 In this video, I want to graph the rational function r of x equals 2x squared minus 5x plus 2 in the numerator and x minus or x square minus 4 in the denominator and I want to graph this without using any technology whatsoever just based upon information of how the function behaves near its asymptotes and how it behaves near its intercepts. So let's start figuring out what those asymptotes, what those intercepts are. To do this, I would begin by factoring the numerator and the denominator. The numerator is a little bit hard. Let's start with the denominator. x squared minus 4, that's the difference of squares. It should factor with x minus 2 and x plus 2. The numerator is a little bit more challenging, but we can reverse foil this thing. We take 2 times 2, that gives us a 4. We need factors of 4 that add up to be negative 5. So we can take negative 4 and negative 1. We then proceed to factor this thing by groups using this magic pair right here. 2x squared minus 4x minus x plus 2. You have the first group and the second group. The first group, you can take out a 2x leaving behind an x minus 2. And with the second group, take out a negative 1 that leaves behind x minus 2 right there. And so as x minus 2 shows up twice, we can factor it out. And we end up with x minus 2 times 2x minus 1 times x minus 2. So we get the factorization there. Now taking this fraction as it is, look at the denominator. What makes the denominator go to 0? We get x is 2 and x is negative 2. That will be the forbidden values from the domain. We're going to come back to this issue in just a second. But we have these two numbers that are forbidden from the domain. Once that's decided, now look to see if there's any common factors on the numerator and denominator. You notice like, oh, there's an x minus 2 on top and bomb that cancels out. And so this rational function can simplify to be 2x minus 1 over x plus 2. So what I'm going to basically do is for the rest of the problem, I'm going to graph the function. I'm going to graph the function y equals 2x minus 1 over x plus 2. But there's going to be a removed point. A removed point at x equals 2, right? Because the x minus 2 is what canceled out. There's a removed point because those have disappeared from the denominator. So we're going to graph the function y equals 2x minus 1 over x plus 2. But we have this removed point x equals 2. We'll come back to that in a second. So how do we determine the y-intercept? We can plug in x equals 0 into this function, or we can plug it into the simpler function. If you plug it into the simpler function here, you're going to get negative 1 over 2, which is our y-intercept that we see right here. The x-intercept is what's going to make the numerator go to 0. So if you take 2x minus 1 instead of equal to 0, you get x equals 1 half, like so. And so that's where our x-intercept comes from right here. Now notice, if you look at the original fraction, you might be tempted to say that x equals 2 is also an x-intercept. But because plugging in x equals 0 gives you 0 over 0, that doesn't tell you you have an x-intercept necessarily. That's actually coming from the removed point. That's why it's better to use the reduced fraction in terms of this consideration. For symmetry, if there won't be any symmetry, it's not symmetric with the origin or with the y-axis. If you look at the denominator of what makes it go to 0 in the lowest terms, that's going to be x equals negative 2. This tells us that we have a vertical asymptote at negative 2. Because negative 2 is outside the domain of the function, but when we put it into lowest terms, it still didn't cancel out. That suggests that we have a vertical asymptote at negative 2. The multiplicity is going to be odd, because it's just a positive 1 right there. This tells us that we are going to cross infinity at negative 2. We're also going to cross the x-axis at 1 half. Those are important things to realize. Well, what do we do about this removed point at x equals 2? If it's a removed point, there needs to be two coordinates, an x-coordinate and a y-coordinate. How do we find the y-coordinate? Well, the y-coordinate is you're going to take the removed point from the domain, so the x equals 2, and you're going to plug it into the reduced fraction. What happens when you plug in 2 here? Because if you plug in 2 into the original thing, you're going to get 0 over 0, which does not exist. But if you plug in 0 into this function right here, you're actually going to, not 0, sorry, if you plug in 2 into this function right here, you're going to end up with 4 minus 1, which is 3, above 2 plus 2, which is 4. So you actually can plug that in. You don't get 0 over 0. You actually get 3 over 4. And so that suggests to us what this removed point is going to be. You get 2 comma 3 fourths, like so. So I'll show you how to graph that removed point in just a second. The last thing I want to mention here is, what is the asymptotic behavior of this function? If we take 2x minus 1 over x plus 2, if we look at just, as x approaches plus or minus infinity, if we look at just the leading terms, 2x over x, we get 2x over x, which simplifies just to be 2. And so we have a horizontal asymptote at x equals 2. Is it possible for the function to cross the horizontal asymptote? We can investigate that very thing. We're going to come over here and do this. Can our function equal its horizontal asymptote? Can 2x minus 1 over x plus 2 equal 2? Again, it's better to use the reduced reaction because with the exception of the removed point, the two functions are identical, but linear polynomials are a lot easier to do than quadratics here. If we want to solve the equation r of x equals its horizontal asymptote 2, we take 2x minus 1 over x plus 2 and solve it. So clear denominators, we're going to get 2x minus 1 equals 2 times x plus 2. The right-hand side becomes 2x plus 4 and then we have 2x minus 1 right here. Taking away the 2x from both sides, we end up with negative 1 equals 4, which is actually a contradiction. This tells us that there is no crossing of the horizontal asymptote. So now we're in a position that we can start graphing this thing. I'm going to do this to the right, first of all, and then we're going to compare our picture in just a second. So let's first graph the x-axis and we'll put the y-axis right here. So what do we know about this function? So we found an x-intercept at one-half, which we'll just squeeze that in right here. Our x-intercept was at one-half. Our y-intercept, which we found was at three-fourths, that's almost one. Sorry, it was a negative. As soon as we know it was a negative one-half, that's what it was. Three-fourths had to do with the remove point. So the y-intercept was negative one-half. The x-intercept was positive one-half. Did we have any vertical asymptotes? Yes, we had a vertical asymptote at negative two. So I'm going to put my vertical asymptote right here. My picture up just a little bit. So I have this vertical asymptote at x equals negative two. We also had a horizontal asymptote, remember, of y equals positive two. So I'm going to add that to the graph as well. Now, in general, it's possible for a function across its horizontal asymptote, but we've proved already by solving a certain equation that that's not possible in this situation. y equals two is our horizontal asymptote. The other thing I want to put on the graph is this remove point. So the point was removed, right? So when x equals two, y should equal three-fourths. So we have this remove point right here. I'm going to draw this as an empty dot to suggest the point was removed. And so from this picture, we have a pretty good idea of how this function is going to behave. So kind of connecting the dots, it seems like the only natural function would be to connect the dots in this way right here. We're playing connect the dots just like we learned in kindergarten, right? So going from our x-intercept, looking at the x-intercept, we have to cross the x-axis because the multiplicity was odd. So we see something like this. We have to go up towards the horizontal asymptote and towards the remove point because if we went down, we'd have to somewhere turn around and we get an x-intercept we don't have. So we have to be going this direction through that remove point. We cross to go the other direction through the y-intercept, but then at that y-intercept, we could turn around and go up or we continue down. We can't go up because we'd have to cross the x-axis and the horizontal asymptote, which we're not allowed. So we get something like this. I should also mention that we know our function can't bend and come back up like this because we can't cross our horizontal asymptote. So we get using the information we have about the multiplicities and the intercepts and the remove point, we have to do something like this. When we hit x equals negative two, we're going to cross infinity and come from the other side. Thus, we're going to have to get a picture that looks like this. We're going to wrap our spaceship wraps around and comes through our black hole on the other side. So it crosses infinity. And then again, we can't cross our horizontal asymptote, but we have to get close to the horizontal asymptote. So it's going to have to approach above like we see right here. So let's zoom out and look at the final picture, which we see right here. It's like, voila. Although I use red instead of green for my asymptotes. That's the exact picture that we produce right here. And so we're able to graph this rational function without any technology whatsoever.