Hey, everyone. Whenever we found vertical asymptotes, we would factor our function and then cancel out any common factors in order to get our function in lowest terms before setting our denominator equal to 0 and solving for x. But having that common factor actually causes something to happen in the graph of my function called a hole. And we need to be able to identify where these holes are on our graph, which we can do by simply taking our common factor and, instead of canceling it out, just setting it equal to 0 and solving for x. So here I'm going to walk you through how to find any holes in your graph starting from exactly where we do with vertical asymptotes and just taking a different path from there. So let's go ahead and get started.
Now with vertical asymptotes, like I said, we would cancel out our common factor, and then we would set our denominator equal to 0 and solve for x. So for this function here, I end up with a vertical asymptote at x equals 3, which we can see right here on our graph, a vertical asymptote at x equals 3. Now, whenever we find the holes in our graph, we're going to start at the same place. So here I have this factored function, but instead of canceling out that common factor to put it in lowest terms, I'm going to take that common factor and set it equal to 0.
So here, my common factor is x + 2 and setting that equal to 0, I then want to solve for x. So solving for x here, I end up with x equals negative 2. And that tells me that I have a hole in the graph of my function right at x equals negative 2, which I'm going to represent with an open circle on my graph. Literally, just draw a circle on the curve at that point. And that's where our hole is. So looking at this graph, now I have both a vertical asymptote and a hole for the same function. Now you might also hear holes referred to as removable discontinuities. And that's just the more technical term for it, but it's referring to the same thing.
So now that we know how to find the holes in the graph of a rational function, let's go ahead and look at some more examples here. So let's look at this first function. I have \( f(x) = \frac{x + 3}{x^2 + 4x + 3} \). So here, we want to go ahead and factor this the same way we would if we were trying to write it in lowest terms. So my numerator can't be factored here; it just stays as x + 3, but my denominator using the a c method can be factored to (x + 3)(x + 1). Now I want to take my common factor, and I don't want to cancel it. I want to take it, and I want to set it equal to 0. So here my common factor is x + 3. Setting that equal to 0, I'm going to go ahead and solve for x, which I can do by subtracting 3 on both sides. So I end up with a hole at x equals negative 3, and I'm done. That's where the hole on my graph will be.
Let's look at one more example here. On this side, I have \( f(x) = \frac{x^2 + 1}{x - 1} \). So let's go ahead and factor this. So my numerator can be factored into (x + 1)(x - 1) and I'm going to divide that still by x - 1 because that denominator can't be factored anymore. Now, again, I want to take my common factor, and I don't want to cancel it. I'm going to take and set it equal to 0. So here, x - 1 equals 0. And then we want to solve for x. So adding 1 to both sides leaves me with x is equal to 1, which is where the hole on my graph is at x equals 1. And we're done.
So now that we know how to find the holes in our graph, let's get a little bit more practice.