Question

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

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

Select EACH correct answer.

Answer 1

Answer: (A) y = 15 (B) y = 0

Explanation:

`f(x) = (15)/(1+4e^(-0.2x))`

The vertical asymptote is the restriction on the x-value. Since the denominator cannot be equal to zero, then

1 + 4e⁻⁰⁻²ˣ ≠ 0

→ 4e⁻⁰⁻²ˣ ≠ -1

→ e⁻⁰⁻²ˣ ≠
`-(1)/(4)`

→ ln e⁻⁰⁻²ˣ ≠
`ln\bigg(-(1)/(4)\bigg)`

Note: ln cannot be negative. So, there is no vertical asymptote.

The horizontal asymptote can be found by finding the limit as x approaches positive and negative infinity.

`\lim_(x \to \infty) (15)/(1+4e^(-0.2x))`

`= (15)/(1+4(0))`

`= (15)/(1)`

`= 15`

So, the horizontal asymptote when x → +∞ is y = 15

`\lim_(x \to -\infty) (15)/(1+4e^(-0.2x))`

`= (15)/(1+4(\infty))`

`= (15)/(\infty)`

`= 0`

So, the horizontal asymptote when x → -∞ is y = 0