Question

The increase in a person’s body temperature T(t), above 98.6ºF, can be modeled by the function T (t) = StartFraction 4 t Over t squared + 1 EndFraction, 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 person’s temperature will approach 98.6ºF as time elapses. The horizontal asymptote of y = 0 means that the person’s temperature will approach 0ºF as time elapses. The horizontal asymptote of y = 4 means that the person’s temperature will approach 102.6ºF as time elapses. The horizontal asymptote of y = 4 means that the person’s temperature will approach 4ºF as time elapses.

Answer 1

Here, the function that represents the increment in temperature above 98.6ºF after t time,

$T(t)=\frac{4t}{t^2+1}$

Recall that if for a function $y=f(x)$,

$\lim_{x\rightarrow \pm \infty} f(x)=L$, then. $y=L$ is called horizontal asymptote of the function.

Here,

$\lim_{x\rightarrow \infty} \frac{4t}{t^2+1}=\lim_{x\rightarrow \infty} \frac{4}{t+\frac{1}{t}}$

                    $=4\times \frac{1}{\infty+\frac{1}{\infty}}$

                    $=4\times \frac{1}{\infty+0}$

                   $=4\times 0$

                  = 0

Thus, $y=0$ is the horizontal asymptote of the graph of the function T(t).

But, y=T(t) =0 shows that there is 0 increment in temperature 98.6ºF.

Therefore, the horizontal asymptote of y = 0 means that the person’s temperature will approach 98.6ºF as time elapses (or time approaches to infinite).

Answer 2

The correct answer is A, or The horizontal asymptote of y = 0 means that the person’s temperature will approach 98.6ºF as time elapses.

Just got it right on edge 2020, hope this helps and have a great day!! :)