 Well class, I'd like to welcome you to the first episode of Math 1050 College Algebra here at Utah Valley State College. I am Dennis Allison in the math department, and College Algebra is a course that's important to a number of disciplines here at UVSC. For example, if you want to major in, say, if you want to major in business before you can take business calculus, Math 1100, you have to take College Algebra. If you're majoring in engineering or physics or even mathematics, let's say, you take College Algebra and then trigonometry before you take calculus one, which is Math 1210. If you want to be an elementary school teacher, you take Math 1050, College Algebra, before you take a two-course sequence in mathematics for elementary school teachers. Then there are a number of disciplines that have College Algebra as their math requirement, but you don't take any other mathematics beyond that unless you just want to. For example, if you're a history major, a psychology major, or something like that, there's a math requirement, and College Algebra is one of the courses that would meet that math requirement. Now, you know, one of the, or the course that's the prerequisite for College Algebra is Math 1010 Intermediate Algebra, and today we're going to be talking about just some of the topics for Math 1010 that students are either a little rusty on or they have a little trouble with in Math 1050, so I thought I would spend just this first episode going over some of those topics. And while I'm at it, I might mention to you some of the differences between Intermediate Algebra and College Algebra. If you think back to when you took Intermediate Algebra, you may remember that that course is mostly about, let's say, solving equations like quadratic equations, factoring polynomials, multiplying polynomials, simplifying radicals. In other words, it's more sort of, shall we say, computational. There wasn't much graphing. In fact, the only graphing that you do in Intermediate Algebra is to graph straight lines, and you probably graphed a few parabolas near the end of the course. Now, in this course, the emphasis is primarily on graphing, and we'll be using all of those ideas from Intermediate Algebra, solving equations, factoring polynomials, but we more or less assume that you know how to do those things now, and we move on to graphing. Graphing becomes very important in later mathematics courses, and sort of at the heart of graphing is the nature of a function. So today, we'll be reviewing those Intermediate Algebra topics, and at the very end, we'll also go over the notion of a function, some of the basic ideas that you need to know about functions. Now, to begin with, let's look at straight lines and the graphs of straight lines. There are actually three fundamental equations of straight lines that you need to know. In fact, you probably remember, I hope, from Intermediate Algebra. For example, there's the point-slope equation of a line. Who knows the point-slope equation of a line? Yeah, Jeff? Why equals mx plus b? Why equals mx plus b, okay? So y equals mx plus b. And Jeff, what do m and b represent? m is the slope, and b is the y-intercept. Yeah, m is the slope of the line, and remember that's the rise over the run, and b is the y-intercept. So let me just make a note of that right here. m is, well, one way to put it is the rise divided by the run. Another way to put that is to say that it's, shall we say, delta y over delta x. That's the change in y over the change in x. And then, if you're given two points, let me just make a note of this over here. If you're given two points, x1, y1, and if you're given another point, x2, y2, then the slope of a line ends up being y2 minus y1 over x2 minus x1. So those are variations of what's called the slope formula. And that's the value of the m up here. b is the y-intercept, so b is where the graph crosses the y-axis. I mean, technically, it crosses at the point 0b, but we'll just say it's the number b, and then we'll put it in right here. Okay, now, in addition to that, it is another equation of a line. In particular, maybe I should tell you, I'm thinking of the point-slope equation of a line where you're given a point that's not necessarily on the y-axis and the slope. Who knows what that equation would be? Let me just draw an illustration over here. Suppose you were given a point x1, y1, and you were given a line that passes through it, and you know that its slope is m. What is an equation I could write for that line? Anybody know? Well, let me tell you. It would be y minus y1 equals m times x minus x1. So in other words, if you know these two constants, x1 and y1, if you plug them in right here, y minus the y-coordinate equals the slope times x minus the x-coordinate. Have you seen that before? Yeah, so maybe you just didn't recognize it by the name I gave it. Now, there's another equation of a line where you put all the variables on one side, and it ends up looking like this, ax plus by plus c equals 0. This is where a, b, and c are all constants. And this is sometimes referred to as the standard equation of a line. This is the point-slope equation of a line, and this is the slope-intercept equation of a line. Now, we have an interview with a student on the street, so to speak, and I asked the student a question about the equation of a line. So let's go to that videotaping. Well, we're just outside the math lab here at UVSC, and I brought along a few questions to ask students as they come out of the math lab. This is Matt. I just met Matt. He's now one of my closest friends. But we're asking Matt a question on the spur of the moment, and I'm glad it's not me being asked a question on the spur of the moment. But the question I'm asking Matt today is this. What is the equation of this line? You notice it crosses the y-axis at one. It crosses the x-axis at two. And Matt's going to show us how to find the equation of that line. Do you want me to tell you how I'd work it out or just do you want me to work it out? Oh, why don't you work it out or? All right. Whatever you feel like explaining. Well, it's going to be the equation of y equals mx plus b. And then your y is one. So you go plug in your one, your m, and your x is two. And hmm. Well, class, I'd like to introduce you to a friend of mine. This is Jessica. I just met her here two seconds ago. And I'm bringing with me some questions, review questions from math 1010. And we're just right in front of the math labs. I'm catching students as they come by. They're my victims today. Jessica, I've got a question for you. Oh, by the way, what's your major? It's elementary education. Elementary ed. Well, good. Okay. I'm wondering what is the equation of this straight line? Okay. First, what you do is you start out with the equation y equals mx plus b. Right. And then what you do is you look at the line. And you find out where the line crosses on the y-axis, which would be the b. And so since it crosses at the one, the b is going to be one. And you need to find the slope. So it goes down one and over two. So then you have a negative 1 half x. And that's the self of one. So the answer, let's see, the minus didn't show up. Maybe you can make that darker. Yeah, so the answer is? Negative 1 half x plus one. Y equals negative 1 half x plus one. That's exactly right, Jessica. Okay, why don't we just look quickly at the problem that I asked Matt and Jessica. Here's an illustration of that same line. I've changed the scales so that I can move the points out a little bit. So one is actually two strokes up above. And two is four strokes to the right. And so the line's coming down. Now right away we know this has a negative slope because this graph seems to be going downward as you move to the left, or move to the right. The y-intercept is one. So Jessica correctly said that b was one. And then she said if you go down one and over two, that means the rise is negative one and the run is two. So that's a negative one over two. So she said the slope was negative 1 half. And therefore she came to the equation that y equals minus 1 half x plus one. Now you know Matt made a mistake there early on. He had the right idea, but he actually ended up putting, if we go back to this illustration here, he put the one and the two together and made it a point. Rather than thinking of this as zero one and the point two zero. As a matter of fact, after we taped that, I pointed that out to him. And then he was able to figure out the equation of a line just fine. OK, now suppose I were to take this ordered pair, zero two, or rather two zero. Imagine there's a zero over there. And now this is no longer the y-intercept. So let's try getting the equation of the same line if I use the point slope equation of a line. So what I'm going to do is take the point two zero. And I'm going to take the slope negative 1 half. And I'm going to substitute that into the point slope equation of a line. And I get y minus, what number is going to go here? Zero. Zero equals minus 1 half times x minus two. Yeah, so you see in this case, this is really a more all-purpose formula because in the first equation, you have to have the y-intercept. Here, you can have any point at all on the line. So we have y equals minus 1 half x plus 1 when I multiply that out. And sure enough, it is the same equation. Now as far as I'm concerned, you could say this is the equation of the line. In fact, you could take out the zero, perhaps. That would help a little bit. Or you could multiply it out. And then in this case, you have the slope-intercept equation. Now if I want to get the standard equation of the line, what I'll do is get everything over on one side. So I'm going to move the negative 1 half x over and move the plus 1 over. So I'm going to add a 1 half x. I'm going to subtract a 1. And I get a 1 half x plus y minus 1 equals zero. And you see now this is in the standard equation form. What's one thing I might do to that to make it look a little nicer? What would you say, Stephen? Get the 1 half off of the x and make them all whole numbers. Exactly. So what would you do to get rid of the 1 half? Times the entire equation by 2. Right. So I'm going to multiply both sides by 2. And this becomes x plus 2y minus 2. And then I'll multiply on the right by 2. But it's still zero. So most people would say this is the standard form answer where you have integer coefficients in that case. OK, let me just ask you now one question that is related to this. So we'll just make up a problem right here on the spot. And what if I tell you that I have a line? I'm going to draw a line on here, on this graph paper. And I'm wondering what is the equation of not this line, but a line that passes through this point right up here that's perpendicular to it. Now let me just draw this in free hands. I don't give away exactly what the slope is. But somewhere along here, there's a line through 3 on the y-axis. I think if we put that under white paper, it'll show better. And I'm wondering what's the equation of this line knowing the line that I've drawn right here? So this one's given. This one I'd like to know. So the first thing I'll do is figure out what's the slope of the given line. Can anyone tell me what's the slope of this line that I've drawn? Yeah, Jeff. Negative 3 halves. Let's see. Well, now it looks like starting from here, we went up to. And over 3. Oh yeah, so you were thinking of the perpendicular. What's the slope of the original line? 2 thirds. 2 thirds, yes. So this line has slope 2 over 3. So therefore, Jeff, what's the slope of the perpendicular line? Negative 3 halves. This line has slope negative 3 over 2. Because the rule is that any line perpendicular to a given line, the slope is the negative reciprocal of the original slope. So this is the negative reciprocal of 2 thirds. So the last thing then is what's the equation of this line? Can anyone tell us? Well, what's the y-intercept? 3. Yeah, it looks like it's 3. And we know the slope. So we're going to use the equation in this case, y equals mx plus b. And the equation is y equals negative 3 halves x plus 3. That's the equation of the mystery line. Now, you notice I never did have to find the exact equation of the given line. What I needed was the slope of it. And that's all we needed in that case. OK, let's go on to another topic. I'd like to remind you about interval notation, which you use more and more in future courses. But you've probably only seen a little bit, so far. If I draw a number line along here, and let's say this is 0, this is 1, this is 2, maybe this is negative 1, here's negative 2, then what if we take the interval of numbers including negative 1 over to 2? I'm going to put a solid dot there, meaning I'm including the endpoints. There is a shorthand abbreviation for that interval. Jesse, do you know how you write that? Yeah, you use a bracket, negative 1, comma 2, bracket. Square bracket. And this square bracket means that you include the endpoint. Now, what if I change the problem, and I made this an open dot here and an open dot here. In that case, somebody else who can tell us how you would represent that, David. You would just change the brackets to parentheses. To parentheses, yeah. So now this becomes parentheses, negative 1, 2. Of course, a person might say, Dennis, that looks like you're plotting the point, negative 1, 2. It doesn't look like the ordered pair. So how are we supposed to know if you're talking about an interval or plotting a point? Well, what you do is take it in context. If we're talking about a portion of the number line, then we know this is an interval. And if we're plotting a point, then obviously this is an ordered pair. One more possibility. What if we start at negative 1, say, including negative 1, and we want to just keep on going? We just keep on going all the way out. How would I represent that using this sort of abbreviation, Steven? A square bracket, negative 1. Negative 1, comma, infinity. Infinity, yeah. So it's going to go all the way out to infinity. And by the way, when I put infinity there, do I put a parenthesis or a square bracket? Yeah, what's the reason for a parenthesis is supposed to a square bracket? It'll never touch infinity, so. Yeah, because there is no last point. See, if you put a square bracket, it means you enclose the last point. And there is no last point if you're going to infinity. So the standard procedure is you always put a parenthesis on here. And if I were going to, say, begin at 0, but not including 0, and take the numbers all the way down without end, how would I abbreviate that? Susan, how would I write that? Parenthesis, negative infinity, comma, 0. Comma, 0. And parenthesis, exactly. This first interval up here with the square, with the corner brackets, that's referred to as a closed interval, because we sort of closed the door. We included the last number. When I use parentheses, and I don't include the last numbers, that's called an open interval. In this case, this is a closed interval, because I did include the last number on this end, and there wasn't a last number. This one is a closed interval, and this last one is an open interval. One more possibility. What if I take these out? Let's go back to this first case where I had negative 1 to 2. And suppose I started at negative 1, including negative 1, and I went up to 2, but I didn't want to include 2. If I use this notation, what's one thing I'll have to change there? The bracket by the 2, and I have to be a parenthesis. Exactly. I'll just put a parenthesis here. So in this case, we would call that a half open or a half closed interval, like is the glass half full or half empty? It's up to you. It's a half closed or half open interval. OK, let's try working a problem that asks us to use these intervals. Suppose we have the absolute value of x plus 2 is less than 4. Now, the way we solve an equation like this, or actually it's an inequality like this, is to say that the number inside x plus 2, if its absolute value is smaller than 4, then that number is smaller than 4. But also, that number is bigger than negative 4. For example, x plus 2 couldn't be negative 5, because its absolute value would be 5 rather than smaller than 4. So now, what I have is a double inequality. And to solve this, I'm going to subtract an x from all three expressions, or rather subtract a 2 from all three expressions. So this becomes a negative 6, just an x in the middle, and I have a 2 over here. And this means I'm talking about solutions that are bigger than negative 6 and smaller than 2. So I'll write that as the interval negative 6, 2, 2. By the way, some of the ways you wouldn't want to write it, well, obviously you wouldn't want to put square brackets on there, because that would mean that you were including the endpoints. Also, you wouldn't want to use set brackets like this. And the difference is this means you're talking about exactly those two numbers, negative 6 and positive 2. And of course, neither one of those numbers are actually in the interval. I want the numbers that are in between those two. OK, now, in a slightly different problem, suppose we take the absolute value of 3 minus 2x is greater than or equal to 7. Now, this one's a little bit trickier to solve, because you see, if I want the number inside to have absolute value, 7 or larger, then there are actually two possibilities. 3 minus 2x could be greater than or equal to 7. Or the other possibility is 3 minus 2x could be less than or equal to negative 7. For example, negative 8. This expression could be negative 8. Its absolute value would be 8. And that would be greater than or equal to 7. So I'll solve these two inequalities separately. In the first case, I have negative 2x is greater than or equal to 4. And if I divide by negative 2, x is less than or equal to negative 2. Or the other possibility is that negative 2x is less than or equal to negative 10. And when I divide by negative 2, x is greater than or equal to positive 5. The numbers that solve the first half of this problem would be the numbers from minus infinity up to minus 2, including negative 2. And the numbers that solve the other portion would be the numbers including 5 up to plus infinity. You can put a plus on the infinity if there's any doubt about what you would mean by that. And I want to put these two sets together, but not the intersection because actually they don't intersect at all. It's the union. So I'll put a union symbol in between here. And then this is the solution to the problem. And when I put a union symbol, it means anything in the first set or in the second set is a solution. OK, let's go to a different review topic. This one deals with exponents. Let's just go over properties of exponents very quickly. Suppose I have 5 to the negative 2. What is another way of expressing that in a more familiar form? Anyone? Yeah, Jeff? 1 over 5 squared. 1 over 5 squared, yeah, 1 over 5 squared are, in other words, 1 over 25. If you have a negative exponent, that just means you invert the 5 squared so you put that on the bottom. What if I had 3x to the 0 power? What does that mean? Or at least, how could I simplify that? Steven? X to the 0 power or anything to the 0 power is 1. So the answer would just be 3 in that case. OK, that'll be 3 times 1 or 3. You know, there actually is 1 qualification on that. Not anything to the 0 power is 1. 0 to the 0 power isn't defined. It has no meaning. If you take 0 to the 0 power, it's not 0. It's not 1. It just doesn't mean anything in mathematics. What if I put parentheses on that and said 3x quantity to the 0 power? What would that be? 1. That would be 1, yes, because now everything's raised to the 0 power. And so assuming x isn't 0, 3x to the 0 power is 1. What about a fractional exponent like 9 to the negative 3 halves power? Now, first of all, what does that even mean when you put a fraction in the exponent? That's a 3 right there. Anyone? Who can explain? Not just what the answer is, but what this even means. OK, I'll explain it. First of all, it's 1 over 9 to the 3 halves, because we have a negative exponent. And then the 1 half portion means take the square root of 9, but there's a 3 on that, so then I have to cube that answer. So the 1 half portion of the fraction means take a square root, and the 3 means to cube it. So this is going to be, let's go down here, this is 1 over 3 cubed, and that's 1 over 27. Now, another possibility is instead of taking the square root of 9 and then cubing it, you could have as an alternative, you could have cubed the 9 and then taken the square root. I didn't do that because cubing the 9 makes it awfully big, and knowing the square root, which would end up being 27, is something I wouldn't have known right offhand. Let's take one more problem that involves exponents. How about this one? 6y squared over 2y to the negative 3. Let's see. To begin that problem, well, who can tell me, what would you do to begin this problem if I want to reduce this? Stephen. I'd distribute the squared into the parentheses. OK, what would you get in the numerator? 36y to the fourth. 36y to the fourth. So what he's doing is he's squaring the 6 and he's squaring the y squared. So he gets 36y to the fourth. What about the 2y to the negative 3? What would you do with that? Now, what's raised to the negative 3? The 2y or just the y? Just the y. Just the y. So what I'm going to do is put the y cubed up on top. This is a multiplication sign right here. But the 2 didn't get to move because the 2 didn't have a negative on it, rather than a dot. I think I'll use a parenthesis on that. So now if I just reduce this, 36 over 2 is 18. And y to the fourth times y to the third is y to the seventh. Yeah, y to the seventh. 18y to the seventh for that. OK, let's move on to simplifying radicals. You know before we do this, let's go to the next graphic there. And let me just tell you a little bit about what's available as a resource for this course. As you probably know from the information that you've been supplied, there's a website that accompanies this course. If you go to the UVSE website, if you go to math the 1050 telecourse web page, and then you go to episode one, there will be a number of pages that you can go through on your computer to see some of the things that we've demonstrated right here. But along with the review discussions on the website, you'll see 18 problems that I've made up along the way. Those are not to be turned in, but they're supposed to be for your benefit. You can sort of practice some of the things that we've been doing here. So if you look on the website, you might try working those 18 problems that I have available for you. And the answers will be in episode two. I won't discuss the answers in the classroom here, but if you look on the website for episode two, you'll find all the answers for those problems. You can check your work. And if you find a few of those a little difficult, that's okay. I don't think that's too serious. But if you have trouble working all of them, I think that means you might want to reconsider whether you want to take math 1050, or you may want to drop back and review math 1010 as a possibility. But some of those problems I think are fairly easy. Some of them are rather complex, and it's supposed to give you a sort of a general review of a number of ideas. Okay, let's go to simplifying radicals. Now, when I mean a radical, I mean a square root, a cube root, something like that. Suppose I have the square root of 75x cubed. Now, this radical isn't simplified because I can find some square factors in each of these numbers. In 75, there's a square that divides it. In an x cube, there's a square. So what I'll do is I'll factor out the 25 and an x squared. And what would be left over is my missing factor here. 3x. 3x, okay. One of the properties of square roots and cube roots and other radicals is that you can split this into the square root of the first factor and the square root of the second factor. By the way, that only works for products. It doesn't work for sums. If I were adding these two things together, I couldn't add the two square roots together separately. This first one I can reduce, that's gonna be 5x. By the way, I guess it should be the absolute value of 5x, but I'm gonna assume that x is positive. I guess x would have to be positive over here, wouldn't it? Because I wouldn't be taking a square root of a negative number. And this would be 5x times the square root of 3x. So as a general rule, if within a radical, you can find a power that you can remove from the square root, then you wanna do so. Another possibility is what if I have, let's say, a over the square root of 2a. Now as a general rule, we say you don't leave a radical in the denominator of a fraction. You try to, if you're gonna leave a radical, leave it in the numerator. And so to get rid of the radical in this denominator, I would multiply, in this case, by another square root of 2a on top and bottom. And so what that's gonna do is to essentially square the 2a, and then I'll take the square root of the square. And so on the bottom, the square root of 2a times the square root of 2a, well, I guess we could put in one step here and say that's the square root of 4a squared. And on top, this is a times the square root of 2a. Now if I reduce the square root of 4a squared, once again, I'm assuming a is positive. So I have a times the square root of 2a all over 2a. Now will these two 2a's cancel out? No, they won't. One's under the radical, one's not. But these two a's will cancel out. I can cancel those two. And so my simplified answer is the square root of 2a over 2. Well, now I have the radical on top. This expression looks quite different from the original expression, but they have exactly the same algebraic values. You know the whole purpose of these ideas of simplifying radicals is let's say I'm in a chemistry lab and I'm working a problem and I get this answer, and let's say my lab partner gets this answer. And so we think, hey, one of us made a mistake. We didn't have the same answer because they look so different. But they're actually the same thing. So this is merely a way of sort of standardizing the answer so that we all have answers that look the same rather than surprisingly different. Another possibility. What if I have the square root of 3 over 8? Now generally we don't leave fractions or rational expressions under a square root. So I want to somehow eliminate that fraction. The first thing I do is separate this into a ratio of two square roots. And you see now I'm actually back in this first category where I have a radical in the denominator. But before I continue, I think I'll simplify the square root of 8. Can anyone tell me a way to simplify that in the denominator? Two times four? Yes, eight is two times four. So when you simplify it, how would you write it? You'd put two times the square root of two. Two times the square root of two. And you know that's actually like this first example we did up here where I was removing a square from the radical. And on top I have the square root of three. So I've reduced this radical somewhat, but I still have to get rid of the square root of two. So I'm gonna multiply, I'll just do it right here, I'm gonna multiply by another square root of two on top and bottom. And this gives me now the square root of six over what? Over four. Yeah, it's over four. Another way you could write that is to say that's one fourth times the square root of six. I mean either one of those forms would be a good answer. Those are all equivalent to the square root of three eighths. They don't look anything alike, but either one of these is said to be reduced and this first one obviously isn't. Okay, I think there's a graphic about reducing radicals that we could put on the screen now that kind of summarizes this. If you're simplifying radicals, here are the various things you do. Number one, simplify expressions that have rational exponents. Let's see, I haven't worked an example of that, but I think there's an example on the website for that one. Then next, remove powers from radicals. That was like that first example I worked. Eliminate radicals from the denominator. That's like the second example we worked. Remove rational expressions from the radical, like the square root of three eighths, three eighths is a rational expression. And then finally, combining rational expressions. Let me just work one example like that before we go on. Suppose I have the square root of 18 C cubed, and I wanna add to it the square root of C over the square root of two. Now, you see at the moment it doesn't look like, I'm gonna put a C out in front of that. C times the square root of C over the square root of two. Now, at the moment it doesn't look like these things are anything alike, so we can't do anything with them. But if I reduce each of them, I think I can combine them. In the 18, I can factor out a nine, so I'll put a three outside. In the C cubed, I have a C squared, so I'll put a C outside. And I'm assuming that C is a positive number here, so the square root of what's gonna be left over? Two C. Two C, very good. Because we took out a nine, that leaves a two. We took out a C squared, and that leaves a C. Now, in the other expression, I have a square root of two on the bottom, so I'm gonna multiply top and bottom by the square root of two. And this first one's already reduced, so I'll just rewrite that. And in the second one, I have C times the square root of two C over two. So if I want to combine these, it looks like I have a common factor, namely I have a square root of two that I could factor out, a square root of two C. And I also have a C that I could factor out, and that's gonna leave a three here. What's it gonna leave right here? One half. One half. Now three plus a half, three and a half, why don't we call that seven halves? Seven halves, C, square roots of two C. So I've been able to add two expressions that look totally different by actually making them look somewhat alike and getting a common factor to factor out there. Okay. One more thing that's gonna come up in this course that I think I should probably mention to you, and that is, we'll be solving quadratic equations. I bet you know how to do that already, but we'll also be solving expressions or equations that are in a quadratic form. Take, for example, this possibility. What if I have a to the sixth plus seven a cubed equals eight? Now at first glance, you wouldn't say this is a quadratic equation because I have a sixth power and a third power in here, but this is in a quadratic form. If I think of a cube as my basic variable, then this is the square of a cube, and so I'm gonna rewrite this as a cube squared plus seven a cube, and then minus eight equal zero. So if I think of this as being sort of my first degree term, then now I have the square of the first degree term, I can factor that into two factors. Each one of them begins with an a cube. So I'm treating it sort of like a quadratic equation, and in fact, this is said to be in a quadratic form. This sign is negative, so that tells me that one of these is positive and one of these is negative. So I'm looking for two numbers to plug in here, whose product is eight and whose difference will be seven. What do you think the numbers will be? Eight and negative one. Eight and one. Do you think I should put the eight here or should I put the eight over here? Put the eight and put the positive. With a plus, yes, because I wanna get a plus seven a cube and a one here. Now, this represents a single number, this represents a single number, and the product is zero. So one of those numbers is zero. Either a cube plus eight is zero or a cube minus one is zero. Now, in the first case, this is the sum of two cubes. I have a cube plus two cube. That'll factor. Does anybody know that formula right off hand? It's not one you see a lot. How to factor the sum of two cubes? Okay, I'll do that one. It's gonna be a plus two times a squared minus two a plus four. And that's equal to zero. Now, either the first factor here is zero or the other factor is zero. If this factor's zero, what would a have to equal? If the first factor's zero? Negative two. Negative two. If the second factor is zero, I'd have to use the quadratic formula. Now, let's see. Let me just erase this and I'm gonna solve that on this screen. That was a squared minus two a plus four. We were wondering if that had any factors or any solutions for it. So in this case, I'd go to the quadratic formula to see if it has a solution. So a would be, let's see, how does the quadratic formula go over here? If I have a x squared plus b x plus c equals zero, then the formula for x is what? Yeah, Steven? Negative b plus or minus the square root of the quantity b squared minus four a c. Minus four a c. All over two a. All over two a, exactly. Okay, it's a little confusing because a happens to be the variable over here. But if I substitute in the numbers in the right places, negative b, that's gonna end up being a positive two, plus or minus the square root of b squared, that's four, minus four times a, well now that's the coefficient of this term, times one, times c, that's a four. All over two a, that'll be two times one. So this is two plus or minus the square root of four minus 16 over two. Now you know as soon as I see that there's a negative under the radical, I'm gonna stop because I'm looking for real number solutions. Otherwise, I'd end up with some complex number solutions. So the only solution I'm gonna get for that one side is a equals negative two. Okay, but there was another half to the problem. Let's see, we got a equals negative two and the other possibility was there was no solution for that quadratic. But the other factor was a cube minus one equals zero. And this is the difference of two cubes. So I'm gonna factor it into a minus one times a squared plus a plus one is zero. Now if the first factor's zero, then a is equal to one. And if I set this other factor equal to zero, I would use the quadratic formula and it'll turn out that that has no solution also. There's no real number solution. So the only solutions I get are negative two and plus one. So the way I'll write my final solution is to use set brackets and say negative two one set brackets and that means I'm listing those particular numbers. Okay, there are more topics that are reviewed on the website and if you'll take some time later today or tomorrow looking at the website, you'll see examples very much like these worked out in detail. So it's not just the problems but some more examples. And then there are those 18 problems that you can try if you have the time to review and sort of check your memory on some of these ideas from Intermediate Algebra. Okay, let's move on to yet another idea from Intermediate Algebra and that's the notion of a function. And let me just tell you what are the two characteristics of a function. Basically a function has a rule and a function has a domain set. So first of all we have a rule and we have a domain. Now basically a function is a correspondence between two sets, the domain and the range so that each member of the domain is assigned to exactly one member of the range. Let me give you an example. Suppose this is the domain set over here, there's various numbers let's say in there and this is the range set over here. I'm gonna pick a few numbers in the domain. I'm gonna tell you how they're assigned and I'm gonna see if you can guess the rule that I'm using. For example, if I choose the number two, let's say that's the number two, it's gonna be assigned by way of f to a number, let's say seven. And if I choose the number five, it's gonna be assigned to 16. And if I choose the number one, it's gonna be assigned to the number four. Now you can see this is already getting crowded. If I put any more numbers in there, all the arrows are gonna sort of get mixed up. So I'm gonna abbreviate it this way. Two is assigned to seven, five is assigned to 16, and one is assigned to four. Does anybody see a pattern yet? Okay, I'll tell you what Steven, let me give you another number in the domain. I want you to tell me what it's assigned to. Suppose I choose the number 10 in the domain. What's it gonna be assigned to? 31. 31, is that what the others of you were thinking? 31. In other words, you have in mind a rule that I had in mind too. What's the rule that you had in mind? Y is equal to three x plus one. Three x plus one. Yeah, you see, if I were to pick just a sort of a general number, x, then what I was doing was tripling the number and adding one. So it's assigned to triple the x and add one. Now this second number is sometimes referred to as y, but for functions, it's usually written this way, f of x is three x plus one. Now that's the rule of the function right there. And I had that rule in mind when I started giving you these examples, and after you saw a few examples, you kind of caught on to the pattern and you recognized what the rule was. But one thing I haven't told you is what is the domain? Now if we, let's go to the next graphic that we have on the screen. If the domain isn't specified, then we generally assume that the domain is the largest possible set for that rule. Now as long as we're working with real numbers here and not imaginary numbers, not complex numbers, it looks to me like you could essentially plug in any number you want, any real number, and you could triple it and add one. So to go with this set, we would assume that the domain would be all real numbers. There's an abbreviation for that and that's sort of a Gothic R. Another possibility is you could use that interval, the symbols I was using. That would be the open interval from minus infinity to plus infinity. You could write that. Or you could just spell it out in words, all real numbers or the set of all real numbers and that would be fine. If I wanted to place a restriction on this domain, for example, someone may think that I'm only picking positive integers because all my examples chose positive integers, I'd have to specify that. Because when you write down the rule, if the domain isn't mentioned, then we assume it's the biggest possible set. Okay, well, let me give you a couple function rules and I'd like for you to tell me what you think is the largest possible set that could be the domain. Suppose I had the function, I think rather than using F, I'll call this one G. Suppose G of X is equal to three X squared minus five. There's no domain mention. So what's the biggest possible set that could be for the domain? Susan? All real numbers? All real numbers, yeah. So it looks like that the domain of G is all real numbers. Okay, now in the next case, I'm gonna call the function, let's say we call it H, and I don't always have to use the letter X, let's say we use the letter T. And in this case, I'm gonna say negative one over T squared. For example, for this particular function, what is H at two? What is H at two? Negative one fourth? Negative one fourth, yeah, it's negative one over two squared or negative one fourth. What is H at five? Negative one 25th. Negative one 25th, yeah, negative one 25th. And what if I choose a fraction? What if I say what is H of negative one 10th? That one's kind of tricky. Let's see, that would be negative one over negative one 10th squared. And that would be negative one over one over a hundred. And? 100? Close, negative 100. Negative 100, yeah, you put the negative on there. Negative 100 in that case. Well, you know, you might say, well, gee, Dennis, it looks like you could pick anything there for T. I don't think so, there's a restriction on the T's. So what is the domain for this function? Susan, what do you think? All real numbers but zero. All real numbers other than zero. Yeah, now let me show you how we could write that. See, you could write that out in words, that would be fine. You could say the set of all real numbers except zero. But another way to write that is to say the set of all real numbers minus the set containing zero. In other words, you're excluding this set from that bigger set. Another possibility is you could use the interval abbreviation and say minus infinity up to zero, not including zero, union the numbers bigger than zero. So that's open parenthesis there, zero to infinity. And of course, you could write it out in words. One more possibility. Suppose I have a function capital F of U. I'll call the function capital F and for the variable I'll use a U. And let's say this is the square root of U minus three. U minus three. Now what would be the domain of this function F? Yeah, Stephen. All numbers three and greater. Okay, tell us how you got that. That's the right answer. Well, underneath the radical, the number can never be less than zero, or else you'll get an imaginary number. So U minus three, this number, U minus three has to be zero or larger. So it comes down to solving this inequality. And so U is greater than or equal to three. So when I go to write the domain, how am I gonna express all the numbers that are three or larger? Well, I guess that'd be square brackets three to infinity. You notice how we're using the parentheses and the square brackets to represent these sets. So it's a convenient way of expressing your answer. So you wanna keep that sort of abbreviation in mind. Okay, now suppose I go to this example. Let's say I have F of X equals two X minus one. In this case, I would call this a linear function because this is equivalent to saying Y equals two X minus one. It'd be equivalent to that straight line. And I can draw a graph of this function by saying it crosses the Y axis at negative one and the slope is two. So if I go over one and up two, I get another point there. And now I have a graph of my function. And if every X and Y coordinate paired together, you've seen this before, paired together make a point, every one of those points lies along this diagonal line. So this is sometimes referred to as a linear function. Now I think what's significant here is we're not always given the rule of the function and the domain of the function. Sometimes we're given the graph. So let's go to the next graphic. And we'll look at a representation for a function, but we don't know its rule. In this graphic, we're given John's weight, W of T at time T, where T is measured in years. So along the horizontal axis, that would be the T axis. And along the vertical axis, that would be the, shall we say, W axis because that gives the weight of John. And what would you say is generally the pattern in John's weight over the years in general? Yeah, David? In general, it's increasing. Yeah, it looks like it's generally increasing, although there is sort of an aberration that went on about what age? 30. Yeah, around 30. Yeah, I mean, if you were gonna make up a scenario as to what happened around age 30, how would you describe that? I mean, we don't know exactly what the story is, but generally what happens to John's weight? It drops. Looks like it drops and then it increases a little bit. And then it jumps way back up and then it just keeps on claiming again. So what happened around age 30? Well, gee, I don't know, maybe he went on a diet and he lost maybe 20 pounds, but then he slowly gained a little bit back and then maybe he gave up after maybe two or three years and he gained some weight back and then it started going back up again. Now, we don't know that that's exactly the story that happened, but you see that graph does suggest a lot of information and we don't even know the rule, but a picture, as they say, is worth a thousand words. I think we have another graph that we can put up there too. Yeah, okay, here's another story told in a graph, a graph of a function. It says this graph gives a salesman's distance from his home as a function of time on a certain day. Describe in words what the graph indicates about his travels on this day. Well, for example, at what time did he start off in his travels? 8 a.m. 8 a.m. The equivalent to the origin. What time did he get home? 7. It looks like about 7 p.m. Now, what happened there between 9 and 10 o'clock? The graph's horizontal, what does that mean? Probably had a meeting or something. Yeah, it looks like he was at some place for an hour, maybe a meeting, maybe he was at his office, whatever, but it looks like he wasn't traveling any further, at least he wasn't getting any further away from home, but then he started traveling further away. Now, what happened at noon? Lunch, probably. Yeah, it looks like maybe he stopped for lunch from 12 to 1, because he's not moving further away. We don't know specifically that he stopped for lunch, but it's hard to suggest that in the graph. And then for some reason, he started coming back toward home, but he turned around and he went back, so about 3 o'clock he headed back out and he went further out. We don't know why, but it happened. From 5 to 6, maybe he had dinner, and then he came straight home. He must have come home in a hurry, because in an hour he came back all that distance. Maybe he's on the freeway. Maybe the traffic's really flowing right now. So he got home pretty quickly. So that graph tells us a story that even if we were given a rule, I think the rule would not convey as well as the graph would. Something about this graph is I mentioned early on, I said this is the graph of a function. How do you know it's the graph of a function, as opposed to something that isn't a function? Yeah. Because it passes the horizontal line test? Not the horizontal line test. Yeah, David, tell us what is the vertical line test? How does that go? Basically if you draw a vertical line at any point along the line and two points, there's not two points within that line, within the vertical line? Yeah, if it doesn't cross the graph at two points or more, then it's a function. Whereas if a vertical line crosses it at two or more points, then it's not a function. Let me just show you here on the green screen how that would work. Suppose we have our two axes. Suppose I have a curve, but let's say the curve looks like that. This immediately we know does not represent a function because if I put a scale on here, one, two, three, four, five, let's say that's five right there. If I draw a vertical line, let's say right here at two, this is the vertical line, x equals two. That vertical line crosses this graph three times. It crosses it here, it crosses it here, and it crosses it up there. Now this point might be the point two, two. Let's say, maybe this is two over here. This point might be, I don't know, what would you say? Two, three, okay, let's say that's two, three. And this point might be, I don't know, two, five. We'll say for the sake of argument, two, five. Now what that means is if I go back to that first illustration I drew, two is a sign to five, two is a sign to three, and two is a sign to two. Two is not a sign to some unique member in the range. In fact, two is a sign to three different members of the range. That tells me this is not a function. It's a graph, it's obviously a graph, but it's not a function. Now when we looked at John's weight, when we looked at the traveling salesman's problem, both of those graphs, a vertical line would cross it at most once. I mean if you go way out further out, like for John's weight, if you go up to 200 years, John's not around after 200 years. So a vertical line wouldn't even cross the graph. There's no graph out there. John has no weight, John's gone now. So it still doesn't cross the graph twice, so that does represent a function. Okay, we have one more graphic let's look at. This is what's called a piecewise function, and it says that f of x is equal to x squared if x is between zero and five. That's including zero and five. But on the other hand, f of x is always four if x is between five and less than or equal to eight. This is called a piecewise function because it's chopped into two pieces, and when you go to evaluate f of x, you have to decide where the x lies before you can compute the function value of it. For example, what would be f of three? Well, three lies between zero and five. And so the rule we use is f of x equals x squared. So f of three would be what? Would be nine, okay. What about f of seven? What would f of seven equal? Four. Four. That's because seven is between five and eight, and so the function value is always four. Okay, here's a trick question. What's f of 8.2? Undefined. It's undefined, why is that, Jeff? Because 8.2 is not in our range of values for us. Exactly, because 8.2 is too big. This function isn't defined at 8.2. And finally, f of the square root of 20, well, let's see, the square root of 20, is that smaller than five or bigger than five? Smaller than five. Smaller than five, and therefore, f of the square root of 20 is? 20. Is 20, exactly. The last question is, what's the domain of this function? All numbers from zero to eight. All numbers zero to eight, would that be with square brackets, or with parentheses? Square brackets. Square brackets, yeah, this is the closed interval from zero to eight, in this case. And 8.2 was outside that interval, and therefore, that's why we didn't have a function value at 8.2. Okay, well, let's see, we have one more graphic. Let's just take a quick look at this one, too. We mentioned earlier that any function of the form f of x equals mx plus b, we call a linear function. And you sometimes hear expressions like, if you go to that middle quotation, it says the cost c of t equals mt plus b is increasing linearly with respect to time. You sometimes hear expressions like things are increasing linearly. What they mean is it can be represented by a linear function. Okay, well, you know, class, I think our time is almost up. Let me just kind of summarize what we've done today. First of all, I mentioned that there's a website that goes with this course. And if you look on that website, you'll find a lot of the information we've discussed here, some other examples, and some problems for you to try working on your own. Those problems are not to be turned in for a grade. That's merely for you to have an opportunity to sort of review intermediate algebra. In coming episodes, we're gonna be talking about the Cartesian plane. We're gonna do quite a bit of graphing, and we'll be looking at some fundamental graphs on the Cartesian plane. In fact, that's what episodes two and three are gonna be dealing with. One more thing, class, before you come to episode two, you might wanna bring a calculator with you because we'll be doing a little bit of computing on the calculator that would be associated with some of the graphs we're gonna draw. And I'll see you next time.