summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of (…

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=lnleft(x^{2}+36\right) ).\n\na. the ( x )-intercept(s) is (are) ( x= )\n(round to one decimal place as needed. use a comma to separate answers as needed.)\nb. there are no ( x )-intercepts.\n\nfind the ( y )-intercept(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the ( y )-intercept(s) is (are) ( y=3.6 )\n(round to one decimal place as needed. use a comma to separate answers as needed.)\nb. there are no ( y )-intercepts.\n\nfind vertical asymptote(s), if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the vertical asymptote(s) is (are) ( x= )\n(use a comma to separate answers as needed.)\nb. there are no vertical asymptotes.

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=lnleft(x^{2}+36\right) ).\n\na. the ( x )-intercept(s) is (are) ( x= )\n(round to one decimal place as needed. use a comma to separate answers as needed.)\nb. there are no ( x )-intercepts.\n\nfind the ( y )-intercept(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the ( y )-intercept(s) is (are) ( y=3.6 )\n(round to one decimal place as needed. use a comma to separate answers as needed.)\nb. there are no ( y )-intercepts.\n\nfind vertical asymptote(s), if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the vertical asymptote(s) is (are) ( x= )\n(use a comma to separate answers as needed.)\nb. there are no vertical asymptotes.

Answer

Explanation:

Step1: Find x - intercepts

For x - intercepts, set (y = f(x)=0), so (\ln(x^{2}+36)=0). By the property of logarithms, if (\ln a = 0), then (a = 1). So (x^{2}+36=1), (x^{2}=- 35). Since the square of a real number (x) ((x\in R), (x^{2}\geq0)) cannot be negative, there are no real solutions for (x). So there are no x - intercepts.

Step2: Find y - intercepts

For y - intercepts, set (x = 0). Then (y=f(0)=\ln(0^{2}+36)=\ln(36)\approx3.6) (using a calculator, (\ln(36)=\ln(4\times9)=\ln(4)+\ln(9)=2\ln(2)+2\ln(3)\approx2\times0.693 + 2\times1.099\approx3.6))

Step3: Find vertical asymptotes

The domain of (y = \ln(u)) is (u>0). For (u=x^{2}+36), since (x^{2}\geq0) for all real (x), then (x^{2}+36\geq36>0) for all (x\in R). So there are no values of (x) for which (x^{2}+36 = 0) (in the real - number system). So there are no vertical asymptotes.

Answer:

B. There are no x - intercepts. A. The y - intercept(s) is (are) (y = 3.6) B. There are no vertical asymptotes.