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)=9 e^{-0.5 x^{2}} ).\na. the function has one horizontal asymptote, ( y = 0 ).\n(type an equation.)\nb. the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is\n(type equations.)\nc. there are no horizontal asymptotes.\nfind any vertical asymptotes of ( f(x) ). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. the function has one vertical asymptote,\n(type an equation.)\nb. the function has two vertical asymptotes. the leftmost asymptote is and the rightmost asymptote is\n(type equations.)\nc. there are no vertical asymptotes.

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=9 e^{-0.5 x^{2}} ).\na. the function has one horizontal asymptote, ( y = 0 ).\n(type an equation.)\nb. the function has two horizontal asymptotes. the top asymptote is and the bottom asymptote is\n(type equations.)\nc. there are no horizontal asymptotes.\nfind any vertical asymptotes of ( f(x) ). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. the function has one vertical asymptote,\n(type an equation.)\nb. the function has two vertical asymptotes. the leftmost asymptote is and the rightmost asymptote is\n(type equations.)\nc. there are no vertical asymptotes.

Answer

Explanation:

Step1: Analyze horizontal asymptotes

For a function (y = f(x)), the horizontal asymptote is found by calculating (\lim_{x\rightarrow\pm\infty}f(x)). For (f(x)=9e^{- 0.5x^{2}}), we know that (\lim_{x\rightarrow\pm\infty}x^{2}=\infty). Then (\lim_{x\rightarrow\pm\infty}-0.5x^{2}=-\infty). Since (\lim_{u\rightarrow-\infty}e^{u}=0) (let (u = - 0.5x^{2})), (\lim_{x\rightarrow\pm\infty}9e^{-0.5x^{2}}=0). So the function has one horizontal asymptote (y = 0).

Step2: Analyze vertical asymptotes

A vertical asymptote occurs at (x = a) if (\lim_{x\rightarrow a^{-}}f(x)=\pm\infty) or (\lim_{x\rightarrow a^{+}}f(x)=\pm\infty). The function (f(x)=9e^{-0.5x^{2}}) is defined for all real - valued (x) (because the exponential function (y = e^{u}) is defined for all (u\in R) and (u=-0.5x^{2}) is defined for all (x\in R)). So (\lim_{x\rightarrow c}9e^{-0.5x^{2}}=9e^{-0.5c^{2}}), which is a finite value for any (c\in R).

Answer:

For horizontal asymptotes: A. The function has one horizontal asymptote, (y = 0). For vertical asymptotes: C. There are no vertical asymptotes.