Plot points between and beyond the Exitir serves in the vertical asymptote evaluate the function at -5, -2,2,5 and 6 Simplify

Step 1: Write out the definition of the function f:
The function f is given by:
[tex]f(x)=\frac{x^2-9}{x}[/tex]Step 2: Calculate the value of the function f at t=-5:
[tex]\begin{gathered} f(-5)=\frac{(-5)^2-9}{(-5)} \\ \text{Hence,} \\ f(-5)=\frac{25-9}{-5}=-\frac{16}{5} \end{gathered}[/tex]Step 3: Calculate the value of the function f at t=-2:
[tex]\begin{gathered} f(-2)=\frac{(-2)^2-9}{(-2)} \\ \text{Hence,} \\ f(-2)=\frac{4-9}{-2}=\frac{-5}{-2}=\frac{5}{2} \end{gathered}[/tex]Step 4: Calculate the value of the function f at t=2:
[tex]\begin{gathered} f(2)=\frac{(2)^2-9}{(2)} \\ \text{Hence,} \\ f(2)=\frac{4-9}{2}=-\frac{5}{2} \end{gathered}[/tex]Step 5: Calculate the value of the function f at t=5:
[tex]\begin{gathered} f(5)=\frac{(5)^2-9}{(5)} \\ \text{Hence,} \\ f(5)=\frac{25-9}{5}=\frac{16}{5} \end{gathered}[/tex]Step 6: Calculate the value of the function f at t=6:
[tex]\begin{gathered} f(6)=\frac{(6)^2-9}{(6)} \\ \text{Hence,} \\ f(6)=\frac{36-9}{6}=\frac{27}{6}=\frac{9}{2} \end{gathered}[/tex]Hence