How do you find the vertical, horizontal or slant asymptotes for $y = \frac{2 e^{x}}{e^{x} - 5}$ $y = (2e^x) / (e^x - 5)$ ?

Question

8468 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T12:09:38

We have a vertical asymptote at $x = ln 5 = 1.6094$ $x=ln5=1.6094$

and horizontal asymptotes at $y = 0$ $y=0$ and $y = 2$ $y=2$

We have $y = \frac{2 e^{x}}{e^{x} - 5}$ $y=(2e^x)/(e^x-5)$

As $x \to - \infty$ $x->-oo$ , $y \to \frac{0}{0 - 5} = 0$ $y->0/(0-5)=0$

dividing numerator and denominator by $e^{x}$ $e^x$ , we get

$y = \frac{2}{1 - \frac{5}{e^{x}}}$ $y=2/(1-5/e^x)$

As such, when $x \to \infty$ $x->oo$ , $y \to \frac{2}{1 - 0} = 2$ $y->2/(1-0)=2$

Hence, horizontal asymptotes are $y = 0$ $y=0$ and $y = 2$ $y=2$

also $y e^{x} - 5 y = 2 e^{x}$ $ye^x-5y=2e^x$ or $y e^{x} - 2 e^{x} = 5 y$ $ye^x-2e^x=5y$

i.e. $e^{x} = \frac{5 y}{y - 2}$ $e^x=(5y)/(y-2)$ or $x = ln (\frac{5 y}{y - 2}) = ln (\frac{5}{1 - \frac{2}{y}})$ $x=ln((5y)/(y-2))=ln(5/(1-2/y))$

and as $y \to \pm \infty$ $y->+-oo$ , $x \to ln 5 = 1.6094$ $x->ln5=1.6094$ ,

we have a vertical asymptote at $x = ln 5 = 1.6094$ $x=ln5=1.6094$

graph{(2e^x)/(e^x-5) [-10, 10, -5, 5]}

How do you find the vertical, horizontal or slant asymptotes for y = (2e^x) / (e^x - 5)y=2exex−5y = (2e^x) / (e^x - 5)?