutes)=5*60/(0.01*3600+3.3)=300/39.3=about 7.6}}}{{{"mcg/mL"}}}
{{{C(2hours)=C(120minutes)=5*120/(0.01*14400+3.3)=600/147.3=about4.1}}}{{{"mcg/mL"}}}
IMHO:
The drug must start being absorbed in the stomach (if not the buccal mucosa), because the rate of increase in {{{C(t)}}} is maximal at {{{t=0}}} ,
and if it were not substantially absorbed before the intestine,
we would see a lower rate at {{{t=0}}} , followed by increasing absorption over some time,
before seeing the absorption rate decrease towards {{{C[MAX]}}} happening at a longer {{{t[MAX]}}} .


D) The function has only an horizontal asymptote: {{{C=0}}} .
{{{C(t)>0}}} approaches zero asymptotically as {{{t}}} increases.
That tells us that the concentration in the blood decreases towards zero,
but theoretically there is some minute (and decreasing) level of the drug in the blood forever.