Question

the increase in a persons body temperature ( t(t) ), above ( 98.6^{circ}f ), can be modeled by the function ( t(t)=\frac{4t}{t^{2}+1} ), where ( t ) represents time elapsed. what is the meaning of the horizontal asymptote for this function?
the horizontal asymptote of ( y = 0 ) means that the persons temperature will approach ( 98.6^{circ}f ) as time elapses.
the horizontal asymptote of ( y = 0 ) means that the persons temperature will approach ( 0^{circ}f ) as time elapses.
the horizontal asymptote of ( y = 4 ) means that the persons temperature will approach ( 102.6^{circ}f ) as time elapses.
the horizontal asymptote of ( y = 4 ) means that the persons temperature will approach ( 4^{circ}f ) as time elapses.

Explanation:

Step1: Find the horizontal asymptote

For a rational function \(y = \frac{f(t)}{g(t)}\) where \(f(t)=4t\) and \(g(t)=t^{2}+1\), since the degree of the numerator (\(n = 1\)) is less than the degree of the denominator (\(m=2\)), the horizontal asymptote is \(y = 0\).

Step2: Interpret the horizontal asymptote

The function \(T(t)\) represents the increase in body temperature above \(98.6^{\circ}F\). As \(t\to\infty\), \(T(t)\to0\). So the person's temperature (original \(98.6^{\circ}F\) plus \(T(t)\)) approaches \(98.6^{\circ}F\) as \(t\) (time) elapses.

Answer:

The horizontal asymptote of \(y = 0\) means that the person's temperature will approach \(98.6^{\circ}F\) as time elapses.