 In this video, we provide the solution to question number 20 from the practice final exam from math 1050. We have to graph the rational function f of x equals x squared times x minus five over x plus three times x minus two. We have to also make sure we label all the intercepts and acetotes with their multiplicities here. So let's start with that. Define the y-intercept, right? That's one of the easier things to find out here. The y-intercept is going to be discovered by looking at f of zero. We just plug in zero for all the x's. So you're going to get zero times negative five over three times negative two. That simplifies just to be zero. So the x-intercept is going to be zero. The y-intercept, excuse me, it also is an x-intercept. And so I'm going to label this on the graph. We have a zero zero right here. All right, let's look for some x-intercepts now. The x-intercepts are going to come from setting the numerator equal to zero. They're going to be the roots of the polynomial in the numerator, which is already factored for us. So when you look at the x-squared, that goes to zero and x equals zero. So you do the x-intercept of zero, like we already said. But you'll also get an x-intercept at five. So we have two x-intercepts, zero and five, one, two, three, four, five. I'm going to put this here on the graph as well, five comma zero. It does also say we need to mention the multiplicities of the x-intercepts. How often do they show up? So x-squared shows up twice. So that's an even multiplicity. And then x minus five shows up once. That's an odd multiplicity. So at x equals zero, we're going to touch the x-axis, but at five, we're going to cross the x-axis. That distinction will be helpful for us. Let's next consider the asymptotes. Vertical asymptotes are usually pretty easy to see. You look at the denominator in that situation. So if we have vertical asymptotes at x equals, well, looking at the first factor, that will be negative three. And we have a vertical asymptote at positive two as well. And so I'm going to add these to the graph as well. So at negative three, we have this asymptote. Just do your best to draw a straight line there. Typically put it as a dashed line. So we get x equals negative three. We also have one at positive two. Excuse me. So x equals two. And two like right there. x equals two. It also mentioned, you have to mention the multiplicities of the vertical asymptotes. So how often do they show up? x plus three shows up once. x minus two shows up once. Those are the multiplicities, one in one. That means we will cross infinity at both locations since they have odd multiplicities. We have to also mention, we should mention the in behavior, right? Does the graph have a horizontal asymptote, right? So if we look at the in behavior for our function here, right, f of x is going to be approximately, just looking at the leading terms, x squared times x over x times x. It's going to be approximately x as x approaches plus or minus infinity. So actually, if you get an x here, this is a situation where we actually have an oblique asymptote. So we actually can do better than x. We need to actually do some long division here. So let's back up a little bit and take a look at that. For the long division, it might be a little bit easier to multiply things out. This one's not too heinous, so don't worry too much about it. So here, if we distribute the x squared, we end up with x cubed minus five x. In the denominator, if you foil it out, you'll end up with x squared plus x minus six, like so. So how many times does x squared go into x? That happens x times. We observed that already. Times the divisor by x, we end up with x cubed plus x squared minus six x, like so. The x cubes will cancel out. Then we get zero x squared minus x squared. That's going to give us a negative x squared, like so. We could do the rest of it, but we're just looking for the oblique asymptote. We only need the quotient. We don't need the remainder, so actually don't need to do that calculation. I mean, if you're dying to know, of course, it's x, but we don't even need it because we want to know next how many times does x squared divided into negative x squared. That's going to happen exactly negative one times, for which again, we don't need the rest of the calculation, not necessary whatsoever. We will have an oblique asymptote at y equals one right here, right? So an oblique asymptote. So let's put this onto the graph. So it's an oblique asymptote, so it'll have its y intercept, be negative one. There's no y scale, so just do the best you can. So you get something like the following. It's y intercept would be negative one. It has a slope of one. So something like this, label this as, I'll put it over here, y equals x minus one. So with this information, we're now ready to start drawing our graph. So let's start graphing this thing. Let's go to the y intercept, which is also x intercept. So going from there, it's like, what are my options? I can go off towards infinity, or I can go off towards negative infinity, right? Just because there's an oblique asymptote there doesn't mean we don't cross it, right? We don't actually know that information. We can try to solve f of x equals to x minus one. That will be a little bit of a challenge polynomial equation. So it'll be a little bit of overkill. So what I'm going to do instead is just actually use a test point. What if we try like x equals one? What happens right here? Not sure yet. We'll try it out. So upon doing so, if we do f of one, again, just trying to pick an easy value, you're going to end up with one times negative four on top. And the denominator, you're going to end up with four times negative one in the denominator. This would simplify just to be one when we're done. And so actually we're going to get a point like this. Use that as a test point. So that tells us we have to curve off towards positive infinity there, right? Because there's no x intercepts after x equals zero before the asymptote. So if it goes up once, it has to go up a whole way. Now at x equals zero, since we touch the x-axis, it has to bend off this way, like so. You cross infinity at negative three, so it's going to have to wrap around and then go off towards its oblique asymptote like so. And then same thing at two. We cross infinity, right? And so then in this situation, we see that we do have to cross. We're going to have to cross our oblique asymptote to get to that x-intercept, which we crossed through on the other side. But then we're going to bend towards the asymptote after that. So it turns out there is at least one point where you do have to cross the oblique asymptote. So if you don't over here, do you over here, you can investigate that one and find out that you don't. But this is a sufficient graph for this one. Pretty advanced question here. But it has all the pieces we need to give us the graph and get the full credit here.