Well, let's see. We're at episode 15, and we've just finished another set of material for our second exam. And episode 15 is a review and a study guide for the second exam. This is Math 1050 College Algebra, and I'm Dennis Allison. I teach in the mathematics department at UVSC. If you remember when we first began the material for the second test, we first introduced some fundamental graphs of some higher degree polynomial functions. And let's look at the first graphic there. And I better mention this before we get into those fundamental functions. Let me remind you that you cannot use a calculator on this exam. Now, if you think back, you've probably seen me use a graphing calculator on just a few occasions thus far during this material. But there won't be any problems on the exam that would require you to use a graphing calculator. And therefore, you cannot have one available to you. You see, on this test, we want to find out, if you know, how to graph these functions without using a graphing calculator. And if you know, the important characteristics of a function as you graph it. The next thing is you cannot, of course, bring any notes. You can't use the textbook during the test. There'll be some scratch paper attached with your exam, which you can use if you need it. But there should be enough room on the exam for you to work most of the problems right there. But you should turn in all of the scratch paper with your test, even the paper that was not used. And finally, be sure to show your work so that you justify your answers. Now, there are some problems that are so short that there's really not much work, if anything, to show. But on the more complicated problems, you wouldn't want to just write down an answer, because then would have no way of knowing how you arrived at your answer if you haven't worked it out in some form. If you do show your work on scratch paper, leave us a note on the exam to tell us to look on the scratch paper, and then number it or circle it on the scratch paper so we can find it, so that it's easier for us to locate. OK, now let's go to the next graphic. And back in episode nine, we introduced several higher order polynomial functions. The first two there that you see listed, f of x equals x squared and f of x equals x cubed, were two fundamental functions from earlier episodes. But now we know how to graph x to the fourth, x to the fifth, x to the sixth, and you remember there are target points for those. Now, let's come to the green screen and let me just remind you how that goes. For f of x equals x squared, there were three target points, and the target points were located at 1, 1, 0, 0, and negative 1, 1. And so when we draw that graph, we use only those points and we graph this parabola. Now, if I ask you to graph this function on the exam, the instructions will say for you to use target points only, don't make a table of values. In class, you recall some time ago when I graphed this for the first time, I did make a table of values, but that was the last time we did for this function. And instead, we located the target points 1, 1, 0, 0, and negative 1, 1. So you're responsible for knowing what the target points are and not calculating or locating more points than that on the graph. Because you remember what we're after here is speed and not accuracy. So I'm not expecting you to get a totally accurate graph, but it should generally have this shape. Another fundamental function was f of x equals x cubed. And this goes back to earlier episodes. In this case, the target points are, let's see, can anybody tell me in the class what it was of the target points for x cubed? 1, 1. 1, 1, yeah? 0, 0. 0, 0. And negative 1, negative 1. Negative 1, negative 1. Yes, now on the exam, I would draw these axes neatly, so you wouldn't have to draw the axes and label them. And if there's no indication of what the scaling is, these are all one unit. So for example, this is 3, and this must be positive 3 up here as well. So these are the three target points. And then you should know that the general shape of the graph is that it comes down like a, well, what they used to call a higher parabola. This is no longer referred to as a higher parabola, it's just a cubic curve. And then it goes down into the third quadrant. So this graph appears only in the first and third quadrants. OK, now for the new fundamental functions that were on that list, let's look at how we would graph, say, x to the fourth and x to the fourth power. And it has three target points, and they're identical to the target points that we had for f of x equals x squared, because you see this is an even function, and it's symmetric about the y-axis. So the target points are at 1, 1, 0, 0, and negative 1, 1. Now when I draw this graph, it may look like a parabola, but technically there are some differences in it and the graph of f of x equals x squared. Can anyone remind me what are the differences between x to the fourth and x squared and, David? Doesn't it flatten out more at the bottom? It's a little flatter in the middle. Yeah, so right here, I may be exaggerating it too much, but it's a little bit flatter there. And what about on the two outer edges? Steven? It goes up much steeper then. It goes up steeper. Yeah, so this is going to rise faster. So it goes up faster. It doesn't go up vertically, and it's not straight. Although it may look straight the way I'm drawing it, because it's just a freehand graph. But this is a bit exaggerated, but that's what the fourth power function looks like. Now, let's just pick a higher degree polynomial, such as f of x equals x to the 13th power. I don't know that we've actually graphed that particular function before, but I think you know enough about these fundamental polynomials to know what they're going to look like. Can anyone tell me the target points for x to the 13th? Negative 1, negative 1. Negative 1, negative 1. Zero, zero. Zero, zero. And 1, 1. And 1, 1. OK, and Jeff, it sounds like you're right on the money here. What's the graph going to look like? It's going to look similar to the x cubed function, but it's going to climb a lot steeper. Much more exaggerated. In fact, it's going to look almost vertical here, and then in the middle it's going to look almost flat. And it looks like it's running along the x axis, but actually it crosses it only at zero. And then it's almost vertical right there. For example, if I were to choose 1 half, which is right there, halfway between zero and one, of course. And if I substituted in 1 half into this function, 1 half to the 13th power is about 1 over 8,000. So the altitude of the graph at a half is about 1 8,000. Well now, there's no way we're going to be able to distinguish that from zero. That shows you how close the graph is already to the x axis. But if I come over here to x to the fourth at 1 half, what is 1 half to the fourth power? 16. 1 over 16. Yeah, 1 over 16. Now we don't think of 1 over 16 as being very big, but that actually sounds huge compared to 1 over 8,000. 1 over 16. So you see this graph has already begun to leave just a little bit from the x axis. And over here it's running very, very close to the x axis. So things get more exaggerated. They're steeper, they're flatter, and they're steeper again as they exit. Another item that was on that, can we go back to episode number nine, that graphic? Another thing that was on that list in episode number nine is to know how to make transformations of these functions. And let me just show you what I mean by that. Suppose I wanted to graph g of x equals 2 times x plus 2 to the fifth power. 2 times x plus 2 to the fifth power. Well, this is basically, over here, if I'm thinking of my fundamental graph, I'm basically going to be graphing g of x equals x to the fifth, but I'll be making two changes. There's a stretch of 2, and I have to shift it to the left two units. So when I graph this, I better put in a few more markings over here on the left, because we're moving it that way. Here's the x axis, and here's the y axis. I'm going to move it over two units, so this is my new origin. And there's a stretch of 2. So if I go over 1, I go up 2. And because this is an odd power degree, if I go to the left one, I should go down 1, but the stretch says I should go down 2. And now, with these three target points, I'll sketch x to the fifth, shift it over. So it's going to come down fairly steep, steeper than the cubic function. It's going to flatten out, and it's going to turn, and it's going to go down here. So this is a rough sketch of the graph. I'll just put a g beside it. So you should be able to make transformations where there are vertical transformations or translations, horizontal translations, stretches and compressions. And if I put a negative on that, that would flip it over. So you should be able to do those things. OK, let's go to episode 10 and look at the review items for episode 10. There we are. Now, I should remind you that if you look on the website, you'll actually see a few more items listed than what I have on these graphics, because there wasn't enough time or even space, actually, on the graphic, to put all the things that could be on the test. But what I've done is to sort of hit the highlights of each episode here. Let's see, you should be able to perform long division and synthetic division. So let's just do an example of each one of those. If I were doing long division, say, if we wanted to divide, can someone give me a cubic polynomial? Let's start off with just an x cube. x cube, and then what? 7x squared. Plus 7x squared? Minus 9x. Minus 9x? Plus 13. Plus 13, OK. And suppose I wanted to divide this by x squared plus x minus 2. I'm just making that up as well. Now, is this a problem that I could use synthetic division on? No. I don't think so, because I'm not dividing by a linear polynomial with variable coefficient 1. In other words, I'm not dividing by x plus a constant or x minus a constant. So this has a quadratic term in it, so I'd have to use long division. And long division would go like this. I would list all of the terms inside. And if I were missing a term, I'd have to leave a space for that degree. But it looks like we have all the terms accounted for here. And then I divide the lead term of the divisor into the lead term of the dividend. This is sometimes referred to as the dividend. x squared goes into x cubed x times. So I'll put an x in the x column. And if you put your x up here at the front, that's OK with me. But I think it looks neater. And it's a little bit easier to follow if you keep all your powers in line with one another. So x times x squared is x cubed. And if I continue to multiply plus x squared minus 2x, then I have to subtract those off. Now, some students go through and change all the signs of those terms. I hesitate to do that because once you change the signs, if you look back at this later, it makes it a little bit harder to follow to see what the problem said. So I prefer personally to put the negative outside and then not change the signs, but do this mentally. x cubed minus x cubed is 0. 7x squared take away x squared is 6x squared. And now, who can tell me what is negative 9x take away negative 2x? Negative 7x? Negative 7x. Because this is actually negative 9x plus 2x. See, if you change that sign, that's plus 2x. Negative 9 plus 2 is minus 7. And because I've eliminated the cubes, I can bring down the next term. And I can divide x squared into 6x squared. And it goes six times. And I'll put the 6 in the constant column. Multiplying, I get 6x squared plus 6x minus 12. And here, I have to subtract again. So we'll see what kind of a problem Stephen has made up for us here as we work this out. See how bad the numbers get. 6x squared minus 6x squared is 0. Now, what is negative 7x minus plus 6x? Negative 13x. Negative 13x. Thank you very much. And in the constant column, 13 take away negative 12 is? 25. Plus 25. Yes, I'll add on a 25 there. And I can no longer do this division because I've gotten a degree too small. So this is my remainder term. And I would express this as a fraction over here on the end. Now, it looks like I don't have room to write that in there. I didn't leave myself enough space. So my answer would be x plus 6 plus negative 13x plus 25 all over the divisor x squared plus x minus 2. Now, sometimes students prefer to move the negative out in front. In fact, frankly, I do too. But if you move this negative out in front, you have to remember to take the negative out of both terms. So an alternative answer would be x plus 6 minus. And this will be 13x minus 25 because I factored the negative out of both terms over x squared plus x minus 2. I hope you can still see all that on the screen. Where in this problem do you think is the most likely place for someone to make an error? Subtraction. In the subtraction, yes. So you have to be very careful of the subtraction. And of course, if you make one subtraction mistake, it throws everything else off after that. When I'm grading these, if you make a careless error with your subtraction, but you generally understand the process after that, I'll take off something for the subtraction error. But I'll generally give you credit for the problem. Now, if you make several subtraction mistakes, then of course I'll have to take off more points. But don't think that you're doomed to miss the problem completely just because you might make a subtraction mistake. I make subtraction mistakes as well. That's probably why my wife balances our checkbook. But that's another story. OK, now let's do a synthetic division problem. And tell you what, could someone make up a synthetic division problem for us? A problem that would be appropriate for synthetic division. You don't have to be too hard on us, but something that would take a few steps. Or shall I make up one?  Well, I'm thinking, do you want both parts of it? Yeah, so we want the dividend and the divisor both. OK, let's do x cubed minus 6x squared plus 2. Plus 2. Oh, OK. What do you notice? She's kind of throwing a little ringer in there for us. What's different about this? We have to be careful of? Missing the x term. We're missing the x terms. We have to be sure to put in a 0 there when we do our synthetic division. And Jenny, what would you like for us to divide this by? Try x minus 1. x minus 1. OK, she's being very kind with x minus 1. OK, so when I use synthetic division, the process is that I put the coefficients of the dividend inside 1, negative 6, 0, and 2. You don't need to put the plus on the 2. And outside, normally, I would just put the negative 1. See, just throw away the x. But if you would like to add rather than subtract, most people would, you change that sign to a plus 1. So I'll put a 1 here. And in other words, this number becomes the root or the 0 of that divisor. 1 is what makes that 0. So by changing the sign, you now have the root. And now we can add. Let me just make a note over here that I'll be adding. If you don't change that sign, if for some reason you want to put negative 1 here, you have to do subtraction in these columns. But of course, subtractions where people tend to make more mistakes, at least I would. OK, so we begin this by bringing down the 1. And then 1 times 1 is 1.  And I add and get negative 5. 1 times negative 5 is still negative 5. And I add and I get negative 3. That's my remainder. So what this tells me is this answer is this quadratic with this remainder. This is a quadratic, not a cubic. Because if that's the remainder, here's the constant term, the linear term, and the quadratic term. So my answer is x squared minus 5x minus 5 with a remainder of negative 3. Now I think I'll put the negative in front and put a positive 3 over x minus 1. And that's the quotient. Now this is exactly what you'd get if you were using long division. But there's no need to go to that. And as a matter of fact, on the exam, if I ask you to use synthetic division here, then I wouldn't want you to be using long division for that. Another item in this section, episode 10, there was the remainder theorem. Let me just remind you what that says. And the remainder theorem says when dividing a polynomial, p of x, like this could be a cubic polynomial, 4th degree polynomial, by a linear factor, let's say x minus c. So if you're dividing this by the divisor x minus c, the remainder is p evaluated at c. All you do is take the root or the 0 of the divisor, plug it into the polynomial, and that'll tell you the remainder. For example, here's a problem that would require that. What if I said find the remainder in this division problem? Let's say it's 2x to the 4th minus 5x cubed plus 1 when divided by x minus 2. OK, find the remainder. But let's say I want to do this without dividing. Without dividing. Now you might say, Dennis, how can we find the remainder when you divide if you don't allow us to divide? Well, if you use this remainder theorem up here, it says the remainder will be the polynomial. This is the polynomial right here. p evaluated at the root 2. So that'll be 2 times 2 to the 4th minus 5 times 2 cubed plus 1. And let's see, 2 to the 4th is 16. So when you double it, that's 32. Minus 5 times 8 is 40 plus 1, and that gives me negative 7. So the remainder will be negative 7. Now if you actually carry out this division and divide x minus 2 into this fourth degree polynomial, you should find that the remainder is negative 7, using either synthetic division or long division. But of course, this said to do it without dividing. So we've been able to find the remainder very quickly. OK, the other theorem that you need to know is the factor theorem. And we use this quite a bit in the next episode. And the factor theorem says x minus c is a factor of the polynomial p of x, if and only if p of c is 0. That means if the remainder is 0. So this is actually, shall we say, a corollary to the remainder theorem that says you have a factor whenever the remainder is 0, and the remainder is 0 whenever p evaluated at c is 0. You'll see us use the factor theorem when we come to the next episode, speaking of which, let's go to episode 11 and look at some things we need to know from episode 11. First of all, we should be able to factor and graph polynomial functions using Descartes' rule and the rational root theorem. And then we should be able to solve applications of polynomial functions. And I put a note in there that no graphing calculator would be needed. If it's an application that requires you to graph a polynomial function, it's something you should be able to graph using the techniques listed just above it. So let's take an example. And let's see, let me find a problem in this book here that we can use to demonstrate. I'm going to take a polynomial like, OK, let's take this polynomial. p of x is 2x to the fourth plus 3x cubed minus 4x squared minus 3x plus 2. Now, what I'd like to do is to factor this polynomial in graph. So for the instructions, I'll put right below it. I'll say factor and graph. OK, well, let's see. First of all, I'd like to decide what are the possible number of positive and negative roots. And I use Descartes' rule of signs for that. So to find the number of positive roots, let's just put positive roots over here. Or you might say positive zeros. What I do is count the number of sign changes. And it looks like there are one, two sign changes. So there are either two or none for the positive roots, either two or zero positive roots. Now, what about the negative roots? What do we do? Substitute in negative x for x. Negative x for x. Now, actually, graphically, what that does when you replace x with negative x, what that does is flip the graph across the y-axis. So when you're looking for negative roots, that moves them over to the positive side. And then you just count the number of positive roots in that flipped graph. So if I substitute in a negative x for a positive x, well, let's see. Negative x to the fourth power is x to the fourth. So this is 2x to the fourth. And if I put in a negative x in qubit, that'll be a minus 3x cubed. What will this term be? Well, I think that's going to be a negative 4x squared.  Of course, the 2 doesn't change because there's no substitution in it. So if I count the sign changes, I get one, two sign changes there. So it looks like we have either two or none sign changes. Steven. All you have to do is change the signs on the odd powers, correct? You know, as a matter of fact, that's right. If it's an even power, taking an even power of a negative x is going to leave it as it was. And if it's an odd power on a negative x, you're going to produce an extra negative in front. So you change the sign. So that would be a shortcut you could certainly use. Yeah. So every odd power changes signs. Every even power remains the same. OK. Now, I'm going to take out the p of negative x so that I have some room to work here. The next thing I need to do is to figure out what are my possible choices for p over q. And to do that, I need to list possible choices for p and possible choices for q. Now, p has to be a divisor of the constant term. And so that would be plus or minus 1 or plus or minus 2. And q has to be a divisor of the lead coefficient of the highest power term, which is also a 2. That's a little ambiguous, because we have 2s in both ends. So the q's have to divide the lead coefficient.  So when I go to list the possible choices of p over q, what would they be? In fact, I'm going to move that back so I have a little more room to write them. Who can tell me what are the possible choices for p over q that could be rational roots? 1, negative 1. Plus or minus 1? Yeah. Plus or minus 2. Plus or minus 2? Plus or minus 1 half. And plus or minus 1 half. And I think that's it, isn't it? Yeah, so we're putting p's over q's so we can get plus or minus 1, plus or minus 2, plus or minus 1 half. Or if I put 2s over 2s, that's plus or minus 1. And we've already got that one listed. Now, what's significant about this set is that of these six numbers, these are the only rational numbers that could conceivably be a root of this polynomial. So for example, if you're wanting about a 3 or a negative 5 or a plus 2 thirds, none of those could be a rational root of this polynomial. And so we've narrowed it down from an infinitely many rational roots to only six. Now, we'll check these one at a time, and we'll do that by synthetic division using the factor theorem. If I'm going to pick a number, I'd try to make it easy on myself. So let's go with a plus 1, first of all. And I'm going to divide 1 into the polynomial 2, 3, negative 4, negative 3, and 2. Now, if 1 divides this, by the way, I'm going to be adding. If 1 divides this, then by the factor theorem, that means that x minus 1 is a factor of the polynomial.  Multiplying negative 2, I got a 0, so we found a root right away, and that tells me that this polynomial factors into, maybe I can just squeeze it in here, x minus 1 times another polynomial. And let's see, that was originally fourth degree, and I'm dividing out a linear polynomial. This should be a cubic.  Well, plus x. O plus x. And then minus 2. Minus 2. So this represents the missing cubic factor. OK, now, you know, 1 could be a root again, so we don't want to overlook a multiple root. So I'm going to try dividing 1 again into 2, 5, 1, negative 2. Because see, now we're looking for factors of this cubic. So 2, 2, 7, 7, 8 doesn't look like we're going to get a 0 here, so we don't even have to worry about calculating that number. Let's go to the negative 1. Let's try a negative 1 here, and we have 2, 5, 1, negative 2. So I get 2, negative 2, 3, negative 3. Adding, I get negative 2, 2, and I get a 0. So we found another root. So this is now going to be x minus 1 times, what's the factor for this divisor? For this? x plus 1. x plus 1 times, and now my missing factor goes from cubic to quadratic, and this is the quadratic that's going to be 2x squared plus 3x minus 2. And what's good about a quadratic is if it factors, we should be able to factor it by sight. So let's see if that will factor. If it does, it should be a 2x and an x. And I think if I put a plus 2 here and a minus 1 there, that's the complete factorization. So what turned out to be the 0s? Out of this list, which numbers turned out to be 0s? Plus 1 was a 0. Negative 1. Negative 1 was a 0. 1 half? 1 half, yeah. And you see what Stephen's doing is if you set this equal to 0 and solve for x, you get 1 half. And? Negative 2. Negative 2. So four of these six numbers turned out to be roots. It's not surprising that the very first number I tried was a root, because four out of six of them were going to be roots. So this is the factorization. Now when I go to draw the graph, let's put the graph right below it here. I want to sketch the graph using principles that we learned earlier. And so I'm going to locate the 0s. And the 0s are at 1 and at negative 1. And at 1 half, I'll squeeze 1 half in here. This is 1. And at negative 2, that's negative 2 right there. And if I look at my polynomial, that was a positive x to the fourth, 2x to the fourth. So that says this side goes up. And what's the multiplicity of every factor? 1. Is multiplicity 1. So the multiplicity 1 that tells me my graph passes through each one of these. So it passes through here. It passes through there. It passes through here. And it passes through there. And it goes back up. And so it has that typical fourth power look. Both sides are going together. In this case, they're both going up. You notice when I drew my graph, I drew this dip a little bit lower than that one. In fact, I probably should have made it even lower than that. That's because I had a span of 1 to wander away and come back. Over here, I had a span of only 1 half. So I had to come back fairly quickly. And in the middle, I had a span of 1 and 1 half. So I went up even more there than I went down here. That's the y-axis and the x-axis. And this is a rough sketch of p. If you were to graph this on a graphing calculator, that's more or less what you would see for the graph. I think it's a fairly close approximation for what little time we spent on finding the factors. While I'm drawing graphs like that, let me just ask you this question. How would I graph this polynomial? Suppose it's already been factored. Let's say it's a negative x minus 2 squared times x plus 1 to the third power times x minus 3 to the first power. What is the degree of this polynomial if I were to multiply it out? Six. This is a sixth degree polynomial. Yes. Sixth degree. Because you see there's an x squared there. There's an x cubed when I expand that times x. So x squared times x cubed times x makes x to the sixth. And the negative says it's going to be inverted. So I'm expecting, for an even degree, both sides go either up together or they go down together. And this is inverted. So I'm expecting to see both sides go down on this graph. And if I locate these x-intercepts, let's say I'll put a point at 2. I'll put a point at 3. And I'll put a point over here at negative 1. And I'm expecting to see the graph come up from below on the right-hand side. So it comes up to 3. And when it gets to 3, it's going to pass right on through because that has multiplicity 1. It's going to turn. Eventually, it'll turn and it'll come back to 2. And at 2, it's going to turn around and go back up because that had multiplicity 2. So it looks somewhat like a parabola right there. Now I expect in this next interval, it's going to go off pretty far before it comes back down. But I don't have a lot of room to show it. So it goes up quite a ways, comes back down. And as it approaches negative 1, you notice this factor has multiplicity 3. So here it looks a bit like a cubic function. It comes in, it levels off, it turns, and it goes back down. And sure enough, I do have both edges going down on the right and on the left. And this is a rough sketch of the function p. So you should be able to sketch graphs of functions like that. Let's go to the next episode. Let's see, episode 12. There we are. Complex roots of polynomials and construction of polynomials with real coefficients if you're given the degree and the roots. Why don't we take an example of that letter part, reconstructing a polynomial given its degree and its roots. Suppose I told you I'm looking for a polynomial that I'll call this one f of x. And suppose I told you that this polynomial is a fourth degree polynomial with real coefficients. That's significant with real coefficients. And suppose I told you two of its roots. One of the roots is 1. And another root is 1. And let me give you a third root. The third root is 2 minus i. In other words, this has a root of multiplicity 2 because 1's listed twice. And then it has a complex number root. Now, to get a fourth degree polynomial with real coefficients, I have to have one more root. And to get real coefficients, it has to be very specific. Can anyone tell me what a fourth root would be here? The conjugate of the complex number. It could be a conjugate of this, so 2 plus i. Someone might say, Dennis, is there anything else that fourth root could possibly be? Well, any non-zero multiple of 2 plus i. For example, if you tripled it, 6 plus 3i. So any multiple of 2 plus i could be a root. But this is sort of the fourth root in its primitive form. So f of x looks like this. f of x is going to be x minus 1 times x minus 1 again. So I'll just square that. Times x minus the root 2 minus i. And x minus the root 2 plus i. Now, backing up here, if you were to ask, why did you know to put in the conjugate? Just because 2 minus i is in there, why not some other number here? Well, if you want to get real coefficients and no imaginary numbers or complex numbers in the coefficients, you have to have complex roots appearing in conjugate pairs. So if you change the minus to a plus, that's its conjugate. OK, now, I need to multiply this out to figure out what the polynomial is. Well, x minus 1 squared is x squared minus 2x plus 1. And over here, when I multiply this out, what I'm going to do is distribute the negative signs, x minus 2 plus i and x minus 2 minus i.  Rather than multiplying these two, shall we say, trinomials together in that form, I'm going to group the x minus 2's together, group those, and group these in this way. x squared minus 2x plus 1 times the quantity x minus 2 plus i times the quantity x minus 2 minus i. Now, what's the advantage of writing these two trinomials in this form so that actually I have a binomial and a binomial of some and a difference? What's the advantage of that? The difference of squares. This is the difference of two squares, so this is easily multiplied. I haven't done anything with this first trinomial. It's sort of waiting in the wings for me to finish this multiplication. So I'm going to multiply the difference of two squares here, a plus b and a minus b.  And b squared is minus, well, is i squared. So now I can reduce that. And I have x squared minus 2x plus 1 times x squared minus 4x plus 4 plus 1. And you notice there are no more imaginary numbers in here. Everything is real, so I'm going to end up with a polynomial with real coefficients. So this is x squared minus 2x plus 1 times x squared minus 4x plus 5. OK, well, we're almost to the polynomial. The only way I can get a fourth power is when I multiply these two terms together. That's x to the fourth. Now, how can I get cubic terms? Well, I can get cubic terms if I multiply these two together. That's negative 4x cubed. And here's another one that makes negative 2x cubed more, makes negative 6x cubed. How can I get x squared terms? Well, let's see, there's a 5x squared. And here's a plus 1x squared. That makes 6x squared. But here's another way. I can get 8 more x squared, makes 14x squared plus 14x squared. Now, what about the linear terms? Well, right here I can get a negative 10x. And here I get a negative 4x, makes a negative 14x. I think those are the only ways you can get x to the first power. And then the constant term is plus 5. So this is the polynomial we're looking for. You notice it is a fourth-degree polynomial. It does have real coefficients. And it has four roots, although they're not all real roots. Now, you notice if I were working this backwards, if I were using Descartes' Rule of Sines, look at the number of sign changes. There is 1, 2, 3, 4 sign changes. So according to Descartes' Rule of Sines, how many possible positive roots could there be? 4, 2, or 9? So when you go to look for positive roots, if we were working this backwards, there would be 4, 2, or 9. And it turns out to be 2. See, we have these two positive roots. And if you calculated f at negative x in the interest of time I won't write it all out, I think you would see no sign changes, which means this has no negative roots. And that's exactly right. We have no negative roots up here. So we have two positive roots, no negative roots, and we have two complex number roots. See, complex numbers are not considered positive or negative. We have no way of comparing one to be larger than the other, so we have no further positive or negative roots from them. OK, let's go to episode 13, no, 13. Episode 13 is next. OK, here we have transformations of two more fundamental functions, f of x equals 1 over x and g of x equals 1 over x squared. And you may recall that in those last two episodes, episode 13 and 14, this was the beginning of a rather prolonged discussion of rational functions. And these functions can look rather complex by the time we get to episode 14, but they're not that difficult to graph.  Let's begin with these fundamental functions. Take, for example, f of x equals 1 over x. It's a graph. Looks like this. There were only two target points. And those were at 1, 1, and at negative 1, negative 1. The reason there wasn't a target point at the origin or for x equals 0 is because this function is undefined at 0. We actually have a vertical asymptote at 0. So then we have to know the shape of the graph, and the graph comes down and looks like this. The other fundamental function is g of x equals 1 over x squared. Now you notice this function can never be negative, because it's 1 over a square. So there's no way we'll ever get a negative number. So we expect from the beginning to see this graph above the x-axis. And when I graph it, there are two target points. And not surprising from all that we've seen and done, the target points are at 1, 1, and at negative 1, 1. The difference is not only are we in the first and second quadrants, but the graph is a little bit steeper coming in, and it approaches the x-axis, the horizontal asymptote, a little bit faster. So this is all relative. I don't know that you can actually make these discriminations very easily when you draw these. And I'm certainly not an artist. But you should know the fact that this function approaches the x-axis faster than 1 over x did. So we have a horizontal asymptote at the x-axis and a vertical asymptote at the y-axis once again. Now there were transformations of these functions. And rather than me drawing the graph of a transformation, let me do this. I'm going to draw the graph of a function, and I want you to tell me a reasonable rule for this graph. So I'm going to call this function, why don't we call it r of x for rational function? And the graph looks like this. And I'll give you a few minutes to think about it class, and then I want you to tell me what you think might be the rule of this function. There is a horizontal asymptote at 2. That's the line y equals 2. There is a vertical asymptote at negative 1. This is the vertical line x equals negative 1. And there are target points here and here. And the function looks like this. Now knowing what we know about these fundamental graphs, I think you can predict a rule for a function that would look like this.  Which one do you think it'll be? It's going to be g of x equals 1 over x squared, because both of these turn down, so they go together. Now the fact that they turn down tells you what about the function rule. There's going to be a negative in front of it. So I'm going to put a negative up there already. And does it look like there's a stretch on this, or did we merely flip it over? We just flipped it over. We just flipped it over, because look, from the new origin, we've gone over one and down one. We didn't go down three, or we didn't go down a half. So there's no stretch on it. So that tells me that the numerator is going to have a 1 in it up there. And I think I'll erase that x equals 1 to sort of in the way. OK, now, instead of putting x squared here, I'm going to have to put a shift in it that shifts it over to the left 1. So what should go in those parentheses? x plus 1. x plus 1, yes. x plus 1, and that's been squared. Now I think we're still missing one thing. What are we still missing here, Jenny? Plus 2. We have to add plus 2, because this graph has been raised to, so I have to put a 2 outside. That is a 2, not a z. My 2 sometimes look like z, so that is a 2 there. So this would be an excellent rule for this function. Now, you know what makes this a little, are there any questions about that from anyone in the class? OK, what makes these functions easily disguised is this rule could also be written as 2 minus 1 over x plus 1 squared. In other words, the 2 could be put in front. And so you may not recognize that that's a vertical shift, because it's written on the left instead of on the right. Another way this can be disguised is we could put this all over a common denominator. So this would be 2 times x plus 1 squared minus 1. I had to multiply this term top and bottom by x plus 1 squared. And if you multiply that out, the numerator would be 2x squared plus 4x plus 2, but minus 1 makes a plus 1 over x squared plus 2x plus 1. What I've done is I've expanded the x plus 1 squared on top and on bottom in collected terms. I was sort of in a hurry, so I saved a bit of time. This is the same function. Now, if I were to give you a graph, this is a perfectly good answer, but all of these other functions have that very same graph. So we have to be careful to recognize these things go by various names, just like you and I go by various names. And somehow we don't get confused, and we refer to each other by, say, if you call me Mr. Allison or Dennis, I know who you're talking about. Well, you want to be comfortable with naming functions by various names. Now, to that regard, let's go to episode 14 and see how these names for various functions can be graphed. We should be able to find the intercepts of rational functions. We'll talk about that in a minute. We should know how to write asymptotes and holes for rational functions and how to test each side of an asymptote. Let me kind of summarize that with a graph here. Suppose I were going to graph this rational function, f of x equals. I'll just make up something as I go along here. Suppose in the numerator, we had x plus 1 times x minus 2. And in the denominator, suppose we have x plus 2 times x minus 3. OK, now, this looks to be a long shot away from those fundamental graphs we were looking at just a moment ago. But we don't have time to do an example of everything that leads up to this, but I think this example will capture a number of important ideas. Let's see, first of all, this function will have several vertical asymptotes. Can anyone tell me where the vertical asymptotes are? At negative 2 and 3. At negative 2 and positive 3, because those are where you divide by 0, and those factors don't cancel. Neither of them cancel. So one vertical asymptote is x equals negative 2. Another one is x equals 3. By the way, Jeff, while we're on the subject, if this had been an x minus 3 over x minus 3, in other words, that would cancel. This wouldn't be a vertical asymptote. What would it be? It'd be a hole in the graph. There'd be a hole rather than a vertical asymptote. So it's very tricky. What I would do is I would cancel off those factors. I'd draw my graph of whatever's left over, and I'd go back and fill in a hole right there. OK, now, is there a horizontal asymptote? I think there is. Is there a horizontal asymptote? Well, to find that, I'm going to need to divide this out. You notice I have a quadratic over a quadratic. This is x squared minus x minus 2 over x squared minus x minus 6.  And it looks like my divisor goes into this only one time, x squared minus x minus 6. And when I subtract that, everything cancels, except I get a plus 4. So this is going to be 1 over 4, 1 plus 4 over x squared minus x minus 6. So let's write that in here as well. 4 over x plus 2 times x minus 3 plus 1. What I've done here is I've factored that denominator and put the plus 1 outside. So what's the horizontal asymptote? This equals this equals this. What's the horizontal asymptote? Was there a vertical shift? Yes, there's a vertical shift up 1. Of 1. So that means the horizontal asymptote moved up 1. And so right here, I'll say the horizontal asymptote is y equals 1. OK, are there any x intercepts in this graph? Yes. Yeah, we'll see. To find x intercepts, we'll let y be 0. So if I let y be 0, then what I'll need to do is multiply both sides by x plus 2, x minus 3, and I get 0 equals x plus 1 and x minus 2. So in other words, whatever are the roots of the numerator become the x intercepts. So this is going to have an x intercept at x equals negative 1 and an x intercept at x equals 2. And finally, is there a y intercept? Well, to find a y intercept, I'll let x be 0. Now, of all of these forms, I think probably the simplest one to plug into 0 is this one right here. Because if I plug a 0 in on top, I get negative 2. And if I plug 0 in on bottom, I get negative 6. You could have found the same information by plugging in 0 over here, but I think this is a little faster. And so we get 1 third. So I have a y intercept at 1 third. Now, let's put all this information together and draw this rather complex graph. But with this information, it should go fairly smoothly. OK. And if I go up and down on the scale, I'm going to put in my vertical asymptotes at negative 2 and at positive 3. OK. And then I'm going to put a horizontal asymptote at plus 1. Now, for a function that's complicated, there are no target points to plot. That's something I should certainly mention to you. Because you see, we don't have just a single factor on bottom that's been squared, but we have several factors all over. And those tend to affect where I would have placed the target point. So instead, what I have to do is go on the basis of the information I've listed over here. x intercepts at negative 1 and at plus 2. And a y intercept that's 1 third, 1 third right here. OK, now, let's look at the graph to the right of 3. This is 3 right here. You notice there's no x intercept. So that tells me the graph has to turn up because it cannot turn down. It would have crossed the x-axis. So my graph has to look like this.  And you may turn further out. Those things are relative. And as long as you have this general idea, I'll accept it. By the way, what happens over on this side? Will it be above or below? Above. It's above because there are no x intercepts over here. If I had had an x intercept, I'd know it had to turn down. But instead, it has to turn up because it can't cross the x-axis. Now in the middle, let's see. I have an x intercept here, multiplicity 1, which tells me the graph passes through. And it's got to go over here. So it must have been passing through this way because it has to get over to that point. So that tells me it must be turning down here. And that tells me that my graph goes up. It goes through this point. It passes through there. And it turns down over here. I haven't drawn that very well. It's probably a little bit more symmetrical than what I've shown. But because it crosses the axis and because it has to go to the positive y-axis, my graph has to turn down. It cannot turn around and go back up because there are no more x-intercepts. So it has to keep going down. So here is a graph of this very complex function. And we did it with very little effort. We did it through long division, but that's not particularly a lot of effort. And we had to know about asymptotes and intercepts. And multiplicity. I knew that the multiplicity of each x-intercept was 1. So the graph passed through. So I think that's quite an accomplishment considering what little we had to go on.


Everybody forced here up in 2022 limited football hobby seven box half case number 15 random team style find these results anytime late sports cards Dot com. Let's go ahead and get started everybody I'm in in seven boxes There's a debo to 125 for the Niners Limitless patch card number 249 Ezekiel Elliott for the Cowboys first they go into the Cowboys Raiders on Daniel Carlson membership autograph to 99 Kick her hit there you go Raiders and Brown's with a jumbo rookie patch auto to 99 David Bell We go Brown's Then more baseball breaks tomorrow with Matt everybody Revolution basketball Wednesday a release day as well some prism WWE undercard, which is kind of like a hobby hybrid That's revolutions the the release there So get your PYT spots on the site Who pretty fun one of my favorites? Louvoo Redemption Randall Cunningham game day aesthetics gold spotlight auto for the I would imagine that's Eagles, right? Have died I don't know there's a few other teams. Yeah, it's most likely Eagles here though day aesthetics card 30 oh It's a Viking really yeah, it is glad I checked Yeah game day aesthetics card 30 is Randall Cunningham Vikings. Yeah actually doesn't have any Eagles auto in this That's the only hit he has that's a Viking. Yep Vikings Randall Cunningham game day aesthetics gold spotlight auto Number to 49 or less Glad we looked that up right Vikings on the Cunningham Randall Cunningham It's got to be like one of his only non Eagles hits eight or 49 Aaron Donald for the Rams Jersey card to 125 rookie Trayvon Walker Jags and a rookie patch auto to 75 Terry Davis price and Niners All right box three now Panthers to 125 DJ Moore rookie Jersey to 125 Bill's James Cook Kings Warriors Kings Oilers Broncos out of Givante Williams that one's 16 out of 25. Hopefully he can stay healthy next year Very good running back and jumbo RPA to 99 Packers Romeo Dobbs Solid head for the packer. Let's see a big RPA now. We got four boxes left here Mark Andrews number to 125 for Baltimore Rookie patch card at 25 chief Skymore. It's like a name plate. It could be rookie Premier patch From a rookie premier jersey patch there Yeah, it should be Mike should be rookie auto for the Eagles is Cam Jergens to 75 and There's an RPA Seahawks Kenneth Walker the third to 75. That's a cool-looking card. Nice Seahawks nice card good patch too like that feather that feather pattern Think so, yeah, I Hope so I just said I've said a lot Got to a 7 to 25 dolphins They Josh Allen stadium star swatches Jersey card Buffalo number to 99 rookie DeMarvin Leal That's an on-card auto game day aesthetics out of the 25 Brian or locker bears So there are on cards in this Go bears cool. You can card too like the design and redemption goes to the Lions rookie jumbo patch auto emerald spotlights Lions that's awesome Good stuff there. You know, it's when emerald spotlight jumbo Is going to be to 10 or less That one that's hand or less on the emerald whole cap Oh, James Williams is five or less five to less five to five or less as well. He's like the only one that's different His his his rps are all lower numbered than everybody else's Interesting. It's a five or less on that. That's what the checklist says He's literally he has like a special like Thing for him. It says his parallels are like all lower than everybody else's Huh pretty sick My home's there The buccaneers mike all-star game day aesthetics out of to 99. There you go bucks. These are cool cards, man a train It is for the giants wandale robinson rookie patch out of gold spotlight Packers patch Aaron rogers playoff brilliance to 99 Lamar jackson's 60 of 99 ravens parallel base parallel and last one Devonte Smith number to 125 eagles bulls Rookie patch to 75 jets breeze hull Come on kings Come on kings. We got a ring of honor auto on card to 25 broncos steve atwater Nice card broncos rookie cave on tibido And on the back is christian watson rpa ruby to 49 looks like a georgia bulldogs card There you go packers very nice had a dobs and a watson now It's a 51 44 kings are up like kings up seven with Two and a half to go in the half Eight to go on six series now very nice Last eight spots tonight everybody last eight spots Again cheaper than it was last week. It's it's five dollars down. We got relics of breeze hull Aaron rogers josh allen sky more james cook trayvon walker Ezekiel elliott autos of uh Watson there christian watson steve atwater wandale robinson redemption giants Mike all-star brian earlocker Kenneth walker the third cam jergens A romeo dob is javante williams tirion davis price randall cunningham vikings redemption. It was a vikings randall cunningham uh david bell Daniel carlson And we had to five it says to five or less emerald emerald spotlight rookie patch auto jumbo james and williams for the lions Big car there and a five or less and that'll do it everybody onto the next half case


Santa knows where all the best toys are, and this December many of the Hunter smartphones are just a penny with service in Appalachian Wireless. Stop by your local Appalachian Wireless to check out all of the great smartphone deals that consistently top the nice lists. This Christmas is truly going to be bigger and better than ever. Better service, bigger savings than is. That's today's Appalachian Wireless. First responders in Pikeville join together today for their annual Christmas with a cop and firefighter program. While most area first responders host a shop with a cop program, the Pikeville group offers something a little different. Throughout the year, the Pikeville police and fire departments raise money through the annual haunted trail and other various programs. Then, during the holiday season, each first responder will receive a name of a child, their wish list, and a gift card with the amount of money they can spend. We have a limit that we set out to buy for each child and some of the police and firemen went beyond that limit out of their own pocket. We also had some of the businesses that went above and beyond this year too. Once all the gifts are bought and wrapped, Pikeville first responders will host a breakfast at Pikeville Elementary School. The children were able to eat breakfast, play games, meet Santa, and then open gifts. When they get to open those bags and presents and see some of those wish lists coming true, it's really worth being able to be present to see it. McDonald's of East Kentucky also volunteered during the event and provided breakfast for the 32 children and their families at today's Christmas with a cop and firefighter program. It goes back to my brother Bob and I, and learning from our parents, giving back to the community. And both Bob and I and our staff were all equally involved within the communities that we do business and support of. We believe, putting back into the community, that it's helped provide us jobs and careers for not only us, but for over 800 of our employees. So it's great to be a great community partner along with being a good personal community supporter. Not only did the children leave with presents, each family was given a food basket provided by Food City and Save-A-Lot with items to make Christmas dinner and a Walmart gift card. Reporting in Pikeville for EKB News, I'm Shelby Steele.


All right, so I'm going to start off by talking about juvenile idiopathic arthritis and really focusing on the management and some of the challenges that we face and decisions that have to be made in managing this disease. We're going to go through this with a case of a patient of Dr. by Tally's, which really is a great case to illustrate a lot of the things that we have to think about in ways to manage this disease. This is a three-year-old boy, and when he first presented, he had the diagnosis of oligoarticular JIA, and he's A&A positive. So he initially was seen by ophthalmology in 2002, just for a screening exam for any ocular inflammation, which was found to be negative. An overview and just some important points about JIA and UVitis associated with JIA, most of the children who are diagnosed with joint disease have that diagnosis first, and then have ocular complications secondary to that or after that. 15 to 47% of pediatric UVitis associate, or 15 to 47% of pediatric UVitis is associated with JIA, and that's a huge range of numbers, but there's differing reports. But the main takeaway is that this disease does account for a lot of the pediatric patients that we'll see with UVitis. 27 to 56% of kids with oligoarticular arthritis will develop UVitis, so it is important in this particular subset of kids that we really follow the screening guidelines and have the eye exams as kind of outlined, which we'll talk about. And then 30 to 50% of patients have structural ocular complications already at the time of diagnosis of UVitis, which is really not ideal, and that pretends a poor prognosis. So there's screening guidelines for when these kids need to be seen. Kind of the important things to remember is that oligoarticular JIA, RF negative polyarticular JIA, and psoriatic arthritis are kind of in the category where they may need to be seen more frequently, and then if they have a positive ANA and a diagnosis at less than six years of age, that kind of puts them in the high-risk category, and then a kid who's had the disease for less than four years is one that needs to be seen every three months, which is what category our patient falls into or it fell into when he first presented. So moving along this boy, he, in June of 2003, was started on methotrexate, 10 milligrams a week, and this was for the purpose of controlling his arthritis. He still hasn't had any ocular inflammation at this time. In August of 2003, he had his first documented uveitis flare, and at this time he had some leg pain, no eye pain, no eye injection, so parents, you know, no indication that he did have any eye problems, but on examination, he had one-plus-cell inflare and the right eye, two-plus-cell inflare and the left eye and some KP. So he was started on topical cortical steroids, and then he is also on methotrexate at this time as well, and he's continued on that. Kind of a, this is a very simplified version of the treatment paradigm, but basically, topical steroids are the first-line treatment, but as Dr. Vitale mentioned, there's complications of using these long-term, and so if this isn't controlling the inflammation, we move to anti-metabolites, most commonly methotrexate, and then adding on biologic therapies or using them in place of the anti-metabolites. And then there's little role, I mean, there's some role for oral or regional cortico-steroids, but that has to be short-term, and it's really not a long-term treatment plan. So the cortico-steroids, I mean, the goal is to get, to eliminate inflammation and to keep them at the lowest dose as possible and taper off as soon as you can without having the intraocular inflammation recur. There is a paper put out by Thorn and colleagues that looked at how what a reasonable long-term dose of low-dose cortico-steroids would be in kids with uveitis and looking at the risk of cataract development, and there's some controversy to this, but they did find that the risk having drops from increasing the drops per day from three to four drops per day was an independent risk factor for the development of cataracts, independent of uveitis, activity of the severity and type of cortico-steroids. So they, in their paper, they kind of concluded that a three-day, three-drop per day or less treatment regimen for what they defined as a moderate, moderate course, which was three to five years might be appropriate. Again, there's some controversy to that, but at least we can think of three-drops per day as less being something we can continue. And then even with this, though, up to 60% of patients will require other anti-metabolites or biologic therapies to control their uveitis. So in terms of anti-metabolites, methotrexate is kind of the first line therapy, and there's been some studies that have been done, and there's good evidence that it does control the uveitis in JIA patients, and for the, for time we'll kind of skip through, but there's evidence to support the fact that this is an effective at controlling ocular inflammation. You, what's been reported as kind of appropriate or a good response to methotrexate is a two-step decrease in AC grading, or AC, anterior chamber cell grading from in about 70 to 80% of patients within three to four months. That's kind of something we can expect, and then of course there's a lot of side, there can be a lot of side effects and any time we have these patients on anti-metabolites or any of these biologic therapies they need to be monitored with labs, and we need to pay attention to all the potential side effects and they need to have supplementation with folic acid. All right, so back to our patients. So in December of 2004, he had another uveitis flare. His methotrexate was increased to 15 milligrams a week, and then over the next kind of year and a half or so he continued to have uveitis flares. He's still on the 15 milligrams per week of methotrexate, and he's been treated for these flares with topical corticosteroids, but really things aren't as controlled as we would like. And then in June of 2007, he has an arthritis flare, and this is in the context of having these kind of recurrent uveitis flares. So his methotrexate was switched to subcutaneous administration. He's still on 15 milligrams a week, and he was started on infliximab at every six weeks. And then in March of 2009, so he was on infliximab for almost two years, and he had another arthritis flare, and this time it was independent of any uveitic flares, and he was switched to adelibimab and started on a very short, a weak prednisone course. And then at this point, their methotrexate dose had kind of been started to be tapered a little bit, but he was on 12.5 milligrams a week. So in 2007, his eye-examely, this is when he was started on infliximab. He just had one and a half plus cell in KP. And then this in March in 2009, kind of the next time he had his arthritis flare and they switched to adelibimab, but of note, he didn't have any ocular inflammation. And so one of the other things that this case does illustrate very well is that the joint disease and eye disease isn't synchronous, and it really is important that we follow very closely with rheumatology and have a really good working relationship and communication so that these patients do get their medications adjusted and changed if necessary and when necessary. So kind of the next step in the treatment after we've tried topical corticosteroids and then the anti-metabolites, the biologic response monobisers and there's a handful that we can use. So there's the tumor necrosis factor, alpha inhibitors, infliximab and adelibimab are two. They infliximab when we use it in JIA will need, we use about five milligrams per keg to 20 milligrams per keg and it's a shorter interval than with the adults in rheumatoid arthritis. It has been shown to have a consistent reduction in intraocular inflammation and it can be effective in permitting systemic immunomodulatory therapy taper and reducing the overall requirement of steroid use. It does have a waning efficacy over time and then adelibimab potentially is effective as well. The problem is just limited data and studies and limited comparisons between the two in this disease. Important things to remember about using these medications, the anti-double stranded DNA antibodies can form so you can have a rare drug induced lupus and then the anti-chimeric human antibodies can also form which can reduce the efficacy and cause infusion reactions and then with all of these there's an increased risk of infectious disease, increased risk of lymphoma but importantly in this category of medications there's an increased risk of demyelination and optic neuropathy and so not related necessarily to GIA but just remember if you're ever treating someone with intermediate UVitis or parisplanitis we need to make sure that they do not have any, that they do not have multiple sclerosis. It's a contraindication to use them. This medication with multiple sclerosis then you can have neuropsychiatric complications. Rutexamab is another biologic response modifier. It's an anti-b-cell mass human monoclonal antibody. Small case series showed some ocular quiescence with Rutexamab and refractory UVitis cases. It's a very small sample. I mean a lot of these studies are fairly small power but it's what we have. So back to our patient again. We're now in November of 2010 and he had a UVitis and an arthritis flare. He was switched to abatosept and he's still on methotrexate 12.5 milligrams. He was actually switched to abatosept six weeks prior to this visit and that was kind of again for his joint involvement but on his exam he still has inflammation more so in the left eye which is kind of chronically what he's had and he was started again on topical corticosteroids. Abatosept is another biologic response modifier. It's a selective two-cell co-simulation modulator. There's again a few very small series that showed some efficacy and clinical remission of UVitis but eight patients in the first series and then two reported cases in the second series and then our patient it didn't really, it wasn't effective in controlling his ocular inflammation. So March 2011 he continues to have low grade ocular inflammation, he continues to have flare ups of arthritis and the decision was made to switch back to infliximab because it did, looking kind of back at the course of his disease, it did seem to be fairly effective and at least more effective in controlling his disease than abatosept and adelimumab and his methotrexate was increased to 20 milligrams a week. So then, and I'll go back here for just, so after that switch he was actually had a pretty good response and was quiescent and has been quiescent since and didn't, hasn't had any ocular inflammation. So the question is how long does he need to stay on all these medications and what's an appropriate length to keep someone who has controlled UVitis on methotrexate. So there's a good paper that came out by ISO and colleagues and in 2011 it was a retrospective chart review between 1989 and 2009. And they basically took patients who had been started on methotrexate for UVitis associated with JIA and looked at what length of time they were on methotrexate and the length when they were taken off of it if they had any relapses and tried to kind of elucidate what the best kind of treatment paradigm for decreasing and tapering off methotrexate would be. They found that 69% of patients that had inactive UVitis at the time of methotrexate withdrawal had a relapse and most were in within the first year. And so looking at all the data they found that a longer period of inactivity combined with an older age or longer treatment with methotrexate might protect against relapses. And kind of just to highlight this the one year increase of duration of inactive UVitis being treated with methotrexate resulted in a decrease of hazard of new relapse in 93%. So really it's important to keep these patients when they're quiet on methotrexate for at least two years before we think about tapering them off. And the duration of inactivity before the withdrawal was an independent risk factor for the relapse of UVitis. So this paper was nice in showing actually providing data to what a lot of UVitis practitioners do in keeping the kids on methotrexate but now we have data to support why we want to keep someone on methotrexate for years after they're actually diseases controlled. And so he has like I said still been had control of his UVitis and so his vision now with 2020 in the right eye, 2025 in the left eye when he was seen most recently in January there was no ocular inflammation. He's on methotrexate seven point five milligrams a week at this point and his infliximab intervals are being extended and he's kind of going through a slow taper. So this case obviously extends and over many years and he's had kind of a chronic course of flare ups and trying to control it with different medications and I think it brings a few important points and how we have to manage this disease. It's a chronic disease and we can see that here and it involves a lot of persistence and just frequent visits and just making sure that everyone is on the same page and the parents understand what's happening. And the pediatric UVitis can have a guarded prognosis but this is potentially modifiable with early and aggressive steroids bearing immunomontulatory therapy and this disease and a lot of other UVatic entities really requires very good coordination and communication between different subspecialists to provide the best patient care and joint disease and eye disease are not synchronous so when there's an ocular flare up we have to communicate to the rheumatologist and vice versa to really get these kids on the treatment that they need. That is it. So thank you to Dr. Vitale, Dr. Shakur, Dr. Bell and Dr. Zimmerman.