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

Apr 16, 2016

Vertical asymptote is $x = 0$ and

two horizontal asymptotes $x = 0$ and $x = - \frac{3}{2}$

Explanation:

Vertical asymptote is given by putting denominator equal to zero.

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

or ${e}^{x} = 1$ or $x = \ln 1 = 0$

For horizontal asymptote, let us find out $L {t}_{x \to \infty} \frac{3 {e}^{x}}{2 - 2 {e}^{x}}$

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

$L {t}_{x \to \infty} \frac{3}{2 {e}^{-} x - 2} = - \frac{3}{2}$ as ${e}^{-} \infty = 0$

Also $L {t}_{x \to - \infty} \frac{3 {e}^{x}}{2 - 2 {e}^{x}} = 0$

Hence we have two horizontal asymptotes $x = 0$ and $x = - \frac{3}{2}$

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