How to find the asymptotes of $f(x)= (3e^(x))/(2-2e^(x))$ ?

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Answer 1 · 2017-10-03T09:01:02

Horizontal asymptotes $y=0$ and $y=1.5$ and vertical asymptote $x=0$

One can find asymptotes of $f(x)$ two ways

one , finding how $f(x)$ behaves or tends as $x->oo$ or $x->-oo$

and here as $f(x)=(3e^x)/(2-2e^x)=3/(2e^(-x)-2)$

as $x->-oo$ , $e^x->0$ and $f(x)->0/2=0$ .

Similarly as $x->oo$ , $e^(-x)->0$ and $f(x)->3/2=1.5$

This gives us horizontal asymptotes as $y=0$ and $y=1.5$

two as $f(x)=oo$ , $2-2e^x=0$ or $e^x=1$ i.e. $x=0$

Hence we have a vertical asymptote $x=0$

graph{(3e^x)/(2-2e^x) [-10, 10, -5, 5]}

How to find the asymptotes of f(x)= (3e^(x))/(2-2e^(x))f(x)= (3e^(x))/(2-2e^(x))?