# How do you find the horizontal and vertical asymptotes to the curve of y =(e^2x +1)/(e^x -2) ?

Jul 14, 2017

The vertical asymptote is $y = \ln 2$
The horizontal asymptote is $y = - \frac{1}{2}$

#### Explanation:

The equation of the curve is

$y = \frac{{e}^{2 x} + 1}{{e}^{x} - 2}$

The denominator is $= 0$, when

${e}^{x} - 2 = 0$

${e}^{x} = 2$

$x = \ln 2$

To calculate the vertical asymptotes , we calculate

${\lim}_{x \to \ln 2 , x < \ln 2} y = {\lim}_{x \to \ln 2 , x < \ln 2} \frac{{e}^{2 x} + 1}{{e}^{x} - 2} = \frac{5}{0} ^ \left(-\right) = - \infty$

As ${e}^{2 \cdot \ln 2} + 1 = 4 + 1 = 5$ and ${e}^{\ln 2} - 2 = {2}^{-} - 2 = {0}^{-}$

${\lim}_{x \to \ln 2 , x > \ln 2} y = {\lim}_{x \to \ln 2 , x > \ln 2} \frac{{e}^{2 x} + 1}{{e}^{x} - 2} = \frac{5}{0} ^ \left(+\right) = + \infty$

As ${e}^{2 \cdot \ln 2} + 1 = 4 + 1 = 5$ and ${e}^{\ln 2} - 2 = {2}^{+} - 2 = {0}^{+}$

Therefore,

the vertical asymptote is $y = \ln 2$

To calculate the horizontal asymptotes , we calculate

${\lim}_{x \to + \infty} y = {\lim}_{x \to + \infty} \frac{{e}^{2 x} + 1}{{e}^{x} - 2} = {\lim}_{x \to + \infty} {e}^{2 x} / {e}^{x} = {\lim}_{x \to + \infty} {e}^{x} = + \infty$

${\lim}_{x \to - \infty} y = {\lim}_{x \to - \infty} \frac{{e}^{2 x} + 1}{{e}^{x} - 2} = {\lim}_{x \to - \infty} \frac{{0}^{+} + 1}{{0}^{+} - 2} = - \frac{1}{2}$

As ${\lim}_{x \to - \infty} {e}^{2 x} = {0}^{+}$ and ${\lim}_{x \to - \infty} {e}^{x} = {0}^{+}$

Therefore,

the horizontal asymptote is $y = - \frac{1}{2}$

graph{(e^(2x)+1)/(e^x-2) [-38.55, 34.5, -19.86, 16.7]}