Question

Let f(x)=20/1+9e^3x .

Answer 1

The answer is the last one

Answer 2

Answer:

y = 0 and y = 20

Step-by-step explanation:

The given function is $\left(\frac{20}{\left(1+9e^{3x}\right)}\right)$

There is no vertical asymptotes because the denominator will never be zero.

Horizontal asymptotes are given by

$y=\lim _{x\to -\infty }f(x)\\\\y=\lim _{x\to \:-\infty \:}\left(\frac{20}{1+9e^{3x}}\right)$

Take the constant 20 out of the limit

$20\cdot \lim \:_{x\to \:-\infty \:}\left(\frac{1}{1+9e^{3x}}\right)$

Now, we can further simplify as follows

$y=20\cdot \frac{\lim _{x\to \:-\infty \:}\left(1\right)}{\lim _{x\to \:-\infty \:}\left(1+9e^{3x}\right)}\\\\y=20\cdot \frac{1}{1}\\\\y=20$

Similarly, the second horizontal asymptote is

$y=\lim _{x\to \infty }\left(\frac{20}{\left(1+9e^{3x}\right)}\right)$

We can simplify this limit as we did the previous one

$y=20\cdot \frac{\lim _{x\to \infty \:}\left(1\right)}{\lim _{x\to \infty \:}\left(1+9e^{3x}\right)}\\\\y=20\cdot \frac{1}{\infty \:}\\\\y=0$

Hence, asymptotes of the graph of f(x) are

y = 0 and y = 20