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Questions: Algebra BusinessCalculus
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Find domain and vertical asymptote of \(\displaystyle f(x) = \ln{\left(x - 1 \right)} - 5\)
The base function \(\displaystyle \ln(x)\) has vertical asymptote \(\displaystyle x =0\) and domain \(\displaystyle (0, \infty)\). \(\displaystyle f(x)\) is shifted 1 unit right and 5 units down. This gives the vertical asymptote as \(\displaystyle x = 1\) and the domain as \(\displaystyle (1, \infty)\)
\begin{question}Find domain and vertical asymptote of $f(x) = \ln{\left(x - 1 \right)} - 5$
\soln{9cm}{The base function $\ln(x)$ has vertical asymptote $x =0$ and domain $(0, \infty)$. $f(x)$ is shifted 1 unit right and 5 units down. This gives the vertical asymptote as $x = 1$ and the domain as $(1, \infty)$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find domain and vertical asymptote of <img class="equation_image" title=" \displaystyle f(x) = \ln{\left(x - 1 \right)} - 5 " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Cln%7B%5Cleft%28x%20-%201%20%5Cright%29%7D%20-%205%20" alt="LaTeX: \displaystyle f(x) = \ln{\left(x - 1 \right)} - 5 " data-equation-content=" \displaystyle f(x) = \ln{\left(x - 1 \right)} - 5 " /> </p> </p><p> <p>The base function <img class="equation_image" title=" \displaystyle \ln(x) " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%28x%29%20" alt="LaTeX: \displaystyle \ln(x) " data-equation-content=" \displaystyle \ln(x) " /> has vertical asymptote <img class="equation_image" title=" \displaystyle x =0 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D0%20" alt="LaTeX: \displaystyle x =0 " data-equation-content=" \displaystyle x =0 " /> and domain <img class="equation_image" title=" \displaystyle (0, \infty) " src="/equation_images/%20%5Cdisplaystyle%20%280%2C%20%5Cinfty%29%20" alt="LaTeX: \displaystyle (0, \infty) " data-equation-content=" \displaystyle (0, \infty) " /> . <img class="equation_image" title=" \displaystyle f(x) " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20" alt="LaTeX: \displaystyle f(x) " data-equation-content=" \displaystyle f(x) " /> is shifted 1 unit right and 5 units down. This gives the vertical asymptote as <img class="equation_image" title=" \displaystyle x = 1 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%201%20" alt="LaTeX: \displaystyle x = 1 " data-equation-content=" \displaystyle x = 1 " /> and the domain as <img class="equation_image" title=" \displaystyle (1, \infty) " src="/equation_images/%20%5Cdisplaystyle%20%281%2C%20%5Cinfty%29%20" alt="LaTeX: \displaystyle (1, \infty) " data-equation-content=" \displaystyle (1, \infty) " /> </p> </p>