# How do you find the asymptotes for  y = log (x + 2)?

##### 1 Answer
Jan 11, 2016

Vertical Asymptote:

$x = - 2$

#### Explanation:

$f \left(x\right) = \log \left(g \left(x\right)\right)$

The Existence Condition is

$g \left(x\right) > 0$

because $\log$ is definited $\forall x \in \left(0 , + \infty\right)$

$g \left(x\right) = x + 2$

$x + 2 > 0$

$x > - 2$

Then:

$F . E .$ (Field of Existence): $\left(- 2 , + \infty\right)$

$x = {x}_{0} = - 2$

Could be a vertical asymptote if

${\lim}_{x \rightarrow - {2}^{+}} f \left(x\right) = \pm \infty$

${\lim}_{x \rightarrow - {2}^{+}} f \left(x\right) = {\lim}_{x \rightarrow - {2}^{+}} \log \left(x + 2\right) =$

${\lim}_{x \rightarrow - {2}^{+}} \log \left({0}^{+}\right) = - \infty$

$\therefore x = - 2$ vertical asymptote

We could looking for horizontal/slant asymptotes

${\lim}_{x \rightarrow + \infty} f \left(x\right) = {\lim}_{x \rightarrow + \infty} \log \left(x + 2\right) = + \infty$

$\therefore$ no horizontal asymptotes

the slant asymptote formula is

$y = m x + q$

with

$m = {\lim}_{x \rightarrow + \infty} f \frac{x}{x}$

$q = {\lim}_{x \rightarrow + \infty} \left[f \left(x\right) - m x\right]$

$m = {\lim}_{x \rightarrow + \infty} f \frac{x}{x} = {\lim}_{x \rightarrow + \infty} \log \frac{x + 2}{x} = \frac{+ \infty}{+ \infty}$

Applying The L'Hopital's rule

${\lim}_{x \rightarrow + \infty} \frac{h \left(x\right)}{i \left(x\right)} = {\lim}_{x \rightarrow + \infty} \frac{h ' \left(x\right)}{i ' \left(x\right)}$

${\lim}_{x \rightarrow + \infty} \log \frac{x + 2}{x} = {\lim}_{x \rightarrow + \infty} \frac{\frac{1}{x + 2}}{1} =$

$m = {\lim}_{x \rightarrow + \infty} \frac{1}{x + 2} = 0$

$q = {\lim}_{x \rightarrow + \infty} \left[f \left(x\right) - m x\right] = {\lim}_{x \rightarrow + \infty} \left[\log \left(x + 2\right) - 0 x\right] =$
$= {\lim}_{x \rightarrow + \infty} \log \left(x + 2\right) = + \infty$

It is not finite, then $\cancel{\exists}$ slant asymptote