Horizontal and Vertical asymptotes of a curve

**Oldspice1212** · October 2nd 2012, 09:00 PM

Find the horizontal and vertical asymptotes of the curve.

y = 4e^x/e^x-6

x=
y= (smaller y - value)
y = (larger y - value)

How do you do this?

**MarkFL** · October 2nd 2012, 11:07 PM

I assume you mean:

$y=\frac{4e^x}{e^x-6}$

To find the vertical asymptote(s) solve:

$e^x-6=0$

To find the horizontal asymptote(s) evaluate:

$\lim_{x\to\pm\infty}\frac{4e^x}{e^x-6}$

**FernandoRevilla** · October 2nd 2012, 11:15 PM

Originally Posted by Oldspice1212

Find the horizontal and vertical asymptotes of the curve. y = 4e^x/e^x-6 How do you do this?

Vertical asymptotes: $e^x-6=0\Leftrightarrow x=\ln 6$ .

Horizontal asymptotes: $f(x)=\frac{4e^x}{e^x-6}=\dfrac{4}{1-\frac{6}{e^x}}\Rightarrow \begin{Bmatrix} \displaystyle\lim_{x \to{+}\infty}f(x)=4\\\displaystyle\lim_{x \to{-}\infty}f(x)=0\end{matrix}$

Edited: Sorry, I didn't see MarkFL2's post.

**MarkFL** · October 2nd 2012, 11:20 PM

Hey, no worries! I make posts not seeing someone else has posted in the meantime quite frequently!

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