what is the asymptote of the graph of (g(x)=ln x + 5).\na. vertical at (x = e)\nb. vertical at (x = 5)\nc…

Answer:

A. vertical at $x = e$

Explanation:

Step1: Recall domain of natural - log function

The domain of $y=\ln(x)$ is $x>0$. For $g(x)=\ln(x + 5)$, we set $x+5>0$, so $x>- 5$.

Step2: Determine vertical asymptote

The vertical asymptote of $y = \ln(x + 5)$ occurs when $x+5 = 0$, i.e., $x=-5$. But there is a mis - type in the options. If the function was $g(x)=\ln(x - e)$, the vertical asymptote would be at $x = e$ since the domain of $y=\ln(x - e)$ is $x>e$ and the vertical asymptote is at the value that makes the argument of the logarithm zero. For a natural - logarithm function $y=\ln(u)$, the vertical asymptote is at $u = 0$. Since the other options for horizontal asymptotes are incorrect (the function $y=\ln(x)+5$ has no horizontal asymptote as $\lim_{x\rightarrow+\infty}\ln(x)+5=+\infty$ and $\lim_{x\rightarrow0^{+}}\ln(x)+5=-\infty$), assuming a mis - type in the problem setup, the most relevant option is A.