I am having some confusion about holes in rational functions. As I'm aware, a hole is where both the

vasorasy8

vasorasy8

Answered question

2022-07-16

I am having some confusion about holes in rational functions. As I'm aware, a hole is where both the numerator and denominator become zero due to some discontinuity. For example,
f ( x ) = ( x + 1 ) ( x 1 ) ( x + 1 )
would have a hole at x = 1.
What is the point of distinguishing between a hole and Vertical Asymptote?
What is the logic behind the numerator needing to having a discontinuity at the same point as the denominator?

Answer & Explanation

Marisol Morton

Marisol Morton

Beginner2022-07-17Added 13 answers

Both the numerator and the denominator being zero is a necessary but not sufficient condition for a hole; see for example the function f ( x ) = x + 1 ( x + 1 ) 2 . The difference between a hole and a vertical asymptote is that the function doesn't become infinite at a hole. See it for yourself: Take your f, evaluate it at points close to x = 1 and compare to what happens with mine.
dikcijom2k

dikcijom2k

Beginner2022-07-18Added 6 answers

Plot the functions x 2 1 x 1 and 1 x 1 . You can easily see the difference between a hole and vertical asymptote. A rational function with a hole means it looks very nearly to be a polynomial except that at one (or more points), it is undefined (recall 0 0 isn't defined). At a vertical asymptote, the function blows up because the denominator approaches zero at some point but the numerator does not (or if it does, not to the same "degree" as the denominator; see Javier's example).

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