The function CCh) 24st models the concentration of a medication in the bloodstre

Question

The function CCh) 24st models the concentration of a medication in the bloodstream (as a per 2h2+5h The functio hours after its injection into muscle tissue. as a percent) h 20. Find the equation of the vertical asymptote of this function. (Hint: a +b-(a+ b)(a (SLO #3, 6) Would this concern a medical professional? Explain. 21. Fin 22. Find 23. How many hours after injection does a maximum concentration of the drug occur in the bloodstream d the equation of the horizontal asymptote. What does this mean in this context? ll of the intercepts of C and interpret their meaning in the context of this problem. Round answer to nearest hundredth 24. Suppose you need to re-administer this injection at the point when the concentration is less than 0.5% If the first injection was given at 8:00 am, what time should the next injection be given?

Explanation / Answer

20)

C(h)= (2h2+5h)/(h3+8)

C(h)= h(2h+5)/(h3+23)

C(h)= h(2h+5)/((h+2)(h2-2h+4))

C(h)= h(2h+5)/((h+2)((h-1)2+3))

for vertical asymptote , denominator =0

=>((h+2)((h-1)2+3))=0

=>h=-2

equation of vertical asymptote is h =-2

this doesnot concern a medical professional since h is negative. (in the context of the question h has to be positive)

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21)

horizontal asymptote,C = lim[h->] C(h)

horizontal asymptote,C = lim[h->][(2h2+5h)/(h3+8)]

horizontal asymptote,C = lim[h->][h2(2+(5/h))/h3(1+(8/h3))]

horizontal asymptote,C = lim[h->][(2+(5/h))/h(1+(8/h3))]

horizontal asymptote,C = [(2+0)/(1+0)]

horizontal asymptote,C = 0

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22)

for C intercept , h=0

C(0)= (2*02+5*0)/(03+8)

C(0)=0

initial concentaration is zero

for h intercept ,C(h)=0

(2h2+5h)/(h3+8)=0

=>(2h2+5h)=0

=>h=0, h=-2.5

concentration is zero at h=0. h=-2.5 has no significance

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23)

C(h)= (2h2+5h)/(h3+8)

C'(h)= -2(h4+5h3-16h-20)/(h3+8)2

C''(h)= (4h6+30h5-224h3-480h2+256)/(h3+8)3

for maximum , C'(h)=0,C"(h)<0

-2(h4+5h3-16h-20)/(h3+8)2=0

=>h=-4.41,1.95

C"(1.95)<0

maximum concentration in blood stream occur 1.95hours after injection

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24)

C(h)<0.5

=> (2h2+5h)/(h3+8)<0.5

=> 4h2+10h<h3+8

=>h3-4h2-10h+8>0

=>h<-2.2,-2<h<0.656069, h>5.54357

5.54357 hours after 8:00 am ,the next injection should be given

the next injection should be given at 1:33 PM

please rate if helpful. please comment if you have any doubt

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