How do you find holes in rational functions?

The term "hole" used here is another name for a removable discontinuity or removable singularity. A rational function  f(x)=\frac{p(x)}{q(x)}  is the quotient of two polynomials.  Of course, when the denominator  q(x)  is 0, then  f(x)  is not defined.  That is to say, if  q(a)=0,  the number  a  is not in the domain of  f(x). However, if you find that (x-a)  is a factor of both the numerator and denominator, and it occurs in the numerator with as great a multiplicity as it does in the denominator, then you can cancel it.   The denominator will no longer have the value 0 at  a, and  f(x)  will become defined at  a.  You've filled in the hole. So, to find and fill in the holes, factor the numerator and denominator, and cancel common factors.

Let p(x) = \frac{f(x)}{g(x)} . You can factor f(x) and g(x) to see the hole. Example : p(x) = \frac{x^2-1}{x^2+4x+3} p(x) = \frac{\boxed{(x+1)}(x-1)}{\boxed{(x+1)}(x+3)} p(x) = \frac{x-1}{x+3} The function has a hole at (x+1) . The vertical asymptote of p(x) is x=-3 because \lim_{x\to\infty} p(x) = \infty or = -\infty . How to find other asymptotes? Maybe, you can read this @https://en.wikipedia.org/wiki/Asymptote, I'm sorry for my bad answer. Thanks!

Well a rational function is in the form of f(x) = p(x)/q(x). It is some expression over another, right. A hole is a special case when q(x) = 0. It becomes this point where our overall f(x) cannot be determined. It is undefined. A hole occurs when the bottom part, the quotient, is equal to 0. To find one identify where a division, or equivalently, a multiplication of an inverse (something to the power of -1), is equal to 0.

One thing you know about a hole is that the function is undefined for that value of x, and one thing about rational functions is that they are undefined when their denominator's value is 0. Put together, those facts mean that any hole in a rational function must be at a zero of the denominator, so you can make a list of the x-values that are eligible to be holes.Once you have that list, all you need to do is eliminate the values of x where the function increases (or decreases) unbounded. What remains, those where the function is undefined but approaches a finite value (i.e. has a "limit"), are holes. Depending on the strictness of your teacher, you may be able to get away with using a graphing calculator to size this up visually, or you may need other strategies such as factoring to find a usually-equivalent function (in which case you should still graph it if you're allowed a calculator, to check your work).