Well class, welcome to episode three of Math 1050, College Algebra. I'm Dennis Allison in the Mathematics Department at Utah Valley State College. Today, we'll be talking about fundamental graphs that you'll need to be aware of, and we'll be using quite a bit this semester. There are eight fundamental graphs that we'll talk about today, and there are several more that we'll see later in the course. Then we'll also talk about graphs of piecewise functions. And then finally, we'll talk about symmetry and graphs, symmetry about the y-axis, about the x-axis, and about the origin. So let's begin by looking at the first fundamental graph. OK, let me list the eight fundamental functions, then we'll look at their graphs. First of all, there's the function f of x equals just a constant, so I'll write constant here. And then there's f of x equals x. That's going to be a straight line graph. I think we pretty much know how to graph that, but we'll look at it in a moment. Then there's f of x equals the absolute value of x. Let's call the absolute value function. Then we have f of x equals x squared. And f of x equals x cubed. And f of x equals the square root of x. That's called the square root function. f of x equals the cube root of x. And then finally, there's a new function you may not have seen before. I'm going to write it this way. This is called the greatest integer function, but we'll get to that one at the very end. Now, what would the graph of each one of these look like? Well, let's begin with f of x equals a constant. So for example, what if I wanted to graph f of x equals 2? For example, this is the same thing as saying y equals 2. And that's the same thing as saying y equals 0x plus 2. And so you notice it's beginning to look like it's in the form y equals mx plus b. So class, what's the slope of that linear function? 0. Slope is 0, OK. The slope is 0. And what's the y-intercept? 2. 2, OK. So when I go to graph it, what I'll do is just go up to 2 on the y-axis. And I'll draw a line with slope 2. So I get a horizontal line. So to kind of summarize this, basically, when you have f of x equals a constant, it's going to be a horizontal graph. If I wanted to graph, say, g of x equals negative 1, I could quickly graph it by just going down to negative 1 and drawing a line right there. So that's the graph of g. And over here, this is the graph of f. OK, so constant functions are fairly easy to graph. Let's go to f of x equals x. Now, the function f of x equals x is called the identity function. Because basically, what you substitute in for x is what you get back out. So there's really no change that occurs to x. So it keeps its own identity. So if I want to graph f of x equals x, that's the same thing. We'll see f of x and y are interchangeable. So that's the same thing as y equals x. We recognize this as a straight line graph. Susan, what's the slope of that line? 1. The slope is 1. And what's the y-intercept of that line? David, what's the slope? I'm sorry, what's the y-intercept of that line? The y-intercept? No, no, not the y-intercept. Anybody? Zero. Zero, right. Because see, there's a zero attached right there that I haven't. I haven't shown because it's not necessary to show a zero. So the y-intercept is zero. And if I graph the identity function, then it's going to cross the y-axis at zero. In other words, it passes through the origin. And the slope is 1, or 1 over 1. So if I go over and up 1, I get another point there. And I get a line like that. I might just point out just for later reference that if I had gone back 1, I would have gone down 1. And it goes through this point right here, too. That has some relevance to the later discussion. So this diagonal line goes to the points 1, 1, 0, 0, and negative 1, negative 1. OK, so that one's not a completely new function for us. But let's go now to the absolute value function. So the function f of x equals the absolute value of x. This one is probably new to us. So since I've never graphed this one before in here, I'm going to make a table of values. And we'll plot a few points that are on the graph. So I'll make a table over here for x and for f of x. Now, I'll just pick a few random values for x, positive and negative, and 0 as well. So what if I choose 0, 1, 2, 3? What if I also choose negative 1, negative 2, and negative 3? Then what I want to calculate for the function value is the absolute value of each one of those numbers. So what's the absolute value of 0 going to be? 0. Is 0, OK? And the absolute value of 1 is 1. The absolute value of 2 is 2. The absolute value of 3 is 3. And what are the absolute values going to be here? All positive. All positive, 1, 2, and 3. So if I think of these as ordered pairs, 0, 0, 1, 1, 2, 2, 3, 3, negative 1, 1, negative 2, 2, negative 3, 3. When I go to draw the graph, I'll just plot those points to get an indication of what the graph looks like. And let's see. I'll locate 0, 0 is right here. I'll locate 1, 1 right there. I'll locate 2, 2 right here. 3, 3. OK, I hope those are lined up. And for the negative values, there's negative 1, 1. There's negative 2, 2. There's negative 3, 3.  The points over here look like they're lined up. And as a matter of fact, if you plot more points in there, you'll verify that this graph has a V shape. And it has two branches that come together at a right angle. These actually meet at a right angle. Each one of those is at a 45 degree angle to the horizontal axis. And this is the graph of the absolute value function. Let's just go to the graphic of the absolute value function. OK, so here's a nicer graph of the absolute value function. And you notice there are three points picked out very near the origin. In fact, one of the points that's highlighted is at the origin. There's another point highlighted at 1, 1, and a third point highlighted at negative 1, 1. Let's come back to the board and let me just show you what is significant about that. Let's say in the future, I want to graph the absolute value function. I don't want to make a table of values because that's obviously rather slow to do that. But on the other hand, if I just sort of freehand the absolute value function, here's what students sometimes draw. They say, well, let's see, these things come in, it makes a V. But they don't get a very good V. So what we need are just a few points to help us sort of line up the graph. So what I suggest you do is to plot three points for this function, which I call target points. And the target points are 1, 1, 0, 0, and negative 1, 1. Now, if you plot those three points, and then if you know the shape of the graph, make this V shape, then you just extend these two rays going out. And you have a very nice graph that way, I think. So let's say this is my graph paper, and I want to graph the absolute value function now. So I want to graph f of x equals the absolute value of x. What I'll do is just remember the three target points, 0, 0, 1, 1, negative 1, 1. And then when I draw these two rays, I have a fairly accurate graph. And so that is the graph of f. I'll put a little name tag on it there and write f beside it to mean it's the graph of this function, absolute value of x. So there are three targets for the absolute value function. And I think we're going to find that for most of the other functions here, there'll be some target points to help me graph them. Let's go to the squaring function, f of x equals x squared. Now, once again, I haven't graphed the squaring function in this course. You may have seen it in a previous course, but perhaps not. So because it's a new graph for us, I'll make a table of values, and I'll plot some points. But then I'm going to reduce it to three target points. And in the future, I'll just quickly plot three target points and I'll draw the graph. So the function is f of x equals x squared. And the table that I'll make looks like this. I'll choose some values for x. Like once again, why don't we choose 0, 1, 2, 3. You could choose more values if you want. In fact, you could choose fractions and decimals if you care to. Negative 1, negative 2, negative 3. Now, I need to know the function value for each one of these. Ginny, can you tell us what these function values will be? It'll be 0, 1, 4, 9, 1, 4, and 9. And 9 again. And once again, I'll think of each of those as ordered pairs. For example, here's the origin. Let's take this one down here. Here's the point, 3, 9. So we have seven ordered pairs that we'll plot. And if you really want more accuracy, you can plot more points. You could say x is 1 half, x is 3 halves. But I think this would be enough to get the idea of the general shape. So when I go to graph this, let me make my scale a little bit smaller now so that I can squeeze these points in. Well, there's seven right there. And then negative values down below. So we'll plot 0, 0. We'll plot 1, 1. We'll plot 2, 4. There's 4. No, that's actually 3. So let's go up a little bit higher. There's 4. And there's 3, 9. Well, 3, 9, let's just say it's going to be up about there. And on the other side, at negative 1, 1, at negative 2, 4, and at negative 3, 9, right about there, I think. So this time, it doesn't look like these branches are coming off straight, but they're actually curved this time. And when I draw the graph, I'm going to draw it so it just comes down and goes tangent to the x-axis, and then it turns back up. Now, what's different about that is this is what you call a smooth curve. It's not pointed like the absolute value function. The absolute value function, you could run your finger along that edge down there. You could actually hurt yourself. This should only be done by someone who's really a pro at this, you see. So this is actually a smooth curve. And so you don't want to make it jagged right there, but it turns. And there's a name for this particular shape, by the way. What is it called? Parabola. Parabola, yeah. This particular shape is known as a parabola. It's very useful in science courses and in various applications of mathematics. And for example, one place that you've seen a parabola is if you throw an object into the air. Like if I throw this marker over to you, it follows more or less a parabolic arch as it comes down to you. It's not quite parabolic because there's air resistance. And if we don't take the curvature of the earth into effect, I mean, obviously this isn't very far away.  But if you assume the earth is flat, if you assume there's no air resistance, then generally this would be a parabolic arch, this flying projectile. Let's see, well I don't want to make a table of values again for this ever again. If I can help it, I'd like to speed this up. So the target points I'll choose will be 1, 1, 0, 0, and negative 1, 1. And in the future, when I want to graph the squaring function, when I graph the squaring function, here's what I'll do. I'm going to locate the points 1, 1, 0, 0, and negative 1, 1. So the graph could be drawn more quickly like this. 0, 0, 1, 1, negative 1, 1. Now you might say, wait a minute, those are the same target points as the absolute value function. How are we going to know the difference? Well you have to remember the general shape. And I know that the squaring function is a parabola. So when I draw it, I come down, go through this point, turn, go back up like that, and that's a rough sketch of the graph. Now you might say, well Dennis, that's not maybe as accurate as you would like. Well of course you get more accuracy if you plot more points. And so what we're sacrificing is accuracy for speed. So I want to be able to quickly sketch the curve. I don't care that it's perfectly accurate.  Over here it wasn't very accurate, and I plotted seven points. But what I got was speed. So I can quickly sketch it, and then I can move on to whatever application I have for this. OK. Now I'm not going to make a table for every one of these, because I think you probably see how this procedure goes. Let me just show you what the other functions look like. By the way, we also have a graphic of the squaring function. Maybe we should show that one. There we go. You see there's the parabola graph for f of x equals x squared. And there are three points highlighted near the x-axis at 1, 1, 0, 0, and negative 1, 1. OK, next function is the cubing function. Now all of these functions are so basic in college algebra that we would just want to know their graphs.  You need to know that 12 divided by 4 is 3. Those are sort of basic arithmetic principles. Well, in college algebra, they're basic graphs. And these are the ones that come up most often. So we want to know how to represent them graphically. And we want to be able to sketch them quickly, but relatively accurately. Let's see, this time, let's go straight to the target points for the cubing function. Now as you might guess, the three target points are going to be the points associated with 0, 1, and negative 1. If I plug in a 0, I get 0 for the function value. So this function goes through the origin. And if I plug in 1, what do I get? We get 1 back again. So we get plus 1, so the point is 1, 1. Now if I substitute in negative 1, what do I get? We get negative 1 back. So there's a change here in that we don't get this triangle of target points, but they seem to line up. So what does the cubing function look like? Well, if I made a table, and if I plotted 7, 10, a dozen points, I would find that it comes down sort of like a parabola, only steeper. And it gets flat, flattens out at the origin. But instead of turning up, it turns down. And back in the old days, the old days, meaning back in the 17th, 18th century, these sort of curves were called higher parabolas because it looks sort of like a parabola coming down. But it's not really. It's steeper than a parabola. It's a little bit flatter than a parabola. But these days, we don't generally call them higher parabolas. It's just the cubic function. Who can explain to me why you would expect the curve to turn down over here rather than turning up when it's the cubing function? Stephen. Negative values plugged into the equation are still negative once they come out. Exactly. When you pick an x that's negative and you cube it, you're going to get a negative y. So you're going to get two negative coordinates. So you're in the third quadrant rather than in the second quadrant. For the squaring function, when you picked a negative value and squared it, you got a positive y. So you had a point of the form negative positive, so you're in the second quadrant in that case. If I had gone over to 2 on the cubing function, how high would I go to get onto the graph? Yeah, Stephen? 8. It'd be 8, exactly. But if I had been using the squaring function, how high would I go to get onto the graph? 4. Before. So you see we'll be twice as high at 2 in the cubing function than we are for the squaring function. So that indicates that it turns up faster. On the other hand, at 1 half, how high would I go up to get onto the cubing function? Susan, what would you say? If I chose 1 half right here, how high is that point on the cubing function? Over 3. No, no, not over 3 right there. If you plug in a half, what's 1 half cubed? 1 eighth. Is 1 eighth. So you only go up 1 eighth right there. Whereas if it were the squaring function, I would go up 1 fourth. So 1 eighth is below. So you see what happens is the cubing function ends up being below the squaring function when you're near the origin. Let's look at two graphs that we have on graphics. First of all, we have the cubing function, I believe, on a graphic. There it is. You see, it goes up rather dramatically on each end. It goes to the origin. And it goes to the points 1, 1, and negative 1, negative 1. Now, the next graphic has a comparison of the squaring function and the cubing function. You see, there's the squaring function in black. It doesn't go up nearly as fast. Then the cubic function is dotted. It goes up faster. And if we could zoom in, I don't think we can, but if we were to zoom in, we would see that the dotted graph is actually closer to the x-axis than the squaring function between negative 1 and 1. It actually flattens out a little bit more than the squaring function does. OK, let me just summarize the graphs that we've seen so far. And then we'll go to the last three over here. So just as a point of summary at this moment, we have f of x equals a constant. And generally, that graph will be a horizontal line crossing the y-axis at whatever that constant is, if it looks like I've indicated a positive constant here. Then we have f of x equals x, the identity function. And its graph is a diagonal line, 45-degree line, through the origin. And remember, there were three points that I highlighted, and I said this is going to be relevant later. The three points I highlighted were 1, 1, 0, 0, and negative 1, negative 1. I was doing that because these become target points for the graphs. Now, if you think of f of x equals x as being so basic, you don't need to plot three target points. You can just sketch it. That's fine with me. But if we look at it from the target point of view, the target point point of view, the graph would look like that. So that's the graph of the identity function. Then we went to the absolute value function. And it had three target points. And let's see, who could remind me? What were the target points for that one, Jenny? 1, 1. 1, 1, yeah? And 0, 0. 0, 0. Negative 1, 1. Negative 1, 1, very good. OK, and then we have to remember the shape. And for the absolute value, we have this right angle shape to raise coming out like that. So this is a quick sketch of the absolute value function. Then we had two more functions so far. We had the squaring function. And it has the same target points as the absolute value. 1, 1, 0, 0. Negative 1, 1. But it's a different shape. So we have to remember the shape as a parabola that comes down, goes to the origin, and looks like that. Hopefully it looks better than that. But you remember, we've sacrificed accuracy for speed, so we're doing this faster now. And the last function that we've considered thus far is the cubing function. And it has target points at 1, 1, 0, 0. Negative 1, negative 1. You notice those are exactly like the target points over here for x. And in fact, the reason is, you notice this is x to the first power, and this is x to the third power. All of the odd powers of x will have these three target points. The higher the power, the steeper it comes down, the flatter it is in the middle, and the steeper it goes down again. But this is the cubic. So the cubic is going to look like generally like that. And you might say, well now, Dennis, how do we know if we're drawing it steeper there than it is here? Because you said the cubic was steeper than the quadratic. Well, it's all relative. I wouldn't be grading you on just how steep you make it. But I just want you to know that as the power gets bigger, the function gets steeper as you go over there. But it's sort of hard to actually make variations in that steepness. OK, three other fundamental functions. We have the square root function next. As you might guess, there are going to be target points that we'll come up with for sketching that. But one of the differences is the square root function only has two target points. Let's see. If I graph f of x equals the square root of x. Now let's see. Once again, I'm going to make a table since I haven't officially graphed this one in this course before. I have to be careful about which x's I choose. What's the restriction on x here? All the x's have to be greater than 0. Do they have to be greater than 0? What about x equals 0 or greater exactly? So we can choose 0. We just can't pick a negative number. And you might say, but I thought the square root of a negative number was an imaginary number. Well, that's true. But see, on these axes, I have a real number x-axis and a real number y-axis. So we're not graphing any imaginary numbers. So we'll restrict ourselves to non-negative real numbers. Let's say I chose 0, 1, 2, 3, and 4 this time. Let's see. The square root of 0 is 0. The square root of 1 is 1. The square root of 2, well, you know actually, the square root of 2, on a calculator, it's about 1.4, more or less, not exactly 1.4. And the square root of 3 is approximately, does anybody know the square root of 3 right of hand? It's about 1.7. And the square root of 4, does anybody know the square root of 4 right of hand? 2. 2. Thank you very much. OK, it's 2. Very good. And we could just keep on going down the line. But I think these are enough points to plot to get an idea. So x-axis, y-axis, 1, 2, 3, 4. OK, so that's as far as we got in our table. 1, 2, let's say there's 3. So the points I'll plot will be 0, 0, 1, 1, 2, 1.4. I'll just have to take a while, guess at that one. 3, 1.7, 4, 2, 4, 2. Well, those obviously don't line up. But then, you know, this isn't a linear function. You can't write this in the form y equals mx plus b. But actually, what it is, it's half of a parabola coming in sideways. And if I could complete the parabola, it would go out underneath. But you see, these function values can never be negative. We're taking the principal square root of x. So I'll never get a negative y. So it's the upper half of a sideways parabola. In fact, this parabola would be identical to the parabola for f of x equals x squared, except it's going out the x-axis instead of up the y-axis. Well, if you were going to pick target points, so you kind of get how this is working. Which target points would you pick, David? What would you choose for target points? 0, 0. 0, 0, OK. Then 1, 1. And 1, 1. And those are the only two we'll pick, because we don't have a function value at negative 1. So we have this down to two target points. And in the future, when I want to graph this function, the square root function, and we'll have lots of opportunities where we'll want to graph this in this course. If I want to graph f of x equals the square root of x, then I'll remember the two target points, 0, 0, and 1, 1. But then I have to remember the shape. I have to remember it's half of a parabola, and it opens out to the right-hand side. So I'll draw it like this. Now, one thing you have to be careful of, when you come into the origin, you want to turn this down so it goes vertically. So at the moment you reach the origin, you're going vertical at that moment. Sometimes students draw the graph like this, where it sort of comes in about a 45 degree angle. But you remember that this turn on the parabola is always smooth, so you don't want to make it look like it could be jagged. I've perhaps overemphasized that in my graph here, but this is the graph of the square root function. One of the things we might point out while we have this graph up here is if you want to find the domain of a function, like take the domain of this function right here, if you imagine just pressing it down onto the x-axis, look at all the x's that are covered, and those are the x's that were used, and so that's the domain. If I press this graph onto the x-axis, it looks like it's going to cover everything from zero on out. So the domain of the square root function would be the closed interval zero to infinity, because I can plug in zero or any number larger, and I can see that in the graph by just pressing it onto the x-axis. I can also get the range by looking at the graph. But the difference now is I press it onto the y-axis, and that tells me all the y-coordinates that were used. So if I were to press this onto the y-axis, what would be covered on the y-axis? Zero to infinity. Zero to infinity again, but we're talking about these values for y, that's the range. So we have the same set for the domain and range, but they actually come from different axes. So every graph for every function, you can find the domain by pressing it onto the x-axis to get the domain, and pressing it onto the y-axis to get the range. I'll do another example of that with a function that's maybe not one of these listed here in a moment, but let's finish our list of fundamental graphs. Next function is the cube root function. So we want to graph f of x equals the cube root of x. Once again, I could make a table of values, but we sort of get the idea of how that goes, and it certainly kind of drags it out. Let's just go quickly to some target points and discuss the shape of this one. Well, let's see, this time, am I allowed to use negative numbers under the cube root sign? Can you take a cube root of a negative number? Yes, yes. Like what's the cube root of negative eight? Negative two. It's negative two, sure. So it looks like this function is gonna have a domain that's the entire x-axis, because I can pick negative and positive and zero values for x. If I substitute in a one for x, what would be f of one, the cube root of one? What's the cube root of one? One. Is one, okay. So that says I have the ordered pair one one. If I plug in a zero for x, the cube root of zero is zero. And if I plug in a negative one for x, the cube root of negative one is what? Negative one. Negative one, okay. And we've seen those three target points before. We saw those for f of x equals x. We saw those for f of x equals x cube, and now we see them for f of x equals the cube root of x. And what happens here, if you were to plot more points, you would find that this function comes in sideways, sort of like the square root function, but it's only more dramatic. It's not quite so steep out here, and it turns more abruptly right there, and then it turns and goes out in the third quadrant in a similar, in a similar manner. So that's the graph of the cube root function. If I press this onto the x-axis, it completely covers all of the x-axis. So the domain is all real numbers. If I press this onto the y-axis, it completely covers the y-axis. You might say, wait a minute, I don't see anything. If you press it onto the y-axis, I don't see that covered. Well, you see this function continues to rise and rise, but very slowly, and eventually you'll be this high, you'll be 1,000 units high, and you'll be negative 1,000 units down, so you'll cover the y-axis there. And my target points here are 1, 1, 0, 0, and negative 1, negative 1. Okay, the last function. This one's a little different. You notice even the symbols that I'm using there are a little different. Now, when I write f of x equals square brackets, and then I put a little x inside, this is referred to as the greatest integer function. Now, what this notation means is, if I take this function and insert a number, for example, if I plug in the number 4.3, then the function value is the greatest integer that does not exceed 4.3. Now, for example, 5 exceeds 4.3, but 4 does not. 4 is the biggest integer that does not exceed 4.3, so this answer's 4. So my answer is always an integer, and it's the greatest integer that doesn't exceed 4.3. What would be the greatest integer value for 1.9? One. It'd be one, yeah. It's almost two, it's almost two, but if we say the answer's two, then we'd have an integer that exceeds 1.9. So one's the best I can do. What if I were to put this? Here's kind of a tricky one. What if I put negative 1.9? What's the greatest integer value of that? Negative one? Negative two. Well, actually negative one's bigger than negative 1.9, isn't it? But negative two is smaller, so this answer is negative two. If I draw this on a number line, here's zero, here's negative one, here's negative two, right about there is negative 1.9. So what's the greatest integer that does not exceed it? I think it would be negative two. So how would I draw a graph of this function? And more importantly, why would anyone ever want to use it? Well, we're gonna see applications of this functions later in this course, but rather than getting off track, let's look at the graph of this function for the moment. Without making a table, I think we can explain what the graph looks like and why it does. First of all, first thing I'm gonna do is I'm just gonna put square brackets there. In your textbook, you'll probably see an extra little bar in there, sort of like I was just using a moment ago, but when I was writing the documents for this course, I didn't have a square bracket with an extra bar inside, so I had to use just square brackets. So I'm gonna use this notation because you'll see that on the website, but if you look in your textbook, you'll probably see the extra bar in there, so I want you to be flexible on both of those. Now, when I go to graph this, suppose I were to choose a number between negative one and one. Suppose I were to choose an X in here. For example, that might be a third, it might be a fourth, whatever. What would be the greatest integer value of let's say one third? Zero. It'd be zero. In fact, all along here between zero and one, the function values are zero, so I'm gonna make that dark right there. And in fact, at zero, the function value is zero. The greatest integer that does not exceed zero is zero because it doesn't exceed it. However, at one, I'm gonna put an open circle because at one, the greatest integer value is one. The greatest integer value for one is one. If I choose a number between one and two, like let's say one and two thirds, the greatest integer value is always one. And then if I choose a number between two and three, like let's say two and a half, what's the greatest integer value of two and a half? Two. It's two. So I'm gonna go up here to two and make my next branch. Well, you see what's happening is these things are going up sort of stair step. In fact, this is sometimes referred to as a step function because it looks sort of like a staircase. It's called step function. And if I were to pick negative values, the negative values would look like this. And then the next one going backwards would look like this. Let's just check that. What if I were to choose right here negative one and a half? What's the greatest integer value of negative one and a half? Negative two. Negative two. And you see right there, we are at negative two. Sometimes in textbooks, you'll see them right negative two on top of that line so you'll know exactly what it corresponds to. They're right negative one here, zero here, one here, two there. In other words, that's telling you the elevation of that step at that moment. Are there target points for the greatest integer function? Well, I'll let you decide if you wanna call them target points, but it does go through zero zero. It does go through one one and it goes through negative one, negative one. So, we do have three points that we could use as a basis to draw a graph. Although this is rather primitive, I would just draw these horizontal segments all the way across there. Okay, so that is the so-called greatest integer function. What I think you've seen as a theme that carries through all of these graphs is that for every one of these, there are usually three occasionally, well this one only I guess, only two target points. And in fact for the constant function, there aren't any target points at all. You just draw a horizontal line at that moment. Now, how would I use these fundamental graphs in mathematics? Well, let me give you an example. Suppose I wanted to solve this equation. X squared equals three X plus four. Now, we know very well how to solve quadratic equations, but let me show you how to solve it graphically. I'm going to separate these two expressions into F of X equals X squared and G of X equals three X plus four. Now, I want to graph this function and I want to graph this function and I want to see where they might cross because wherever they cross, then these two values are equal and therefore that would be a solution to the equation. Well, if I sketch F of X equals X squared, I'm going to use my target points and the squaring function goes through one, one, zero, zero and negative one, one. So my squaring function looks like this, okay? Now, if I want to graph this function, that's not a fundamental graph. It's a linear function. How would I graph it? G of X equals three X plus four. How would you go about graphing that, Jeff? It goes through the y-axis at y equals four. Okay, it crosses at four right here. Then it has a positive slope of up three and over one. Okay, so if I go up three over one, I'd get another point right there. Down three is the left one. I could go back, I could go down three and back one. In fact, I'd be right on top of that point right there and if I draw that line through there, well, let's see, this is going to go off the graph but you're going to get the picture of what's going to happen. I think there's going to be an intersection right there because the parabola's turning up but the linear graph is continuing in a linear fashion. So I think there are going to be two intersections. And so it looks like one of the places where the intersect is right here at negative one, the other place where the intersect is going to be the other solution of this equation. Now let's use some algebra to solve this and see how we would go about finding those values if I were using algebra. I could set this equal to zero and then I could factor this polynomial into let's see x and x, x minus four and x plus one. And one of those numbers must be zero. Either the number x minus four is zero or the number x plus one is zero. So either x minus four is zero or x plus one is zero. So x is four or x is negative one. Now here's negative one so here's one of the solutions to our equation on the x-axis and the other solution which is off of our graph must be what? Must be at four, yeah must be at four. So although I can't see it up there it must be right above four. So you see what I'm doing is I'm blending and graphing with algebra. So I have straight lines and I have curves which are geometrical. When I graph them I'm using analytic geometry and then when I actually solve this quadratic equation over here I'm using algebra. So we're using a variety of mathematics to solve a problem. Let me take one other equation and see how we would solve it and see it using graphs.  Now you solved equations like this back in intermediate algebra. If I solve this graphically I'm gonna graph two functions. There's the function, the absolute value of, well let's say I tell you what actually we don't know how to graph that one yet so let me change that portion. Let's make this, we know how to graph the absolute value function so let's take that and set it equal to x plus four. Okay well I'm gonna graph first of all the absolute value function and I'm gonna graph this linear function x plus four. Now when I graph those the absolute value function we have some target points for that. Zero zero one one and negative one one. So it's the absolute value function so I know to make this v shape like so. The other function I wanna graph is x plus four so here's four on the y-axis and the slope is one so if I go over one and up one I get a point right there and when I draw that line in I get that graph so let's say this is the graph of f this is the graph of g. How many times do they intersect? Only once, the intersect right here now someone may say well will they intersect further out if you extend the graph? You remember in the last illustration the parabola and the straight line were gonna cross but I just didn't have enough room to graph it but I think these two are parallel because this has slope one and this portion also has slope one you see I went over and up one so these two are parallel they're not gonna cross they only cross here. Anybody wanna take a wild guess what that intersection is? Just two? Looks like it could be negative two let's try solving it now using algebra and see if that's the case. The absolute value of x equals x plus four to solve that in algebra I'd break it up into two cases. One case is that x is equal to x plus four and I'm assuming here that x plus four is bigger than zero because the absolute value couldn't equal a negative number. The other case is the negative of x is x plus four and once again I'm assuming that x plus four is bigger than zero. You see if x inside is a negative number the negative of that is a positive number and that's equal to the x plus four. Solving this first equation that says zero equals four well of course that's impossible so there's no solution there. And in the other case I have negative two x is four and so x is negative two. And let's just check if I plug that number in right here will x plus four be positive? I should say positive or zero that another possibility is zero. Yes it is if I plug in a negative two I get two is in fact greater than zero. So the only solution is a negative two and what we've done is we've drawn a picture of it and the only reason why it may not look like it's directly above negative two is because this is a human drawing it's not a perfect drawing so in an ideal world that would be directly above the negative two. Okay now we have all of our fundamental graphs let's go to a piecewise function and just see how we would graph it. You remember in the last episode we talked about a piecewise function that was divided into two or more parts and its domain was split and we looked at function values of the piecewise function this time I wanna graph a piecewise function. So let's say that this function is x squared for x less than or equal to two and let's say it's equal to, oh let's say make it less than or equal to one and let's say it's equal to four for x greater than one. Well if I wanna graph this piecewise function then I need to think of it in two parts. Here's one and basically I need to draw a graph to the left of one and I need to draw a graph to the right of one. To the left of one I'm gonna draw the squaring function which has three target points zero zero one one and negative one one and my squaring function looks like that but I have to stop right there because when I go beyond one it looks like the graph looks like four. So on this side I have the squaring function but on the right hand side I wanna graph f of x equals four what does that look like? f of x equals four. A horizontal line. Yes it's a constant function so what I do is I go up to four and I draw a horizontal line but I begin at one and the graph looks like that. Now these two together represent the graph of f. I'll put the f over here but actually both of these are part of the f and so I've spliced together two functions like taking a pair of scissors and just gluing them together so those are spliced in there. Okay let's move to a different topic now and this is the notion of symmetry in a graph. Now not all graphs have symmetry but when they do it becomes a very useful idea and there are three types of symmetry that you can see in a graph. At least three types that will be of interest to us. First of all is consider symmetry about the y-axis. Now let me give you an example. Suppose I showed only half of a graph and let's say the graph looked like this. It looks like this is a function so far because it passes the vertical line test so I'm gonna call this function f. Now if this graph has symmetry about the y-axis then if you take it and if you flip it over to the y-axis you see a mirror image of that graph on the other side so the mirror image would look like this I suppose and so now this graph does have symmetry about the y-axis. Now if I had drawn something completely different on the other side there'd be no symmetry at all. Now a function that is symmetric about the y-axis is called an even function so f is an even function and I'll explain why in just a moment but so far we have an even function and I wanna point out a notion to you about this graph. I'll make this just a little bit lower here. If I were to move over here let's say a distance x and if I were to go the same distance the other direction I would be at negative x so this distance equals that distance I just went in different directions and if I go up onto the graph right here I get a point and the height that I go up is the function value at x so I might represent that as f of x. On the other side at negative x I would go up onto the graph and the height there would be f of negative x because it's negative x that I'm substituting in. Well you see if this is perfectly symmetrical then the height over here should be the same as the height over there and so f of negative x is equal to f of x and this is the ultimate test maybe I should say the algebraic test for determining if a function is even that is if it's graph will be symmetric about the y-axis is does f of negative x is it the same thing as f of x? Let me just give you an example take this function f of x equals x squared now you notice if I substitute in a negative x then that's going to be negative x squared and negative x squared is x squared well I get the same thing so f of negative x equals f of x so therefore this equality is satisfied for all x's and therefore if I graph this function it will be symmetric about the y-axis and lo and behold it is because the graph of it is a parabola and you remember that the graph looks like let's see three target points I'll quickly plot there yeah it looks like it is symmetric about the y-axis so we can tell it's going to have the symmetry without ever graphing it we just merely run this little algebraic test on it other functions that are symmetric about the y-axis are these let's see there's f of x equals a constant because there's no x in there so if I replace every x with a negative x it's still the constant so this function or any function of this form would be symmetric about the y-axis this function f of x equals the absolute value of x because if I replace x with negative x I get the same absolute value and sure enough that v-shape is symmetric about the y-axis also f of x equals now we haven't graphed this one yet but we will later but if you take for example x to the fourth if I put any even power on x it'll be symmetric about the y-axis and as a matter of fact if I add or subtract combinations of even powered polynomials for example five x to the sixth minus three x to the fourth plus two x squared every one of those powers is even and therefore this graph will be symmetric about the y-axis we don't know how to graph that yet but just stay tuned and we'll find out more about that and you can see then that's why they're called even functions because many of the functions symmetric about the y-axis are even powered functions okay uh... another type of symmetry there is symmetry about the x-axis symmetry about the x-axis however in this case we don't have functions but we do we still have graphs uh... let me explain why suppose I have a graph which above the x-axis looks like looks like that let's say uh... if I flip that over the x-axis I get another half that looks like well it looks like sort of like that I guess um... now that that couldn't possibly be a function can anyone explain why doesn't pass the vertical line doesn't pass the vertical line test to see if I draw a vertical line through there let's say if I draw a vertical line through at five then I get a function value at five up here and I get a function value at five down here which means I have five associated with two different members of the range so it can't be a function uh... but it's still symmetric about the y-axis and the way I can tell is if I go up a number y and if I go down negative y I go over the same the same amount what I'm suggesting is if you replace a y with a negative y and the equation reduces to the very same thing it was before then you have a graph that's symmetric about the x-axis let me give an example suppose I have x minus three squared plus y squared equals sixteen where have we seen that those sorts of equations before in here what is that the what is that the equation of circle it's a circle exactly what's the center of the circle well i think it's three zero exactly three zero and what's the radius of the circle four four right yeah so you you're not exactly thinking about circles right now because we've been doing all these other things but this is a circle and if i graph it i would go over to three and i would draw a circle whose radius is four now it looks to me like that circle is symmetric about the x-axis it's not a function but it's a metric about the x-axis and if i replace y with negative y i get the very same equation i can tell by looking at this that it's going to be symmetric about the x-axis because if i replace y with negative y i get i get the same result okay one more type of symmetry and this is symmetry about the origin so symmetry about the origin and you notice all of this information is very visual because we're drawing graphs and that's one of the big differences as we pointed out earlier between intermediate algebra and college algebra there's such an emphasis on graphing and it it makes college algebra a very visible discipline for the most part we don't do graphs all the time but much of the time uh... okay now for a graph symmetric about the origin suppose i have a uh... what if i have a function that looks like that in the first quadrant and what i'm going to do is flip it over the y-axis and then flip it over the x-axis so it looks like uh... this we'll say hopefully something like this now if i were to go over a distance x and go back a distance negative x like i did before if i go up onto the graph here i get a point in the altitude would be f of x and if i go down onto the graph here the altitude would be f of negative x looks like this is going to be a negative altitude that's going to be a positive altitude but this one should be the negative of that one because we're going in opposite directions so f of negative x is the negative of f of x now you might say oh i see Dennis what you did is you factored out the negative one now the it's not that the negative one factored out we're talking about two different numbers we have a negative x we have a positive x but if you throw away the negative inside you have to replace it with a negative on the outside these sorts of functions are called odd functions is an odd function and that's because typically x raised to an odd power will give you uh... a graph symmetric about the origin uh... for example f of x equals x cube has that property symmetry about the origin f of x equals x has that property and there will be other functions that we come across uh... also i'll call this one g just for a change if you take a combination of odd functions you still get an odd function like if i take ten x cube minus eleven x uh... that's still an odd function it's going to be kind of curvy on one side and it'll be the the the reverse curve on the other side flipped over the y-axis flipped over the x-axis and you might say well i don't see why it's called symmetry about the origin well if i go from this point directly through the origin the same distance i arrive at the other point uh... the one at negative x so these points are always opposite each other through the origin you could you could see one another across the origin okay well let's kind of summarize what we've done today we've talked about uh... eight fundamental graphs the constant function f of x equals x and then we moved on to some more elaborate things like the square root function the cube root function and even the greatest integer function and those functions had target points uh... well all of them except the constant function had target points uh... either three or two target points what you'll want to know is each one of those fundamental graphs and know how to plot those target points for each one of them uh... then we talked about how you could solve equations graphically and finally we summed it up by looking at symmetry uh... three types of symmetry we'll see you next time


Hi everybody, Joe for Jaspy's casewrakes.com. This is actually not our last filler. I forgot to cross that out, but this is the second to last one. We got to do a 2021 Panini Select Baseball Serial Box, random team number seven, giving away 10 of those flawless baseball spots. Big thanks to this group for getting into the action. Appreciate it. There's all the teams right there. So let's do this break very quickly and then a second dice will be for the giveaway part. Four and a one, five times for names and teams. One, two, three, four, and a one, fifth and final time. Four and a one, five times for the teams. One, two, three, four, and a one, fifth and final time. Pirates down to Yankees. No trade windows and stuff like this. We just want to fly through this break and then do the giveaway a bit. So here's the first half of the list right here. There's the second half of the list right over here. Right over here, damn, right over here. I'm grabbing that left, right scroll button there. All right, and let's print and rip. All right, here's the final print out, hot off the presses. These little cereal boxes are pretty cute. Looks like the ones you get at like a hotel or something. Hotel breakfast bar, the variety packs. All right, there's Whitmaryfield, Louis Robert, Peterson, Adele, Kyle Lewis, et cetera, et cetera. All right, nothing too crazy there, but hey, let's flip back to the screen here. Gather everybody's names for the second time, second dice roll. Top 10, top 10 after seven, one and a six seven times. One, two, three, four, five, six, and seventh and final time, after seven. Now, Dusty has the odds, so there's everybody right here, but let's see who's joining Dusty in that flawless break. I appreciate everybody here giving this a shot, but only the top 10 can be in that flawless break. Dusty, Dusty, Kevin's in. Dusty, Dusty, Dusty, Dusty, Dusty, Dusty, and Dusty. Kevin, you snuck in though. Thanks, Dusty. Nine of the 10 spots for you as well, good draw. I'll see you in the next video. For the next break, jasmiescasebreaks.com. Bye.


Now, let us look at what happens for diagonalizable matrices. So, first of all, we have the following fact, which is that suppose A is a matrix that is diagonalizable with A equal to S lambda S inverse and lambda being a diagonal matrix. And suppose E is a perturbation matrix of size n cross n. Now, if lambda hat is an eigenvalue of A plus E, then there is some eigenvalue lambda i of A for which this difference between lambda hat minus lambda i is at most the L infinity norm of S times L infinity norm of S inverse times the L infinity norm of E, which as we defined it earlier is K infinity of S times the L infinity norm of E. So, K infinity of S is the condition number of S under the L infinity norm. So, we see the condition number showing up in when we try to analyze the stability of eigenvalue computations. But what matters here in the case of diagonalizable matrices is the condition number of S, which is the matrix that diagonalizes A, not the condition number of A itself.  Now, lambda is a diagonal matrix. And so, by the Gershkorin's theorem, there is some eigenvalue lambda i such that if lambda hat is an eigenvalue of A plus E, then there is some lambda i such that lambda hat minus lambda i is at most S inverse E S the L infinity norm. And now the result follows from the submultiplicativity property of the matrix norm. That is it. We can actually extend this result to a more general class of norms, which are norms satisfying this property that the norm of D is equal to the maximum diagonal entry when the matrix D is diagonal. Some examples of such norms are the L1 norm, L2 norm and L infinity norm. So, this is the extension. So, suppose A is a diagonalizable matrix of size n cross n and A can be written as S lambda S inverse where lambda is a diagonal matrix containing the eigenvalues of A on the diagonal. And let E be a perturbation matrix of size n cross n and let this norm be a matrix norm such that norm of D equals the maximum diagonal entry for all diagonal matrices. If lambda hat is an eigenvalue of A plus E, then there is some eigenvalue lambda i F of A such that lambda hat minus lambda i magnitude is at most K of S times the norm of E, this norm of E where K is the condition number with respect to this particular matrix norm. So, let us see how to show this. So, starting point is the same as that of the previous result. Obviously, S inverse A plus E times S is equal to lambda plus S inverse E S. Now, if lambda hat is an eigenvalue of lambda plus S inverse E S, then what we know is that if I take lambda hat times the identity matrix minus lambda minus S inverse E S, okay, what can I say about this matrix? Correct. So, basically eigenvalue satisfy the determinant of lambda i minus A equal to 0 and so lambda hat i minus this that determinant is equal to 0. So, this matrix itself is singular, okay. Now, if this matrix itself is singular, this means that lambda hat equals lambda i for some i and then there is nothing to prove. That is this inequality will be anyway satisfied, okay. But if so, I think we can safely assume that lambda hat is not equal to lambda i for any i so that lambda hat i minus lambda is non-singular, okay. Then I will consider the matrix lambda hat i minus lambda inverse times lambda hat i minus lambda lambda minus S inverse E S, okay. And then I will just expand this out. This is, oops, the identity matrix minus lambda hat i minus lambda inverse times S inverse E S, okay. So, this matrix is singular, okay. Now, recall a result we showed a long time ago which is that A in C to the n cross n is invertible if there is a matrix norm such that norm of i minus A is less than 1, okay. So, what that means is that if this matrix is singular no matter which norm I consider the norm of i minus this matrix should be greater than 1, okay. So, I will just write that here implies norm of i minus A is greater than 1 for every norm, for any norm if A is singular, okay. By the way, how did we show this result? Just to recall, if norm of i minus A is less than 1 then we consider the series k equal to 0 to infinity i minus A power k. This converges to C because the radius of convergence of summation z power k is 1. Then what we do is we look at A times sigma say k equal to 0 to n i minus A power k and this is equal to, we write this as i minus i minus A times sigma k equal to 0 to n i minus A power k and when you expand this out, it becomes a telescoping sum and you are left with only the first and last terms which is i minus i minus A power n plus 1 and this goes to the identity matrix as n goes to infinity because this spectral radius of this is less than 1 and so or rather this matrix converges to the all zero matrix and so we conclude that basically this matrix whatever this converges to is the matrix A inverse A is actually I should write it the other way A is invertible C equals A inverse, okay. This was just an aside to recall how this goes but now we will come back to our the proof we are trying to write out. So we will apply this result and that is to be applied to the matrix i minus lambda hat i minus capital lambda inverse times S inverse E s and so i minus that matrix is so if i minus that matrix is just lambda hat minus i lambda hat so okay so so thus i minus lambda hat i minus lambda inverse S inverse E s singular implies i minus this matrix which is lambda hat i minus lambda inverse S inverse E s norm is greater than or equal to 1 it doesn't matter which norm i pick and so we have that 1 is less than or equal to this norm actually what i'll do is i'll simplify this a bit and write it in this way 1 is less than or equal to this which i'll use my sub multiplicativity and write it as norm of S inverse E s times the norm of lambda hat i minus lambda inverse and now i'll use the property that this norm it returns the largest diagonal entry whenever the matrix is diagonal and this is a diagonal matrix so that so that this thing is actually equal to the right hand side here is equal to norm of S inverse E s times the max 1 less than or equal to i less than equal to n of mod lambda hat minus lambda i inverse it's the largest i can value i largest magnitude diagonal entry which i can also write as norm of S inverse E s to be varied by min 1 less than or equal to i less than or equal to n of mod lambda hat minus lambda i so we have min 1 less than or equal to i less than or equal to n mod lambda hat minus lambda i'm just taking this to the other side so 1 is over here and that multiplied by this just gives me this is less than or equal to the norm of S inverse E s against a multiple kt kativity S S inverse norm of E which is actually equal to k of E k of S times norm of E which is what we wanted to show okay so what we have done is that we have shown the importance of the condition number with respect to solving finding eigenvalues of the matrix but there is an important difference between what we saw just now and what we saw earlier when we were looking at the importance of the condition number in solving linear systems of equations so when you are solving ax equals b it is the condition number of a k of a that mattered here it is the condition number of s k of s that matters not k of a directly of course s depends on a s is the matrix that diagonalizes this matrix a but it is k of s that matters not k of a directly so therefore if k of s is a small number then small changes in a lead to small changes in the eigenvalues but if k of s is large then small changes in a could lead to large changes in the eigenvalues in particular if s is unitary then the condition number of s is equal to 1 with respect to the spectral norm and in this case the eigenvalues of a are actually well conditioned because k of s equals 1 and also recall that a matrix a can be unitary diagonalized if and only if it is a normal matrix okay so we conclude that so I will just write it this way normal matrices can be unitary diagonalized and second point is that unitary matrices have condition number equal to 1 with respect to spectral norm okay which implies that normal matrices are perfectly conditioned are perfectly conditioned with respect to eigenvalue computation okay so we have the following corollary so let a be in c to the n cross n and it is a normal matrix with eigenvalues lambda 1 up to lambda n and let e be an n cross n matrix if lambda hat is an eigenvalue of a plus e then there is some eigenvalue lambda i of a for which lambda hat minus lambda i is less than or equal to norm e spectral norm okay now in the case where a and a plus e are both Hermitian matrices we can actually use Weyl's interlacing theorem to get an even better bound so that's the next theorem if a and e in c to the n cross n are Hermitian are Hermitian matrices normal yes sir yes sir yes okay so if they're Hermitian are normal matrices Hermitian they need not be correct okay so if a and e are both Hermitian and lambda 1 less than or equal to lambda 2 lambda n are the ordered eigenvalues of a and lambda hat 1 lambda hat 2 lambda hat n are the ordered eigenvalues of a plus e then lambda 1 of e less than or equal to lambda hat k minus lambda k less than or equal to lambda n of e and this is true for k equal to 1 2 n and mod of lambda hat k minus lambda k is less than or equal to rho of e spectral radius which is equal to in this case because it's Hermitian the l to norm of e okay so basically this is a this is a better bound compared to the bounds we've seen earlier because it's really comparing the kth eigenvalue of a plus e with the kth eigenvalue of a so it's telling us which eigenvalue of a lambda hat k will be close to okay so there are a couple of different paths I can take from here and I was yet to decide what I will cover in the remainder of this course so I'd like to stop here for today and I'll figure out what I want to do next and continue in the in the next class


Welcome to JSA TV, the newsroom for tech and telecom professionals. I'm Jamie Scott Okutaya. Here with my friend Eric Ballard. He is the Vice President of Network and Cloud for Stream Data Centers. Eric, welcome back to JSA TV. Hey, Jamie. Thank you. Thanks for having me. It's always a pleasure. Every time I run into you at PTC, ITW, I make sure you're so sweet. Pull you over to JSA TV. Thank you again for joining us. And thank you for being here at Telecom Exchange Pop-Up in Dallas. You're hometown. It makes sense that Stream would be here in full glory. So tell us what's the latest with Stream? Oh, well, it's glad to have you guys here in Dallas. It's nice to have you guys here in our hometown. You know, hey, what's going on new for Stream is we are expanding and growing like crazy. You know, Phoenix, Arizona, for us out in Goodyear, Arizona, we have just made a purchase of a very large data center facility or a building that we're going to turn into a data center. It's 418,000 square feet. And we have an extra 126 acres to carry us over to 2 million square feet on a campus out there. So we're pretty excited about that. I was out there last week, in fact, and we're getting going with that in a hurry. You know, so here in Dallas area in Garland, where our building is going vertical out there, we got walls going up, we got steel going up, a lot of progress being made out there. And we're going to be live just after mid-year with that facility. It's really exciting. Oh, my goodness. Stream is just one of those amazing companies that just, the growth story is unbelievable. Phoenix, Garland, what's next? Any little hints or? Yeah, so there's definitely a couple of new markets maybe this year that we'll announce. You know, one for sure, but hopefully maybe two even. You know, good year I think will come out of surprise for a lot of folks that, you know, we went in out there, but we continue to look all the time and we continue to expand and grow and, you know, we continue to build bigger and bigger facilities and continue to drive into new marketplaces. And I'll tell you your people power as well. Just really incredible, good people doing what they love, doing it with their heart, and you can tell every step of the way. So thank you for doing what you do. Thank you, Stream. Thank you for tuning in and tell our viewers where can they go to find out more about Stream. So you go to streamdatacenters.com. We have all of our facilities on there and all the information and how to get ahold of us. Streamdatacenters.com. Thank you, Eric. And thank you viewers for tuning in to JSA TV. Happy networking.


What's going on everyone? Sam here with late sports cards. We've been 22, 23, Leaf, best of basketball, six box keys break, number two, random team style. There's a look again at your random results. As always, you can find those on the website, under the results tab, as well as at the link in chat. So good luck everybody, let's do some Leaf. I don't think at the moment, KJ, I think we have enough of them where I think we will do some more group breaks of them though for sure. That's kind of, that's really nasty.  That is 2000, 2001 tops, SGC 95, Kobe Bryant. On the Quantum Leafs, card number five, I'm assuming. Yep, Quantum Leafs five. Nice early Kobe, that's Frobee still. Frobee with the number eight jersey, nice black Mamba. That's really sick dude, that is so cool. God, these late 90s, early 2000s inserts are just unbeat dude, unbeat. Awesome man, such a cool card dude. Congrats Lakers, go to James S. Here you go Jim, don't take a picture of that one. Right, next one, great first card. Next up, pretty solid one here. PSA nine, 2018 Panini Prism, Shay Gilgis Alexander. On the rookie, going to the Clippers. Nam queue, congrats Nam queue, nice Shay. PSA nine, Silver. And there's the Prism on the back, again that is a Shay. Probably outside of Luca, the second biggest guy you're looking for in 2018 now. All-star this year, really nice time to get that one as well, Shay is a stud man. Shay, Gilgis Alexander, Silver. I had one of those at one point, no longer do unfortunately. Do you wish you would allow that? Yes, I do. Right, next one, good case so far man. That is gonna be nice, 2018 Panini Hoops. Luca Donchich, on the rookie, congrats on that one there, Dallas. Dallas is gonna be, who's got the MABS? Dallas, Dallas, Damien C, congrats Damien. Nice PSA 10 Hoops, yeah. Two back-to-back cases with a PSA 10 Luca rookie too, one from Select, this one being from Hoops. Nice one there, 2018, beautiful, beautiful, beautiful. Yeah, Brunson's up there too, I would say. That's a good point, Jim. Yeah, I would say Luca, Shay, Trey Young. What is happening over there? Bruce Brown, BB6. Hey, what's up Steve? Also another spot came in for 10 bucks, Steve's down to eight to go on that now, everybody. Nice dude, hell yeah, we've got an optic. Signature Series, 2021, Larry Bird. Let's go into the Celtics. There you go, Adam W with the Cs. Larry Bird on the Signature Series, pretty solid one there as well. Larry Bird, that is base, auto. Landry Shamet, best player from that draft.  Can you imagine all the potential of these boxes we had at Ingalls? Like a one-on-one Joe Ingalls from like Noir, like a high-end, like the best Joe Ingalls card out there.  That'd be rough. Nice, dude, for the Cavs. What year is that? 2016, 17, Panini Court Kings, Kyrie Irving on the Jersey Auto. Again, for the Cavaliers, Cleveland is gonna be, where's Cleveland at? Curt D, nice one there, is that game used? Game-worn Jersey Auto, 37 of 60, nice Kyrie. World B flat, Kyrie, very cool. Okay guys, let me drop you a link to Star Wars really quick. And we got three to go in a random box style. This is Calabrini 101, that'd be pretty sweet. I know, I mean, for the value, man, at a $50 buy-in, you are doing well if you hit, for sure. To end it, man, that's pretty solid, actually. That's gonna be rookie, Joe Morant on the on-card auto from Absolute for the Grizzlies, 2019 Joe. Heck yeah, dude. Wow. It's got Memphis, Memphis, back-to-back cases, rich age, with a job. This time it's a rookie job, too. Not bad, dude. Hell yeah, man, let's do it. That's awesome. Heck yeah, dude, what a case. That was an awesome case. That will do it for the case, guys. Thanks again for getting this filled up. Next up, we've got case number three.