# Thread: All asymptotes for this function.

1. ## All asymptotes for this function.

How can i calculate all asymptotes (horizontal and vertical) for this function.

(6) / (1+2e-5x​)

2. ## Re: All asymptotes for this function.

To find vertical asymptotes, equate the denominator to zero, and solve for x.

To find horizontal asymptotes, compute:

$\lim_{x\to\pm\infty}\frac{6}{1+2e^{-5x}}$

What do you find?

3. ## Re: All asymptotes for this function.

Horizontal:

6 /(1 + (2/∞))= 6

vertical:

1+2e-5x=0

2*(-5x)=-ln1

x= ln1/10 = 0.

These are the results i got when i tried earlier, however with f(x) = 0 i get x as 2, so the vertical cant be 0. Was wondering if there might be two horizontal

4. ## Re: All asymptotes for this function.

You have found one horizontal asymptote, try the limit as x goes to negative infinity as well.

You are incorrectly solving for x to find any possible vertical asymptotes:

$1+2e^{-5x}=0$

$e^{-5x}=-\frac{1}{2}$

Without taking logs, we should know that:

$0

for all real $x$, hence there are no vertical asymptotes.

5. ## Re: All asymptotes for this function.

I dont really see how using -∞ will change the results, but again i'm asking here as i dont know how to do this.

6. ## Re: All asymptotes for this function.

Since the exponent is negative, the denominator goes to 1 as goes go to infinity and goes to infinity as x goes to negative infinity.

6/1 = 6 so y = 6 is the horizontal asymptote on the right.

6/∞ = 0 so y = 0 is the horizontal asymptote to the left.