# How do you find the domain, x intercept and vertical asymptotes of f(x)=3lnx-1?

domain $x \setminus \in \left(0 , \setminus \infty\right)$, x-intercept ${e}^{\frac{1}{3}}$ & asymptote $x = 0$

#### Explanation:

Given function: $f \left(x\right) = 3 \setminus \ln x - 1$

Since, Logarithms $\setminus \ln x$ is defined for all positive real numbers $x > 0$ hence the domain of the function is

$x \setminus \in \left(0 , \setminus \infty\right)$

The curve $y = 3 \ln x - 1$ will intersect x-axis at which $y = 0$

$0 = 3 \ln x - 1$

$\setminus \ln x = \frac{1}{3}$

$x = {e}^{\frac{1}{3}}$

The curve $y = 3 \setminus \ln x - 1$ will touch the y-axis $\left(x = 0\right)$ at infinity hence the asymptotes of givem curve is $x = 0$