How do you find the vertical, horizontal or slant asymptotes for $f(x)= (3e^(x))/(2-2e^(x))$ ?

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Answer 1 · 2016-04-16T09:30:54

Vertical asymptote is $x=0$ and

two horizontal asymptotes $x=0$ and $x=-3/2$

Vertical asymptote is given by putting denominator equal to zero.

Hence it is given by $2-2e^x=0$ or $2e^x=2$

or $e^x=1$ or $x=ln1=0$

For horizontal asymptote, let us find out $Lt_(x->oo)(3e^x)/(2-2e^x)$

Dividing numerator by $e^x$ this is equal to

$Lt_(x->oo)3/(2e^-x-2)=-3/2$ as $e^-oo=0$

Also $Lt_(x->-oo)(3e^x)/(2-2e^x)=0$

Hence we have two horizontal asymptotes $x=0$ and $x=-3/2$

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

How do you find the vertical, horizontal or slant asymptotes for f(x)= (3e^(x))/(2-2e^(x))f(x)= (3e^(x))/(2-2e^(x))?