Question

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

What are the asymptotes of the graph of f(x) ?

Select each correct answer.



y = 3

y = 0

y = 9

y = 20

Answer 1

y = 0 and y = 20

Explanation:

The given function is
`\left((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((20)/(1+9e^(3x))\right)`

Take the constant 20 out of the limit

`20\cdot \lim \:_(x\to \:-\infty \:)\left((1)/(1+9e^(3x))\right)`

Now, we can further simplify as follows

`y=20\cdot (\lim _(x\to \:-\infty \:)\left(1\right))/(\lim _(x\to \:-\infty \:)\left(1+9e^(3x)\right))\\\\y=20\cdot (1)/(1)\\\\y=20`

Similarly, the second horizontal asymptote is

`y=\lim _(x\to \infty )\left((20)/(\left(1+9e^(3x)\right))\right)`

We can simplify this limit as we did the previous one

`y=20\cdot (\lim _(x\to \infty \:)\left(1\right))/(\lim _(x\to \infty \:)\left(1+9e^(3x)\right))\\\\y=20\cdot (1)/(\infty \:)\\\\y=0`

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

y = 0 and y = 20