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

(6) / (1+2e^{-5x})

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- Nov 14th 2012, 03:09 PMGardsVisionAll asymptotes for this function.
How can i calculate all asymptotes (horizontal and vertical) for this function.

(6) / (1+2e^{-5x}) - Nov 14th 2012, 03:27 PMMarkFLRe: All asymptotes for this function.
To find vertical asymptotes, equate the denominator to zero, and solve for x.

To find horizontal asymptotes, compute:

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

What do you find? - Nov 14th 2012, 03:43 PMGardsVisionRe: 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 - Nov 14th 2012, 04:00 PMMarkFLRe: 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:

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

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

Without taking logs, we should know that:

$\displaystyle 0<e^{-5x}$

for all real $\displaystyle x$, hence there are no vertical asymptotes. - Nov 14th 2012, 08:45 PMGardsVisionRe: 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.

- Nov 14th 2012, 08:51 PMMarkFLRe: 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.