Answer:
y = 0 and y = 20
Step-by-step explanation:
The given function is [tex]\left(\frac{20}{\left(1+9e^{3x}\right)}\right)[/tex]
There is no vertical asymptotes because the denominator will never be zero.
Horizontal asymptotes are given by
[tex]y=\lim _{x\to -\infty }f(x)\\\\y=\lim _{x\to \:-\infty \:}\left(\frac{20}{1+9e^{3x}}\right)[/tex]
Take the constant 20 out of the limit
[tex]20\cdot \lim \:_{x\to \:-\infty \:}\left(\frac{1}{1+9e^{3x}}\right)[/tex]
Now, we can further simplify as follows
[tex]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[/tex]
Similarly, the second horizontal asymptote is
[tex]y=\lim _{x\to \infty }\left(\frac{20}{\left(1+9e^{3x}\right)}\right)[/tex]
We can simplify this limit as we did the previous one
[tex]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[/tex]
Hence, asymptotes of the graph of f(x) are
y = 0 and y = 20
