Answer:
[tex]f(x)=\dfrac{x+2}{x^3-5x^2-24x}[/tex] has denominator [tex]x^3-5x^2-24x[/tex]
Option 1 is correct.
Step-by-step explanation:
Given: The excluded values of a rational expression are –3, 0, and 8
The excluded value of any function is vertical asymptotes.
We get vertical asymptotes when denominator becomes 0.
If we find a function of excluded value it must we equivalent to denominator of function.
Function of excluded value, (x+3)(x-0)(x-8)
Now we will simplify it
[tex]\Rightarrow x(x^2-5x-24)[/tex]
[tex]\Rightarrow x^3-5x^2-24x[/tex]
Now, we will check each option those has denominator [tex]\Rightarrow x^3-5x^2-24x[/tex]
Hence, Option 1 [tex]f(x)=\dfrac{x+2}{x^3-5x^2-24x}[/tex] has denominator [tex]x^3-5x^2-24x[/tex]
