Question

the increase in a persons body temperature ( t(t) ), above ( 98.6^{circ} mathrm{f} ), can be modeled by the function ( t(t)=\frac{4 t}{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 ( 0^{circ} mathrm{f} ) as time elapses.
the horizontal asymptote of ( y = 4 ) means that the persons temperature will approach ( 4^{circ} mathrm{f} ) as time elapses.
the horizontal asymptote of ( y = 0 ) means that the persons temperature will approach ( 98.6^{circ} mathrm{f} ) as time elapses.
the horizontal asymptote of ( y = 4 ) means that the persons temperature will approach ( 102.6^{circ} mathrm{f} ) as time elapses.

Explanation:

Step1: Find the horizontal asymptote of the function \(T(t)=\frac{4t}{t^{2}+1}\)

For a rational function \(y = \frac{f(t)}{g(t)}\) where \(f(t)=4t\) (degree \(n = 1\)) and \(g(t)=t^{2}+1\) (degree \(m=2\)). When \(nightarrow\infty}T(t)=\lim_{t
ightarrow\infty}\frac{4t}{t^{2}+1}\). Divide numerator and denominator by \(t^{2}\): \(\lim_{t
ightarrow\infty}\frac{\frac{4t}{t^{2}}}{\frac{t^{2}}{t^{2}}+\frac{1}{t^{2}}}=\lim_{t
ightarrow\infty}\frac{\frac{4}{t}}{1 + \frac{1}{t^{2}}}\). As \(t
ightarrow\infty\), \(\frac{4}{t}
ightarrow0\) and \(\frac{1}{t^{2}}
ightarrow0\). So \(\lim_{t
ightarrow\infty}T(t) = 0\). The function \(T(t)\) gives the increase in temperature above \(98.6^{\circ}F\).

Step2: Interpret the result

If the increase in temperature \(T(t)\) approaches \(0\) as \(t
ightarrow\infty\), then the person's temperature (original temperature \(98.6^{\circ}F\) plus \(T(t)\)) approaches \(98.6^{\circ}F\) as \(t
ightarrow\infty\)

Answer:

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