Question

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

Answer 1

Step 1: Write out the definition of the function f:

The function f is given by:

`f(x)=(x^2-9)/(x)`

Step 2: Calculate the value of the function f at t=-5:

`\begin{gathered} f(-5)=((-5)^2-9)/((-5)) \\ \text{Hence,} \\ f(-5)=(25-9)/(-5)=-(16)/(5) \end{gathered}`

Step 3: Calculate the value of the function f at t=-2:

`\begin{gathered} f(-2)=((-2)^2-9)/((-2)) \\ \text{Hence,} \\ f(-2)=(4-9)/(-2)=(-5)/(-2)=(5)/(2) \end{gathered}`

Step 4: Calculate the value of the function f at t=2:

`\begin{gathered} f(2)=((2)^2-9)/((2)) \\ \text{Hence,} \\ f(2)=(4-9)/(2)=-(5)/(2) \end{gathered}`

Step 5: Calculate the value of the function f at t=5:

`\begin{gathered} f(5)=((5)^2-9)/((5)) \\ \text{Hence,} \\ f(5)=(25-9)/(5)=(16)/(5) \end{gathered}`

Step 6: Calculate the value of the function f at t=6:

`\begin{gathered} f(6)=((6)^2-9)/((6)) \\ \text{Hence,} \\ f(6)=(36-9)/(6)=(27)/(6)=(9)/(2) \end{gathered}`

Hence