Given the function
[tex]f(x)=\frac{x}{x^2-9}[/tex]Notice that
[tex]x^2-9=(x-3)(x+3)[/tex]Therefore,
[tex]\Rightarrow f(x)=\frac{x}{(x-3)(x+3)}[/tex]Notice that, if the denominator of f(x) is zero,
[tex]\begin{gathered} (x-3)(x+3)=0 \\ \Rightarrow x=-3,3 \end{gathered}[/tex]Therefore, the two vertical asymptotes are x=-3 and x=3.
The domain of the function includes all the real numbers except those excluded by the vertical asymptotes; thus, (the domain is shown below)
[tex]\text{domain}=\mleft\lbrace x\in\R|x\ne-3,x\ne3\mright\rbrace=(-\infty,-3)\cup(-3,3)\cup(3,\infty)[/tex]Finally, notice that the degree of the numerator is 1 while the degree of the denominator is 2. Since 2>1, the degree of the denominator is greater than that of the numerator. The horizontal asymptote is y=0.
