# How do you find the domain, x intercept and vertical asymptotes of f(x)=-log_6(x+2)?

Feb 26, 2018

$x$-intercept: $\left(- 1 , 0\right)$ Vertical asymptote: $x = - 2$

#### Explanation:

The $x$-intercept of a logarithmic function $f \left(x\right) = {\log}_{a} \left(x\right)$ occurs wherever $x = 1$ (because ${\log}_{a} \left(1\right) = 0$); therefore, we want to solve whatever is in the argument (parentheses) of the logarithmic function for $1 :$

$x + 2 = 1$

$x = 1 - 2 = - 1$

Thus, the $x$-intercept occurs at $\left(- 1 , 0\right)$.

The vertical asymptote of a logarithmic function $f \left(x\right) = {\log}_{a} \left(x\right)$ occurs at $x = c$ where $c$ is the number at which the argument (parentheses) of the logarithmic function becomes $0$, because we know $f \left(x\right) = {\log}_{a} \left(x\right)$ does not exist for $x < 0$. This means we need to solve whatever is in the parentheses of our logarithmic function for $0.$

$x + 2 = 0$

$x = - 2$

Thus, $x = - 2$ is the vertical asymptote.