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 ( 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.
find horizontal asymptote(s), if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the horizontal asymptote(s) is (are) ( y= )
(use a comma to separate answers as needed.)
b. there are no horizontal asymptotes.

Explanation:

Step1: Find the y - intercept

The y - intercept is found by setting \(x = 0\) in the function \(y=f(x)=\ln(x^{2}+36)\).
When \(x = 0\), we have \(y=\ln(0^{2}+36)=\ln(36)\approx3.6\) (using a calculator, \(\ln(36)=\ln(6^{2}) = 2\ln(6)\approx2\times1.792\approx3.6\)).

Step2: Find the vertical asymptote

For a function \(y = \ln(u)\), vertical asymptotes occur when \(u=0\). Set \(u=x^{2}+36\). Solve the equation \(x^{2}+36 = 0\), which gives \(x^{2}=- 36\). Since \(x^{2}\geq0\) for all real \(x\), there are no real solutions. So, there are no vertical asymptotes.

Step3: Find the horizontal asymptote

We use the limit as \(x
ightarrow\pm\infty\). We know that \(\lim_{x
ightarrow\pm\infty}\ln(x^{2}+36)\). As \(x
ightarrow\pm\infty\), \(x^{2}+36
ightarrow\infty\), and \(\lim_{t
ightarrow\infty}\ln(t)=\infty\). So, there are no horizontal asymptotes.

Answer:

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