# How do you find the domain, x intercept and vertical asymptotes of f(x)=ln(x+2)?

Nov 11, 2017

See below.

#### Explanation:

$\ln \left(x\right)$ is only defined for $x > - 2$ for real numbers.

So the domain is:

$\textcolor{b l u e}{\left\{x \in \mathbb{R} | - 2 < x < \infty\right\}}$

$x$ axis intercepts occur when $y = 0$:

$y = \ln \left(x + 2\right)$

$\ln \left(x + 2\right) = 0$

This has no solution, because ${e}^{x}$ can never equal 0.

No x axis intercepts.

Y axis intercept occurs when $x = 0$

$\ln \left(0 + 2\right) = \ln \left(2\right) \approx \textcolor{b l u e}{0.69314718}$

Vertical asymptotes occur where $\ln \left(x + 2\right)$ is undefined:

$\ln \left(x + 2\right)$ is undefined when $x = - 2$

So the line $\textcolor{b l u e}{x = - 2}$ is a vertical asymptote. graph{ln(x+2) [-22.81, 22.8, -11.4, 11.41]}