Answer:
Step-by-step explanation:
Given function is,
[tex]r(x)=\frac{3x^3-7}{x^2-6x+9}[/tex]
For x = 0, substitute the value of x in the given function.
[tex]r(0)=\frac{3(0)^3-7}{(0)^2-6(0)+9}[/tex]
[tex]r(0)=\frac{-7}{9}[/tex]
For r = 3,
[tex]r(3)=\frac{3(3)^3-7}{(3)^2-6(3)+9}[/tex]
[tex]r(3)=\frac{81-7}{9-18+9}[/tex]
[tex]=\frac{74}{(9-18+9)}[/tex]
[tex]=\frac{74}{0}[/tex]
Function is undefined at x = 3.
For x = -3,
[tex]r(-3)=\frac{3(-3)^3-7}{(-3)^2-6(-3)+9}[/tex]
[tex]=\frac{-81-7}{9+18+9}[/tex]
[tex]=\frac{-88}{36}[/tex]
[tex]=-\frac{22}{9}[/tex]
