Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=lnleft(x^{2}+36
ight) ).
oa. the domain is all real ( x ), except ( x= ).
(use a comma to separate answers as needed.)
ob. the domain is all real numbers.
find the ( x )-intercept(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
oa. the ( x )-intercept(s) is (are) ( x= )
(round to one decimal place as needed. use a comma to separate answers as needed.)
ob. there are no ( x )-intercepts.
find the ( y )-intercept(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
oa. the ( y )-intercept(s) is (are) ( y= )
(round to one decimal place as needed. use a comma to separate answers as needed.)
ob. there are no ( y )-intercepts.

Explanation:

Step1: Find the domain

For the function \(y = \ln(u)\), the argument \(u=x^{2}+36\). Since \(x^{2}\geq0\) for all real \(x\), then \(x^{2}+36\geq36>0\) for all real \(x\). So the domain is all real numbers.

Step2: Find the \(x -\) intercept

Set \(y = 0\), so \(\ln(x^{2}+36)=0\). Using the property \(\ln(a)=b\Leftrightarrow a = e^{b}\), we have \(x^{2}+36=e^{0}=1\). Then \(x^{2}=1 - 36=- 35\). Since \(x^{2}=-35\) has no real solutions, there are no \(x -\) intercepts.

Step3: Find the \(y -\) intercept

Set \(x = 0\). Then \(y=\ln(0^{2}+36)=\ln(36)\approx3.6\) (using a calculator, \(\ln(36)=\ln(6^{2}) = 2\ln(6)\approx2\times1.792 = 3.584\approx3.6\))

Answer:

For the domain: B. The domain is all real numbers.
For the \(x -\) intercept: B. There are no \(x -\) intercepts.
For the \(y -\) intercept: A. The \(y -\) intercept(s) is (are) \(y = 3.6\)