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Answer to Question #45306 in Calculus for Jonathon

Question #45306
Problem: f(x)= (x^2-5)/(x+3)

f(x)= (x^2-5)/(x+3)=x-3=4/(x+3) so, there is oblique asymptote y=x-3.

Lack of the horizontal asymptote means that the concentration increases indefinitely.
The vertical asymptote x=-3 means that to obtain concentration of drug equals f(0)=-5/3 at the start time x=0, we should
apply an undefined concentration at time x=-3 (3 time units before start time).

****If the numerator and denominator were switched, would there be a horizontal asymptote? Why or why not?
Expert's answer
This function does not have horizontal asymptotes, even if we change the numerator or denominator. Because  function tends to a certain finite number when x tends to plus infinity or to minus infinity if there is a horizontal asymptote. Nevertheless, this case is impossible for this function, because we deal with the ratio of two polynomials, and degree of numerator (i.e. 2) is greater than degree of numerator (i.e. 1), that is why this function f(x) tends to plus (or minus) infinity as x tends to plus (or minus ) infinity, it does not tend to a finite number. When (x^2-9)/(x+3)=x-3 or (x^2-5)/(x+Sqrt(5))=x-Sqrt(5), the asymptotes disappear. There is a mistake in the statement of question. Instead of "f(x)=  (x^2-5)/(x+3)=x-3=4/(x+3)" we regard "f(x)=  (x^2-5)/(x+3)=x-3+4/(x+3)".

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