Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=lnleft(x^{2}+36
ight) ).

a. the ( x )-intercept(s) is (are) ( x= )
(round to one decimal place as needed. use a comma to separate answers as needed.)
b. 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.

a. the ( y )-intercept(s) is (are) ( y=3.6 )
(round to one decimal place as needed. use a comma to separate answers as needed.)
b. there are no ( y )-intercepts.

find vertical asymptote(s), if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the vertical asymptote(s) is (are) ( x= )
(use a comma to separate answers as needed.)
b. there are no vertical asymptotes.

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.