Answer
Domain: $x\gt-6$
x-intercept: $(-5.632,0)$
$x=-6$ is the vertical asymptote.
Work Step by Step
$f(x)=1+\ln(x+6)$
Domain:
$x+6\gt0$
$x\gt-6$
x-intercept:
$f(x)=1+\ln(x+6)=0$
$\ln(x+6)=-1$
$x+6=e^{-1}$
$x=-6+\frac{1}{e}\approx-5.632$. So, the x-intercept point is $(-5.632,0)$
$\ln x \to\infty$ when $x\to0^+$. So:
$\ln(x+6) \to\infty$ when $x+6\to0^+$
$\ln(x+6) \to\infty$ when $x\to-6^+$
Thus, $x=-6$ is the vertical asymptote.
