 I'm Dennis Allison in the Mathematics Department at Utah Valley State College, and this is episode eight of Math 1050 College Algebra. The first seven episodes have covered material that was the beginning of this course. Episode eight is strictly a review of material that will be on the first exam. Now let's call up the first graphic and I'll show you a few items that you need to be aware of. First of all, you cannot use a calculator on this exam. We have used calculators occasionally in class to draw graphs and we've used the graphing calculator for that. That has been supplemental and you will not be allowed to use a calculator on this test. There is a test later where you will be allowed to use a calculator but I'll advise you of that as the time comes. Also, you cannot use notes and you cannot use your text, of course. And also all of your scratch paper will need to be turned in. There will be some scratch paper that accompanies your test and you should turn all of that in with your exam when you're done. Now episode number one was a review of material from Math 1010 Intermediate Algebra and none of that review material is on this test. That was merely to sort of interface this course with the prerequisite course to give you a chance to just look over some ideas and some information that will be useful in this course. So this exam begins with episode number two and let's look at the graphic for episode number two. I have listed here several items that you need to be aware of from episode two and we'll talk about these in just a moment. I should mention that if you go to the website you'll see more items than this listed for episode two but I just picked out a few that I wanted to talk about today. So in episode number two you should know the distance formula and the midpoint formulas. There were two midpoint formulas. You should know the center radius equation of a circle and you should know how to write the general equation of a circle and you should be able to sketch the solution of a circle of a circle inequality. So let's come to the greenboard here and let's just go over some of this information. You remember there's a distance formula D equals the square root of X2 minus X1 squared plus Y2 minus Y1 squared. That's called the distance formula and then there were two midpoint formulas to find the midpoint between two points X1, Y1 and X2, Y2. So X bar is equal to X1 plus X2 over two. In other words it's the average of the X's and Y bar is equal to Y1 plus Y2 over two. Now these formulas won't be given on the test and you're expected to know these and be able to use them. So let me work a couple examples that where I have to use these formulas and then we'll look at the circle formulas. Suppose I have two points. Let's say a point P whose coordinates are seven, four and a point Q whose coordinates are one and negative six. And I would like to find the distance between those two points. So I'll choose one of these to be point number one, one of these to be point number two and I will substitute in X2 and X1 into the distance formula Y2 minus Y1 into the distance formula and the distance will be the square root of one minus seven squared plus negative six minus four squared. And that's equal to the square root of, let's see, negative six squared is 36 and negative 10 squared is 100 and so this is the square root of 136. And of course we don't have calculators available on the test so you can't approximate this but you could simplify the radical. For example, it looks to me like a four should factor out of this and so if I put a two on the outside this would be the square root of, let's say I have to divide by four. So that's gonna be three, four, 34. So two times the square root of 34 and the square root of 34 can't be reduced any further so this is the distance between the two points. If you get your answer to the square root of 136 and you don't reduce the radical I may take off a point or something for that. It wouldn't be major but I would want you to reduce the radical when you can. Now the other question related to this is what is the midpoint between these two points? So I'm gonna use the formula over here and to get x-bar to find the midpoint I'm gonna average the two x's that would be seven plus one over two is four and to get y-bar I'll average the two y's and that will be four plus negative six over two. That's negative two over two or one. So the midpoint is the point four one. If you were to plot point p, point q and the midpoint you would see that the point four one is on the line on the line segment connecting p and q and it's exactly halfway between p and q. Okay, now for the circle equations there are two fundamental equations for a circle that you'll need to know. The first one is called the center radius equation. Suppose the center is at the point hk and suppose the radius is equal to r then we can immediately say the equation of the circle is x minus h squared plus y minus k squared equals r squared. So if you were given h and k and r you could substitute that in and that's the center radius equation and you could leave your answer in this form you don't even have to multiply out the squares and that would be an acceptable answer. Now if you do multiply out all the terms you would get an x squared term from here. You'd get a y squared term. You would get a multiple of x that would come from this term when you multiply it out and you'd get a multiple of y that would come as the middle term of the square of this plus a number of constants which when added together will call c and I'll set that equal to zero. For example, h squared is gonna be a constant k squared is a constant and if you subtract off the r squared that all goes into affecting the value of c. This is called the general equation of a circle. It's also called the standard equation of a circle. Now let me just take a couple examples of problems where we would use these equations. Suppose the center of a circle is at the point four negative two. And suppose the radius of the circle is five. What's the equation of the circle? Well the center radius equation would be x minus four squared plus y minus negative two so I'll write y plus two squared equals 25. That's the equation of the circle. Now if I multiply this out and collect terms and set it equal to zero I'll have the standard equation of the circle. So this would be x squared minus eight x plus 16 plus y squared plus four y plus four equals 25. Now if I rearrange terms I'll have x squared plus y squared minus eight x plus four y and looks like we have plus 20 equals 25. But I like to set it equal to zero. So I'll write this as x squared plus y squared minus eight x plus four y minus five is zero. And if you compare that with the standard equation that I wrote here a moment ago it looks like a is negative eight. It looks like b is equal to four and c is negative five. So here is the standard equation and here is the center radius equation of a circle. I may ask you to write either one of those equations given information like this. Now suppose the information is given to you in a different form. And I think you've had some homework problems that are something like this. Suppose I told you I'm thinking I would like to know the equation of a circle that has a diameter between the origin and the point four six. Four six is right here. So in other words, if I were to draw a line segment connecting those two points that's a diameter of a circle. Now let me just sort of sketch in the circle lightly. So the question is what's the equation of that circle? Well let's see what do I need to know in order to find the equation? I need to know what is its center and what is its radius? Now its center would be the midpoint of the diameter. And we have a formula for calculating the midpoint. You remember x bar is x one plus x two over two. Well let's see x one plus x two that's four plus zero over two is two. And y bar is y one plus y two over two. And that would be six plus zero over two is three. So the center is at the point two three. Now what is the radius? Well to find the radius I need to use the distance formula. I need to find the distance from the center out to the point four six. Or if you prefer find the distance from the center to the point zero zero. I think I'll find the distance from the center to four six. I'll do that over here. The distance will be the square root of, let's see now let's write that in here two three. So that will be four minus two squared plus six minus three squared. This is the square root of two squared or four plus three squared or nine. And that's the square root of 13. So if that's the distance between these two points that's the radius. So the radius is the square root of 13. Therefore I can write the equation of the circle in center radius form. X minus two squared plus y minus three squared equals the square root of 13 squared. Well when you square the square root of 13 you get 13. And so I'll take this to be the answer because if I just say find an equation for the circle the center radius equation is fine. If I say find the standard equation of the circle you'd need to multiply this out and collect all the terms and set it equal to zero. Okay let's go to episode number two. Excuse me episode number three and we have a graphic for that. In episode number three we covered a wide variety of topics and here are a few of the things that I thought we would talk about here. Finding the domain and the range of a function based on its graph. Being able to sketch the eight fundamental functions using target points only. That is not making a table but only the two or three target points. Being able to graph piecewise functions. Knowing that there are three types of symmetry. Symmetry about the y-axis, about the x-axis and symmetry about the origin. And then knowing that some functions are called even or odd functions based on their symmetry. Well let's take a few of these items. To find the domain and range of a function let's say I have the graph of a function and suppose the function graph looks like this. I'm gonna call this function f. Now it begins at negative two and it ends at plus two. And it goes up as high as two. Looks like three is even higher and it goes down maybe to negative one. So the question I might ask you is what is the domain of this function? Now we indicated in class that if you take the graph of a function and if you press it onto the x-axis you would cover the interval from negative two to two. Those are the x's that are used and so the domain would be the closed interval from negative two to two. Because that's what's covered along the x-axis. How could I find the range from this graph? What I would do would be to compress the graph onto the y-axis and it looks like it would cover the interval from negative one to two. So this would be the closed interval from negative one to two. So I would have to know where that highest point is or at least be able to approximate it from the graph and where this lowest point is or at least be able to approximate it from the graph. So we can find the domain and the range of a function by looking at it to graph. Okay, now there were eight fundamental functions that you should know and there are target points for each one. Now just to turn the tables and make this seem a little different than the way we explained it in class I'm gonna draw the graph of the function and I want you to be thinking about what function is this. So the first function I might draw looks like this. There are target points at the origin, at one one and at negative one one and the graph looks like this. And I think I'll call this function capital F and this is the x-axis and this is the y-axis. Now we've seen that function in class, what function is it? I think it's the absolute value function. So on the exam I may draw the graph of one of our eight standard functions, eight fundamental functions and ask you to name what function it is or I may give you the function and ask you to sketch the graph. Now if you sketch the graph I want you to graph it using the three target points in this case and not making a table. For example, I don't wanna see you graphing the point two, two, three, three, four, four. That's a rather primitive way to graph it and we're trying to speed this up by knowing the target points and knowing the general shape and being able to graph it quickly. Let's take another example. What function is this? This one only has two target points and that probably gives it away already if you remember your fundamental graphs. It goes through the origin and it goes through one one and the graph of the function looks like this and I'm gonna call this function g. What function is that? It's the square root function. It's g of x equals the square root of x. It has a limited domain. The domain is the set of all numbers from zero to plus infinity including zero and the range, if I press it onto the y-axis, is all real numbers from zero up to plus infinity along the y-axis. Okay, now to ask a more straightforward question like we saw in class, what if I give you the rule of the function and ask you to sketch the graph of it? So let's say this is the function g of x equals x cubed. Well, in this case, I would mark off my scale and I would have to remember the target points for x cubed and they are zero zero, one one, and negative one, negative one. And then I have to remember the shape of the function and if you remember it comes down steeper than a parabola. It turns and it levels off at the origin rather than going back up. This one turns and goes down. The reason for that is when you cube a negative number, you get a negative result. So when you pick a negative x, you get a negative y and this is the graph of g. So you notice I didn't make a table of values, I just remembered what the target points are and I drew the graph based on that. Okay, another idea from episode number three was the notion of piecewise functions. So suppose we were given this piecewise function and this one has three pieces to it. Suppose it's equal to two whenever x is less than or equal to zero. But suppose it's equal to x whenever x is bigger than zero but less than or equal to four. And suppose it's equal to x squared. I'll tell you what, let me make that not x squared. Let me make that, let's say the square root of x whenever x is bigger than four. So whenever x is bigger than four. Now you see, this is gonna require that I graph pieces of three functions. There's the function f of x equals two. There's the function f of x equals x and there's the function f of x equals the square root of x. And somehow I'm gonna graph all of the, let's say the square root's in the wrong place. I'm gonna graph all of these in the same graph and I'm gonna just take splices of each one and put them together. So here's my axis system. There is a portion of the graph that I draw when x is less than zero. There's a portion of the graph that I draw between zero and four. And then there's a portion of the graph that I draw for x is bigger than four. Now, for x is less than or equal to zero, I wanna draw the constant function f of x equals two. That's a horizontal graph that would go right through two and it would keep on going, except this says I only draw it for x is less than or equal to zero. So I put a solid dot there to indicate I'm including that last point. Between zero and four, I wanna graph the function f of x equals x. Now that's a 45-degree line through the origin. I have to put an open circle here because you notice this is only for x is bigger than zero and I draw a 45-degree line up to the point four, four and I'll put a solid dot there. And then after four, I graph the square root function. Now if you remember the square root function, we just drew this a moment ago, curls, turns out and goes out this way. And I believe at four, that function would have value two, four, two. I'm gonna put an open circle there and imagine that you see the square root function here and here's where it appears at that point. That's the square root function continuing, but I don't see anything in this portion because my rule said the square root function is turned on when x is bigger than four. So if you put this all together, this is the graph of f. It's called a piecewise function. These functions are very useful in applications of mathematics because in the real world, things aren't as simple as having one function that's used all the time, but on occasion you use one function, on occasion you use another function. Okay, let's see. Something else in this episode I believe was intercepts. Let me just check my list here. Yes, intercepts, no? Oh, symmetry, symmetry was the other thing. Here's a typical question that I could ask you on the exam. By the way, let me just say once more, I think I mentioned this at the beginning. If you look on the website, you will see this list of topics for each episode and then a few more things. I've picked out only a few of the items that are on the website to discuss in this one hour with you. So for a more extensive list, look on the website and you'll see this and more. Now, suppose I have a graph that looks like this and I tell you that this is only half of the graph, there's another half on the other side and the graph is symmetric about the y-axis. So let's say f is symmetric about the y-axis. So I would ask you to complete the graph and draw the other half of it. Well, when we say symmetric about the y-axis, that means that if I take it and just flip it over, it has the mirror image on the other side. So it's gonna come down to a point right about here and it's gonna go up to a peak right about there and it's gonna have an x-intercept right about there. I'm just sort of estimating it, obviously. In fact, that point's probably a little bit too far out. I'll move it back a little bit. So the rest of the graph should look something like this and you may say, well, Dennis, that's not an exact copy of the right-hand half. Well, it's close enough. If you can draw it this well, I would certainly accept it. So I just want to see that you understand what it means for a graph to be symmetric about the y-axis. You may remember that the function f, in this case, is said to be an even function. An even function is a function whose graph is symmetric about the y-axis. Okay, let's take another example. This time, what if I draw half of the graph and I'll draw that much of it. I'm gonna call this function g and what if you're told that g, or the graph of g, is symmetric about the origin. About the origin. Now that has special meaning and it has important applications to later courses in mathematics. So what I do is I take this graph and I flip it over the y-axis and then I flip it over the x-axis. So what's gonna happen is this function is going to turn and let's see, it's gonna turn and it's gonna go up and it's gonna go down sort of like that. Now, if you have a little trouble drawing that, I would say plot the lowest point here, go through the origin and there should be another point just like it on the other side. So put a dot there. And if there's an x-intercept here, go through the origin to the other side and put a point at the same distance away. And let's just pick a random point up here. If I go through the origin, the same distance on the other side, I should get a point right about there. And now, if I go back and connect these points, I have the other half of this graph that's symmetric about the origin. Okay, so you notice this course deals a lot with graphing. I wouldn't say that every episode is gonna have as much graphing in it as we've seen so far, but that's one of the basic features of college algebra is that it discusses graphs to a great extent. Just back to this function one more time, we might say that G is said to be a blank function. And I'm wondering if you can think, those of you at home, if you can think of what kind of a function this is, the answer is it's an odd function. An odd function is a function whose graph is symmetric about the origin. Okay. Now, what if I give you the rule of a function and I ask you whether it is symmetric about the X or the Y axis or neither one? Let's take this function. F of X equals three X squared plus the absolute value of X. Now, frankly, I have no idea what the graph of that looks like, but I can tell if it's an even or odd function or neither one, if I run a simple test. If you remember, for an even function, F of negative X should equal F of X. And if it's an odd function, F of negative X should be the negative of F of X. So I should check to see if this function satisfies either one of these rules. So I'm going to calculate F of negative X and see if I come up with either of these answers. F of negative X is, I'm going to put in a negative X here and a negative X here. That'll be three times negative X squared plus the absolute value of negative X. Well, you know, when I square negative X, that's no different than squaring X. So I could just call that three X squared. And when I take the absolute value of a negative number, that's the same thing as taking the absolute value of a positive number, so I can call that the absolute value of X. But wait a minute, that's the same thing as F of X. So what I have here is that F of negative X equals F of X and therefore this is an even function. Now you might say, well, what is the significance of that? Well, it means that if I draw the graph of this function on the right-hand side of the y-axis, whatever it might look like, if I just flip it over, I'm going to get that same graph or the mirror image of that graph on the other side. In this course, I can't go into all the applications this has, but this becomes a very useful shortcut knowing if a function is even or odd in the context of certain applications of mathematics. If you find out that F of negative X does not equal F of X and that F of negative X does not equal the negative of F of X, then the function is neither even or odd and there are lots of functions that are neither even nor odd and those functions have no symmetry about the y-axis or about the origin. It's just like some numbers are not even or odd. Four is even, seven is odd, but 3.2, 3.2 is a decimal that's not even or odd, so there are lots of numbers that aren't even or odd. Okay, let's go to episode number four and look at a few topics for that episode. In episode number four, I think probably the biggest thing we looked at was transformations of the fundamental functions. Now what I mean by transformations is vertical shifts, horizontal shifts, stretches and compressions vertically and horizontally, and then there were the reflections where you flip a graph upside down. You should be able to name the rule of a function based on its graph, that is a fundamental function or a transformation of a fundamental function based on its graph, and finally you should be able to find intercepts of a curve. Now, let's just look at some examples of this. Suppose you were to be asked on this exam a question like this and you know frankly, I bet you will. Not this example, but something like it. Suppose I wanna graph f of x equals three x squared minus two. Now we don't wanna make a table to graph this, we wanna use target points and what we know about fundamental graphs and transformations. The negative two tells me this graph will be shifted down to, and the three tells me there's gonna be a stretch. Now how can I say that? I'll just say stretch below that. So there's a stretch and there's a vertical transformation downward. So I'm gonna locate my target points for the fundamental function. And of course a good question is, what is the fundamental function? Well the fundamental function, let me kind of put it in a cloud like I'm trying to think of the fundamental function. The fundamental function before all these changes were made, before I shifted it down and before I stretched it, it was just x squared. And that has a graph that's a parabola and it has three target points, zero, zero, one, one, and negative one, one. Now I'm gonna shift that down two units, one, two. And then from this, what we've been calling the new origin, I'm gonna stretch the graph vertically. So if I go over one and up three, one, two, three, I get a point right there. And then if I go to the left one, I go up three, one, two, three, I get a point right here. So now I'm gonna draw this parabola, but it's a taller parabola is the way we described it. It may look thinner, but actually it's taller because we've stretched it. And this is the graph of F. It looks thinner than a typical parabola or the fundamental parabola over here would have looked. But that's just because we've stretched it three units. Okay, now you've seen me do lots of these examples in class, so let me change this. I'm gonna draw the graph and I wanna see if you can think of a rule like this that would represent that graph. Okay, suppose the graph, let's see, let me try another marker here. That one seems to be kind of fading out. Here we go. Okay, the question is, what function am I graphing here? What familiar fundamental function or transformation of it am I graphing? This function looks like this. I'm gonna plot a point here. I'm gonna plot a point here and plot a point here. And this function comes down like this. It levels off there and it turns and it goes down here. Suddenly it got really wide there at the bottom. And let's say this is the x-axis, y-axis. And let's say I call this function H. So my rule is gonna say H of x equals something. Okay, well, first of all, we have to decide what's the fundamental graph that we're looking at before it was transformed. Well, you know it looks to me like it's the cubic function. If you remember, the cubic function had three target points and the cubic function looks like this. And that's sort of the general shape of this one. Now, I would say it's been flipped over for one thing. It's been moved to the right one unit. And has it been stretched? I don't think so. When I went over one, I just went down one. And when I went back one, I went up one. It's been flipped over, but it hasn't been stretched. So now, how do I write that in the form of a rule for a function? Well, if I move it to the right one, that means I need to put x minus one. I need to subtract one directly from the x. And it's the cubing function. So I'll put a cube on it. And it's been flipped over. So I'll put a minus in front of this. And that's all you have to write. Negative the quantity x minus one cubed. Let's do another one like that. This time, I'll draw a function that looks like this. And this function has a point right here. And it goes down through the origin. And it goes down through, this is two right here. And this is three up here, that point is three. And I'm gonna call this function a little f. Oh, by the way, let's say we call this the t-axis. And this the y-axis. So that means I'm looking for a function that I'll call f of t. F of t. Well, I think you would probably agree this looks like the graph of an absolute value function but it's been flipped over. Just like that cubic function just before this was flipped over. So I'm thinking this is basically gonna be f of t equals the absolute value of t with a few bells and whistles added to it to make those changes. Let's see, it looks like we've moved to the right one. And it looks like we've moved up too. If I move to the right one, just like in the last example, that's gonna be the absolute value of t minus one. That moves the absolute value function to the right one. And I wanna move it up too. Okay, oh, let's see. By the way, this has been inverted so I'm gonna put a minus on here. And not only has it been inverted, it's been stretched. Look, if I go over one, I go down two to get to this point. So it's been inverted and stretched. I need to put a negative two on there. And then finally, it's been raised up two units. I'm gonna put a plus two out here. So I don't multiply the whole thing by two, but I multiply this portion by the negative two and then I add the two on at the end right there. Okay, so this is a function, or this is a rule of a function whose graph would look like this. It's the only one we can think of that would look like this. So that's the rule that we're looking for. You should be prepared to work problems like this on the exam. Okay, another notion that we had in the same episode was to find intercepts. Let me just mention this briefly so we can move on to episode five. If you're looking for x-intercepts, the way you do that, the way you find x-intercepts is you let y equals zero. If you're looking for y-intercepts, then you let x equals zero. And what I mean by this, if you're looking for x-intercepts, you let y be zero and you get an equation and you solve for x and any solutions becomes an x-intercept. If you're looking for y-intercepts, you let x be zero and any solutions become a y-intercept. Okay, let's go to episode number five and look at a few items for episode number five. In this episode, we talked about how functions are used as models of real-life applications. This is one of the fundamental ideas about all of mathematics and why you take college algebra, for example, in college, is because many other majors use material in college algebra in particular functions used as models. Then you should be able to interpret the graph of a function when it's used as a model and you should know that there are three types of variation. There's direct variation, inverse variation, and then there are compound or combinations of those compound variations. Okay, functions used as models. Let's take an example. Suppose I had a roll of fencing and I were gonna make a garden out of it, but a little different than the garden that you saw me make in episode number five. This time, I wanna make a rectangular garden and I wanna divide it down the middle in two places so that I divide it up into three pins like this. And let's say that I have 300 feet of fencing. Well, with 300 feet of fencing, I'm looking for a function that would tell me what is the area of this garden as a function of X? And you might say, well, what does X represent? Well, let's say X represents the length of this side over here. So if that's what X is, and if this is a rectangle, and if I've divided it in the middle with some of this fencing into two more sections, so I get a total of three sections altogether, then I would like to find a function that represents the area of that pin. I'm gonna call the function A because it represents area, so generally you name a function with a letter that reminds you what it's supposed to stand for. Well, I'm thinking this would have to be X, this is X, and this is X. So how much fencing is left over for these two sides? Well, I guess there would be 300 minus 4X left over and I have to divide that in half and put half above and half below, so I have to divide that in half and that gives me 150 minus 2X. So I put 150 minus 2X here and I put 150 minus 2X here. Now if you were to add up X and X and X and X and 150 minus 2X and 150 minus 2X, if you add those all up, you'll get exactly 300 because that represents all the fencing that I had to begin with. Now what is the area of this garden? Well, if I look at the big rectangle, it should be length times width. The length is 150 minus 2X and the width is X and that's the total area inside the rectangle. So I would say the area is 150X minus 2X squared and that would be measured in square feet. You don't have to multiply this out. If you wanna leave your area function as this product, that's fine with me, or you could multiply it out like this and by the way, you don't have to even put the square feet on the end. I just did it because it seemed to be appropriate to put units on the area function but this is a function that models the area of that pin. For example, what if I told you that X, that X here was going to be let's say 10 feet? 10 feet, 10 feet, 10 feet, 10 feet. What's the area of the entire garden? If I just plug in 10 into this, I'll have an answer. Plugging in 10 here, I get 1500. Plugging in 10 here, I get minus 200. So I get 1300 square feet. So that would be the area of the pin if this is chosen to be 10. Okay, along a different line, you may remember that we talked about a projectile problem in which we throw an object off the roof of a building and we find out how high the object goes and what time it hits the ground and so forth. That application used a formula that I would give you on the exam if I should ask you about this. This is a formula that was first discovered by Galileo around 1600 and the formula is minus 16 T squared plus V sub zero T plus H sub zero. Let me just draw a little illustration of this problem. Suppose we're standing on top of a building and the building is H sub zero feet tall. This is given in feet and we throw an object straight up into the air and it comes back down down the side of the building and eventually it's gonna hit the ground. And let's say that initially we gave it a speed, the initial speed, well actually velocity, the initial velocity that we gave it was V sub zero. The difference between speed and velocity is speed can't be negative, velocity can't be negative and if V naught were a negative number what that would mean is we actually throw the object straight down rather than straight up. So V naught could actually be negative in this problem. So H sub zero or H naught is the height of the building. V sub zero or V naught is the initial velocity and the minus 16 is always minus 16 on the earth. This has to do with the acceleration of gravity. This number would not change unless we went to outer space if you went to another planet where the acceleration of gravity were different, this number would change but for us it's just gonna be a constant. And this formula calculates the height of the object in feet. Okay, so if I were to ask you a problem like the example that we worked in class about a projectile being thrown off a building I would give you this formula so that's not one that you have to create. Now graphs of functions that represent real life situations. Let's take an example here, similar to some we did in class. In class we talked about the height of the grass as if you mow the lawn every Wednesday and you remember the height of the grass was sort of sloping upward, it would drop back down, slope upward, drop back down. Another problem we had was the traveling salesman as he travels away from home and then he eventually returns to home at the end of the day and you can read the graph to see roughly how far away from home he is at any given time. Let's take another problem. Suppose this time we have an oven. Let's say this is an oven here and here's the oven door and I'm gonna put a baked potato in the oven. So this is the baked potato. And so we leave the baked potato in the oven for quite a long time, it's really hot and we bring it out into the room and we allow it to begin to cool. I wanna draw a graph that represents the temperature of the baked potato over time once I bring it out into the room. So the graph would look like this. This is T measured in hours, time measured in hours. And this is capital T for temperature measured in degrees Fahrenheit. And I'm not gonna draw a negative T axis because this problem begins at the moment that I take the potato out of the oven and I take it out of the oven at the moment T equals zero. So that's when the problem begins. And I'm not gonna draw a negative T axis because that would have it going G well below freezing. I don't think that's gonna happen to the potato. It's gonna be very hot and it's gonna cool. Now how do you think it would cool off? What would the graph look like? Well I think it would begin at a rather high temperature. And let's say when I first take it out of the oven let's say it's at 220 degrees Fahrenheit. Now you know if you bake a potato for let's say an hour in a 350 degree oven that potato will not be 350 degrees. It'll be cooler, a little bit cooler inside. But it's cool, it's hot enough to cook it. So it starts off at 220 degrees. And then the temperature begins to cool off. Cool off and it cools off and it cools off. And you know what it begins to level off. And it begins to level off at a line that I'll draw along here. And it gets cooler and cooler and cooler. But it never drops below the dotted line. And you know what the dotted line represents room temperature. Now let's say the room temperature is 75 degrees all around the oven. So this is 75 right here. And the potato approaches 75 degrees but it never theoretically, it never quite gets to 75 degrees. So here is a graph that represents a real life problem. And as I read this graph I can see what's happening to the potato. But I drew the graph based on some assumptions that I sort of know about the potato ahead of time. Okay. Let's see, let's go to episode six and look at some things we wanna review in episode six for this exam. You should know that there are two ways to find the vertex of a quadratic function. I didn't have room to put quadratic function on there. We'll talk about this in just a moment. You should be able to solve applications of quadratic functions. You should know about the algebra of functions. How do you add, subtract, multiply and divide functions. And finally, you should know how to compute the composition of two functions. Okay, let's consider those topics quickly here as the time remains. Suppose I have a quadratic function like this. f of x equals three x squared plus 12 x minus, minus 11. And this is what we refer to as a quadratic function because the expression on the right hand side is a quadratic. You're familiar with quadratic equations. So this is a quadratic function. And let me just ask you a question to be thinking about at home. What does the graph of a quadratic function look like? Well, it looks like a parabola. And this particular one is a parabola that opens up. And I can tell that because there's a positive on the square term. Now, where is the vertex of this parabola? Well, there are two ways to find the vertex. One of them is to complete the square. The other one is to use a formula. If I complete the square, I would do it this way. And if you want to use this method on the test, it's perfectly fine with me. I will not specify whether you should use the formula or complete the square. So I'm going to factor out the three and put the minus 11 outside. So I factor the three out of the x squared and I factor the three out of the 12x. And then I complete the square. Now, what number should I add on here to make that a trinomial square? The rule is once this coefficient is a one, I take half this number and square it. Half of four is two. Two squared is four. So I have to add on a four. Now you might say, is that legal to just add a four in the middle of a problem? Well, it is if you balance it. You see, I've actually put in three fours, three times four. So I've really added on 12. So over here, I have to subtract 12 to balance it. So this says f of x equals three times x plus two squared minus 23, wow. Now this tells me that basically I started off with the quadratic function f of x equals x squared. I shifted it two to the left and down 23. Well, if you go two to the left and down 23, that means the vertex is at negative two minus 23. That's where the vertex is for this function. But you know, there's another way to do this and you're perfectly welcome to use the alternative method on a test. And let me just write it over here. I'm gonna rewrite my function, f of x equals three x squared plus 12 x minus 11. There's a formula that will calculate the vertex. The x coordinate of the vertex is found by taking negative b over two a. I've used this several times in the episodes that we've watched. And that would be negative 12 over, let's see, two times a is six. And so x is negative two. Hey, look over here, x is negative two. That's exactly right. And to get the y coordinate, what I do is substitute in negative two into the function to calculate the y. That'll be three times negative two squared plus 12 times negative two minus 11. And that's four times three is 12. 12 minus 24 minus 11. Let's see, now that's negative 12 minus 11 and that's minus 23, there you have it. You can use either one of these methods if I ask you to find the vertex of a quadratic function. That's a fairly standard question in college algebra, so you'd certainly wanna be prepared to do that. Okay, another question is applications of quadratic functions. Let me just go back to that Galileo problem and remind you of this. Suppose I told you that I were gonna launch a projectile from the top of a building and its initial speed, initial velocity was, well, let's see, let's say 60 feet per second. 60 feet per second, that's the initial speed going upward. Now of course it starts slowing down, slowing down and it comes back down to the ground. And let's say that the building is 100 feet tall. So the question is how high, how high will this projectile go? Well, I'm thinking that I should use our function for the height of a projectile and that will be minus 16 T squared plus 60 T plus 100. I'm just substituting V naught and H naught into this problem. Now to find out how high it goes, I wanna find the vertex and I'm gonna find it as quickly as I can. So I'm gonna use the formula that T equals negative B over 2A. And that's gonna be negative 60 over negative 32. Negative 32, because I'm doubling negative 16. Now let's see, I can divide top and bottom by negative four and I get 15 over eight, 15 over eight. Now that's gonna be in seconds. What that says is the projectile goes up to a maximum of height and it reaches it after almost two seconds, 15 eighths of a second. How high will it go? Well, if I substitute in 15 eighths and calculate that answer, if I just plug in that number here, here, add that all up, I'll find out how high the object goes. I'm not gonna work that out because I'd like to move on to some other things before we run out of time. Okay, let's go to episode seven. And the things you'll wanna know from episode seven, which we just talked about the last time, we met here in this course, you wanna know about one-to-one functions and the horizontal line tests, those are related to one another. You wanna know about the inverse function property, which says if you take the composition of F and F inverse of X, then you get X as a result. And that's if you take the composition in either order. You should know two ways of determining and then inverse functions and that is by taking the composition of two functions to see if they're inverses or to calculate the inverse algebraically. And finally, you should be able to sketch the graph of the inverse function of a function when you're given the graph of the original function. Okay, let's just go through a couple of those ideas quickly here. Suppose, for example, that I draw a graph of a function and let's say it looks like, it looks like this. I'm gonna draw a straight line and then I'm gonna draw a curve that goes up like that. That's pretty weird-looking function, but I'll call that function F. So let me ask you about this. Does this function pass the horizontal line test? And I think it does because if I draw horizontal lines through the graph, I think they intersect the graph at only one point. Therefore, this is a one-to-one function. What's significant about that? Well, it has an inverse. So there is an inverse function that I should be able to graph on the basis of this. And the way I find the inverse function is I draw a 45-degree line in here and I just imagine flipping this graph across the 45-degree line. So I would get a point right about here. I would get a point right about here. It looks like that portion would be straight right there. And then this portion, I'm guessing, would look about like that. Now, this new graph that I'm drawing in is the graph of F inverse. This is just one of many things you need to know about inverse functions and one-to-one functions. Okay, just to take one more idea about inverse functions, you should be able to calculate an inverse, the inverse of a one-to-one function by using the algebraic method. Suppose F of X was equal to six X plus three. And I'm wondering what is the inverse function for that? What's the inverse function? Now, there is an inverse function because if I graph F of X equals six X plus three, this is a diagonal line and it passes the horizontal line test, so it has an inverse. Now, what is the inverse? I'm going to replace F of X with Y. I'm going to solve for X. I'll have to divide by six here. So I divide by six. And then if I turn it around, X is equal to Y minus three over six. Now, if I interchange the X and the Y, Y is equal to X minus three over six and therefore F inverse of X is X minus three over six. Now, this was a rather brief example of using the algebraic method to find an inverse function. Okay, let me just go over a few items that I think you need to pay particular attention to as you study for this test. This is not a multiple choice exam, but you will need to show your work to solve these problems. So there'll be some space provided after each problem for you to fill in the details. You don't want to just give an answer when there's obviously some work to be done, so you should show that work on your test and then it's for your benefit, you get partial credit for the any work that you do that's correct. Whereas if you just give an answer, you see that your answer might be wrong and I couldn't give you any partial credit for that. You should know that you can't use a calculator, that you can't use notes and of course you can't use your textbook. So you wanna know that as you go into the test and so I wish you the best of luck on this exam and I hope you do well and I'll see you next time for episode number nine.