the increase in a persons body temperature (t(t)), above (98.6^{circ}f), can be modeled by the function…
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?\no the horizontal asymptote of (y = 0) means that the persons temperature will approach (98.6^{circ}f) as time elapses.\no the horizontal asymptote of (y = 0) means that the persons temperature will approach (0^{circ}f) as time elapses.\no the horizontal asymptote of (y = 4) means that the persons temperature will approach (102.6^{circ}f) as time elapses.\no the horizontal asymptote of (y = 4) means that the persons temperature will approach (4^{circ}f) as time elapses.
Answer
Brief Explanations:
First, find the horizontal asymptote of the function $T(t)=\frac{4t}{t^{2}+1}$. As $t$ approaches infinity, divide both numerator and denominator by $t^{2}$: $\lim_{t\rightarrow\infty}\frac{\frac{4t}{t^{2}}}{\frac{t^{2}}{t^{2}}+\frac{1}{t^{2}}}=\lim_{t\rightarrow\infty}\frac{\frac{4}{t}}{1 + \frac{1}{t^{2}}}=0$. The function $T(t)$ gives the increase in body - temperature above $98.6^{\circ}F$. A horizontal asymptote of $y = 0$ means the increase in temperature above $98.6^{\circ}F$ approaches $0$, so the person's temperature approaches $98.6^{\circ}F$ as time elapses.
Answer:
The horizontal asymptote of y = 0 means that the person's temperature will approach 98.6°F as time elapses.