Answer:
Option C
Explanation:
In the function below:
[tex]f(x)=\frac{(x-4)(x-5)}{(x-4)(x-2)(x-6)}[/tex]
Vertical Asymptote
Canceling the common term reduces the fraction to:
[tex]\begin{gathered} \frac{(x-5)}{(x-2)(x-6)} \\ \implies\text{Vertical Asymptote at x=2 and x=6} \end{gathered}[/tex]
Hole
The common term is x-4, therefore, there is a hole at x=4.
Zero
If the reduced fraction is set equal to 0.
[tex]\frac{(x-5)}{(x-2)(x-6)}=0\implies x-5=0\implies x=5[/tex]
The zero is at x=5.
Horizontal asymptote
Since the degree of the numerator is less than degree of denominator, the horizontal asymptote is at y = 0.
The function that fits the criteria is Option C.
