## Algebra and Trigonometry 10th Edition

Domain: $x\gt-6$ x-intercept: $(-5.632,0)$ $x=-6$ is the vertical asymptote.
$f(x)=1+\ln(x+6)$ 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.