 In this video, I wanna show us how we can graph a rational function without any technology whatsoever. We're just gonna use information about the asymptotes and about the intercepts and how the function behaves near those intercepts and asymptotes to graph this thing. Consider the rational function x to the fourth plus one over x squared. If at all possible, I would suggest we factor this, the numerator and denominator. The denominator is already factored because it's x squared plus, or it's just x squared. The numerator, we also really can't factor x to the fourth plus one. If we factor, then we have to use a difference of squares, which because there's a plus right there, there's really no sum of squares factorization with real numbers. We can factor with imaginary numbers, but we won't see imaginary numbers on our graph, so we only focus on real numbers. So the domain of this function is gonna be everything except for zero, because the denominator being zero, x squared equals zero, that implies that x equals zero. So that forbids x equals zero being in the domain of the function. Now, notice this function's already in lowest terms, and so when your function's in lowest terms, if there's something that makes the denominator go to zero, that means it's a vertical asymptote. So this problem with the domain x equals zero suggests there's a vertical asymptote at x equals zero, so the y-axis will be a vertical asymptote. Now, if there's a vertical asymptote at x equals zero, in other words, if x equals zero is outside the domain, that means there's no y-intercept. The y-intercept would be r of zero here, which that would look like one over zero, which does not exist. This function doesn't have any y-intercept, great. But what about x-intercepts? When you take x to the fourth plus one, like I mentioned earlier, the x-intercepts are what will make the top of the fraction go to zero, but you get x to the fourth equals negative one, which would actually say that x equals plus or minus the fourth root of negative one, which is actually plus or minus the square root of i, for which I mean, I can tell you what that is. I mean, that's gonna be, we don't need to worry about all that. I mean, we get these, we're gonna get some, I mean, there's rx-intercepts per se. You get like one over the square root of two, plus or minus, plus or minus one over the square root of two, i, but the thing is these are non-real numbers. These are non-real numbers, complex numbers, yes. But when we're graphing, it needs to be a real number. And so as such, we don't have any x-intercepts. The numerator is always positive. That's important thing there to notice here. So coming back to x squared plus one, it can never equal zero. It's actually strictly greater than zero. It's always positive. So we have a function with no y-intercepts or x-intercepts. That might seem depressing, but that actually is very useful information. We know that there's a vertical asymptote at the y-axis, but we also know that the function never, never crosses the x-axis because there's no x-intercepts there whatsoever. Another thing to consider is that if you wanna do symmetry, notice if you plug in negative x here, you're gonna get negative x to the fourth plus one. The even power will eat the negative, but also the even power of the bottom eats the negative. All the negatives disappear. We end up with the original function r of x. That suggests our function has even symmetry. That is, it's symmetric with respect to the y-axis. We can utilize that at the very end. All this check to see if we have, if we have the correct symmetry when we're done. In terms of the end behavior, this function, you'll notice that the top is x to the fourth, the bottom is x squared. So as x goes to infinity, this will be approximately the same thing as x to the fourth over x squared, that is x squared. And so asymptotically, this will be a parabola. We can do a little bit better than that because after all, since the denominator is a monomial, this looks like x to the fourth over x squared plus one over x squared. So this actually gives us x squared plus one over x squared. And so as x approaches plus or minus infinity, we see that this part right here will vanish towards zero. And so this graph has a parabolic parabola. And so essentially it's gonna be a parabola going forward from there. And so we do have an x-intercept, we don't have an x-intercept, excuse me, we do have a vertical asymptote. It came from the x squared of the denominator, but the power there is two. So the vertical asymptote has an even multiplicity. This tells us that at the vertical asymptote, we're gonna touch infinity. We won't cross, we'll actually touch it. And so I wanna show you, without actually looking at the picture yet, we'll compare that in just a second, what type of picture can we construct here? Okay? So let's graph, let's, let me emphasize where the x-axis and the y-axis is gonna be. So what we know about this graph already is that this function has a vertical asymptote at the y-axis. This is outside the domain of our function. We also have this parabolic asymptote at y equals x squared. So if you think about that, so y equals x squared goes from zero, zero, one, one, it will go through two, four, something like that. So our parabola roughly will look like this, right? And then we go through the other side. Let me fix that. One, two, three, four. Trying to draw this reasonably to scale. So we get this like parabolic parabola, y equals x squared, y equals x, like so. And so these are things, well, the vertical line is something we have to avoid. Does our function cross its oblique asymptote? Well, if there's a crossing there, do we cross, and I shouldn't say oblique asymptote. Does it, is there a crossing of y equals x squared? Well, that means we take the function r of x and set it equal to x squared and we solve that. So we take x to the fourth plus one over x squared. Can this ever equal x squared? We don't know. Clear the denominator in times both sides of x squared. We get x to the fourth plus one is equal to x squared times x squared, that's x to the fourth. Notice the x to the fourth is canceling at one equals zero, which is false. So this tells us that in fact, there will be no crossings of this oblique asymptote. So our function's either gonna, if we're looking at like what's beyond zero, right? We're either gonna be in this sector right here or we're gonna be in this sector right here. Now because we don't have any x intercepts or y intercepts, we don't have any points. We gotta get started somewhere. We need an initial point, a seed, to grow this picture from. So I'm gonna actually consider what happens at r of one. I need a point to get started here. And so if I take r of one, one seems like a good point to use if I can't use the y intercept. You're gonna get one to the fourth plus one over one squared that ends up being a two. So you get the point one comma two, which we can put on the graph. That would be right here, one comma two. By symmetry, I also know that since this graph is reflective across the y axis, I'm also gonna get a point right here negative one comma two, I also computed that. So what's gonna happen if we start at one, two and we move to the right, we have to get close to this parabola. So we either gonna approach it from like above or we're gonna approach this parabola from below. But we can't cross the parabola. Remember, that's what we determined a moment ago. So it turns out it's gotta be this first picture. As we go off towards infinity, our function's gonna look like this parabola. Well, what happens moving from this point to the right? Oh, or to the left, excuse me. Do we approach, we have to approach the asymptote. Do we go up towards infinity or do we go down towards negative infinity? Well, if you went down towards negative infinity, there's two problems. You would need an x-intercept which we don't have. You'd have to also cross the parabola which we can't do. So that second option's not possible. It has to be the first option that our function will approach, our function's gonna approach infinity as x approaches zero. So we could say something like as x approaches zero from the right, y will approach infinity, positive infinity. Now, like we discovered before, that as we get close to our vertical asymptote, we're gonna touch infinity. So that means we won't cross to the other side, like over here. We actually have to come back from the side we came from, like so. And then at this point, one comma two, well, we have to go towards the asymptote but we can't cross this. So we're gonna have to do the mirror image, which remember our function's supposed to be symmetric with respect to the y-axis. And so this is my picture, which I claim is in play right here. Let's look at, here's a skeleton that you could construct using the information from above, but if we compare this to the computer-generated image for which we can see that one right here, you can see that, yeah, that's pretty good. That's the picture. I didn't quite extend mine as far up as the computer did, but we see that these two pictures are in fact one and the same thing, that we were able to very accurately graph this function right here.