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 ( 98.6^{circ} mathrm{f} ) as time elapses.
• 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 ( 102.6^{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.

Explanation:

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

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 \(n

Step2: Interpret the meaning of the horizontal asymptote in the context

Since \(T(t)\) represents the increase in body - temperature above \(98.6^{\circ}F\). As \(t
ightarrow\infty\), \(T(t)
ightarrow0\). So the person's temperature (original temperature \(98.6^{\circ}F\) plus \(T(t)\)) will approach \(98.6^{\circ}F\) 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.