Answer:
The function f(x) has a hole at point x = - 5.
Step-by-step explanation:
We are given a function as
[tex]f(x) = \frac{x^{2} + 7x + 10}{x^{2} + 9x + 20}[/tex]
Now, we have to describe where the function has a hole.
Now, [tex]f(x) = \frac{x^{2} + 7x + 10}{x^{2} + 9x + 20}[/tex]
⇒ [tex]f(x) = \frac{(x + 5)(x + 2)}{(x + 4)(x + 5)}[/tex]
Therefore, the common factor of the numerator and denominator is (x + 5).
Now, (x + 5) = 0
⇒ x = - 5
Therefore, the function f(x) has a hole at point x = - 5. (Answer)
