 50 college algebra. This episode is on transformations of fundamental functions, and then some related information. If you remember back in episode three, the last episode, there were eight fundamental functions that we graph. There was the constant function, the identity function, f of x equals x, and more functions, the square root, the cube root functions. Today we're going to look at how we can transform those functions into other shapes. But to begin with, I think we should look at how we use the graphing calculator to graph some of these functions. Then we'll look at the transformations. Then we'll talk about finding x and y intercepts near the end of the hour. And finally we'll look at graphical solutions to inequalities. So let's begin with the graphing calculator. I'm using a TI-82. There are lots of graphing calculators on the market, and you may or may not have a graphing calculator, but I want to just show you how you can use some technology to graph some of the functions that we've looked at. So, for example, let's see if we turn it on here, I'm going to go to the y equal button, and I want to graph the squaring function, x squared, so I'm going to enter x, and then I'm going to square it, and I want to draw a graph of the function y1 equals x squared. But before we can draw the graph, we have to decide on where on the graph we want to look. Do we want to look up high or down low or where? So I think we should look somewhere around the origin. So I'm going to choose the window where to view it. So the first thing I see is the minimum x, negative 4 is on here. Let's just stick with negative 4, and the maximum x is positive 4. What that means is on the x-axis we'll be viewing this graph between negative 4 and plus 4 on the x-axis. Then the x scale, that says where there will be a tick mark. Where would I like for each interval denoted? So let's say every unit, so there's a 1 in that position. For the minimum y, well, instead of negative 20, I think we should pick like, how about negative 2? I don't think the squaring function ever goes negative, but let's show a little bit of the third and fourth quadrants. And for my maximum y, let's just stick with 20 like we have there. And for the scale, I think I'll choose every five units, like it says, because if I choose a tick mark at every unit from negative 2 to 20, there'll be lots of tick marks. So let's just go with every five units. Okay, the window's been chosen, the function's been designated, so let's draw the graph. And this is the parabola. Now, you notice it does go through the origin. It goes through the point 1, 1, although my scale is different on the x-axis and the y-axis, so the point 1, 1 is sort of suppressed. You remember, this is at 5, this is 10, this is 15, and this is 20 up here. So there is a point at 1, 1, and there's also a point at negative 1, 1, but it's a little difficult to see those in this particular graphic. And this is the graph of the squaring function. There's a button here called trace. And if I push trace, what that does is to highlight the middle point on the screen left to right, and that's right at the origin. And you see that the two coordinates are designated here, x equals 0 and y equals 0. Now, if I push the right-hand button, it moves that trace marker along the curve. Now, you notice it begins to rise right there. And at this moment, we're at about 1.2 for x and about 1.4 for y. If I back up just a little bit, we should be able to get close to the point 1, 1. There's 1.02 and there's 1.04. So we're pretty close to 1, 1, but now if I click it left again, I think I'll probably skip over it. Yes. Now we're at 0.94 roughly and 0.88 roughly. So we don't get to come across the point 1, 1 exactly, but I could just keep moving up the graph like this on the trace. Now you might say, well, why is it jumping from one point to another? Well, I'm going to show you how this graph was actually drawn. If I push the mode button right here, instead of highlighting connected, I'm going to go to dot. So let's click down to connected, switch over to dot, and I'm going to push enter in the lower right-hand corner of my calculator. And now let's draw the graph. And you see what happens is there are actually a sequence of little dots there. We don't have a nice continuous curve, but we have 95 dots from left to right. There are 95 dots located. And when I push connected under the mode section, what it does is it joins those with straight line segments. And what we see looks like the curve, y equals x squared or f of x equals x squared, but it's actually 95 points on that graph that are connected by straight line. So it makes it sort of a jagged thing. But there are so many points that it looks fairly smooth when we draw the graph. Now why does the calculator only plot 95 points? Why doesn't it plot all of the values from negative four to plus four? Well, that would be infinitely many points. It would require infinitely many calculations, and the calculator just can't do it. So it picks 95 values. It divides the interval negative four to four into 95 segments, sub segments, and it plots a point at every end point because it can do that and it can do it fairly quickly. And if I go back to the connected mode, I'll push enter. I'll go back to the graph again. And now we don't see the dots, we actually see an attempt to connect those points. Now you notice it does look a little jagged along here. It's not really a smooth curve. That's because these little cells or pixels are lit up, and it's difficult to arrange them into a smooth curve. So it does the best it can and lights up the little pixel closest to it. Let's take a different function. I'm going to go back to y equals, and let's take x cube. In fact, let's put x cube on a different line. So I'm going to enter x, and I want to raise it to the third power, so I'm going to push this little exponential key over here to the third power. You might say, wait a minute, you didn't push that key when you did x squared. Well, I actually have a button down here for the square because the square is such a common operation, but the cube is not quite so common, so you have to designate your entering an exponent. And I want to turn off the squaring function. So if I go up to the equal sign and push enter, you'll notice that equal sign got turned off, but the equal sign here is turned on. So when I go to graph, I'm only going to get the cubic function now. And when I draw the graph, there's the cubic function. Oh, yes, my window didn't go down very far. Let's adjust the window so we can see further down. So let's go back to the window. I'm going to change the lower bound on the window from negative 2. Let's make it negative 10. And I think I'll make the y max positive 10 so that I see an equal portion above and below the x axis. Now we draw the graph. And here's the cubic function. And you notice it comes down sort of like a parabola, but a little steeper in the middle. It's flatter than a parabola would be. And now it goes down in the third quadrant, whereas the parabola went up in the third quadrant. Now let's superimpose both of these. So if I go back to y equals, I want to turn on both functions. So I'm going to enter, push enter on the equal sign for y1. And now you'll see both of those, both of these are highlighted. So I'm going to get both graphs so they'll be superimposed. It graphs them first y1 and then y2. So we're going to graph x squared, then we'll graph x cube, because I have it set in the mode operation to graph these in sequence. So first we see the squaring function. And now we see the cubing function. Look how the cubing function goes up faster. We talked about that last time. And in the middle, it's very close. Hard to tell. But the squaring function is actually above the cubing function right here. I can zoom in if I push zoom. Zoom number two is in. Zoom number two. And I'm going to enter that. And now I'll see those two graphs a little more closely. And you see there's the cubing function. We can begin to see a space between the two functions. But I have a much smaller interval on the x-axis. This goes from, let's push trace and just see how far over it goes. If I trace over to the edge, over to the edge it looks like we're over about .87. So it goes close to plus one. And over here it goes fairly close to negative one. And class, can you tell on which of those curves is this trace moving? Is it moving on the squaring function or the cubing function? Can you tell? It might be hard to see on that screen. It's actually on the squaring function because that was the first function entered. And if I push the down button, it'll jump to the other function. Now I'm on the other function. Go up. I'm back on the on the on the squaring function. Then I'm on the cubing function, squaring, cubing, back and forth. So if you push left and right, you actually move the tracer back and forth. If you push up or down, you jump between the two functions back and forth. So let's go over here to one half, about one half. This is getting pretty close to one half. It's .51 for x. There's .48. Let's go back to .51. The value on the cubing function is about .33. That's because one half cubed is .125, but we have a little bit more than one half. So this is a little more than .125. It's .133. If I jump to the squaring function up above, x is still at .51. But if I square that, I get about a fourth or .26. So you can see that even though it's difficult to see the difference between those two, there is a difference. And as a matter of fact, I could zoom in more and I could make that difference more pronounced, but I don't think I'll do that right now. Okay. Well, I'm going to come back to the graphing calculator in a minute to look at some variations of our fundamental graphs. But first, let's discuss those variations. Let's go over here to the marker board. Let's see. We were discussing in episode three the fact that there were some fundamental graphs with some target points that I pointed out to help us graph these. And I want to look at now how we can sort of mold or change those functions into an infinite variety of other possibilities. Let's take, for example, the squaring function. Now, the squaring function f of x equals x squared. You remember when we graphed this on the x, y coordinate system, there were some target points to plot. I'm going to keep the scale fairly big here so that this is three and this is three. And there were three target points that I would plot to graph this. There's zero, zero. There's one, one. And there's negative one, one. And then I just remember the general shape. And I draw the parabola coming down like this, coming down like this. And you remember right at the origin, it's nice and smooth and it just turns and goes back up. So it doesn't come in and make a sort of a pointy edge like the absolute value function would do. Now, what would happen if I wanted to graph this function? I think I'll call it g of x so we can keep them straight. This is the graph of f right here. Suppose I want to graph x squared plus two. Now what that does is it adds two to every y value. You see the y value here was x squared and now I've added two to every y value. So that means every y value moves up to. And so what I'm going to do is move my three target points up to. So if I label my axes, instead of going through the origin, if I raise it up to it's going to go through this point at zero two. And instead of going through one one, it's going to go through one three because I raise it up to. And if I back up one, I go up three there also. So I have my same three target points that I had here. I'm going to draw the very same shape. It's just my graph is two units higher. So I get a parabola right here. Now when I take a fundamental graph and I place it in a new position, this is referred to as a transformation. It's a transformation of the basic function f of x equals x squared. In this particular case, this is called a vertical transformation. So there are vertical transformations. That's where you move it either up or down. But we're also going to talk about horizontal transformations. We'll get to that in a moment. There are also stretches and compressions that we'll discuss. And then there are reflections where I actually flip a graph across an axis, flip it across the x axis, flip it across the y axis. So there are reflections. So vertical transformations, horizontal transformations, stretches and compressions, and then these last reflections. Before we get into all the others, let's do another vertical transformation. Class, what do you think the graph of this next function would look like? Let's call it h of x. And this is x squared minus three. How do you think I would graph that? I think it's going to look like x squared, but what change is going to be made on it, Lynne? We're going to lower it three. Just like if you add two, you raise it two units. If you subtract three, you lower it three units. Whenever you add or subtract a constant outside the function, it's a vertical transformation. So when I graph this, instead of passing through the origin like we did over here, I'm going to lower this three units. And you might say this is the quote new origin. It's the new, it's called the vertex. It's called the new vertex of the parabola. So I may refer to this as the new origin. Of course, this is always the origin back here, but I've lowered it three units. And if I go over one and up one, I get a point. And if I go back one and up one, I get a point. And now all I have to do is remember the shape of the parabola. And the curve comes down like this. And this, this is the graph of h. Okay, what if we change the fundamental function? Let's take a different one. Suppose we take the absolute value function. I'll call this capital F of x, so we don't use little f all the time. The absolute value function. Now, that function was one of our fundamental graphs. And if you remember, it has the same three target points as the cubing function or as the squaring function. And that would be at zero zero one one and negative one one. But I have to remember the shape. And this time I remember it makes this right angle with two rays coming together. There's a right angle between those. They're each at 45 degrees with the horizontal axis. Now, how would I graph f of x equals the absolute value of x plus four? Let's see, Matt, how would I graph that? Well, you just graph it like the absolute value function, but raised up for spaces. Exactly. We're just going to take the absolute value function. We're going to raise it up four. Now, rather than just making a quick sketch of what it is, I think we can get a more accurate graph if we plot these target points. So my origin points moves to the so-called new origin at zero four. And instead of going over one and up one down here, I'll go over one and up one from that new origin. And I'll go back one and up one from the new origin. And now I remember the shape looks like this. And you know, we have a fairly accurate graph considering we only plotted three points. But we have to remember these fundamental shapes and where their target points are. Okay, these are vertical transformations. Let's take a different transformation. Let's look at a horizontal transformation. Suppose we take the function f of x equals x squared again. But I'm going to add a number inside the parentheses, x plus two squared. Now that has a different effect than if I added two on the outside. A moment ago when I put a two on the outside, it moved the graph up to. This one's going to move it sideways. Here's why. You notice f of x is a square. Therefore, this quantity can never be negative. But it could be two. Excuse me. It could be zero. And for which x will this be zero? At negative two. So let's say f of x equals zero when x is negative two. That's because if you plug in a negative two right here, you have zero squared or zero. Now that tells me that the lowest the graph can be is zero. And it will be that at x equals negative two. So if I come over here to negative two, x axis, if I come over to negative two, that must be the lowest point on the graph because we said this could never be negative because it's a square. And if it's zero there, that must be the lowest point. So that's where my new origin is. And if I go over one and up one. And if I go back one and up one, there are my three target points. I'm ready to draw my quadratic function. The quadratic function will look like this. And you see what's happened is we've actually moved it over to the left. I'm going to put a little negative two since I moved it in a negative direction. You say, well Dennis, that's easy to remember. If you add a two on the outside, you move it up. If you add a two on the inside, you move it sideways. Well here's the tricky part. If I put a plus two on the outside, plus two, I move it in a positive direction. If I put a plus two on the inside, I actually move it in a negative direction. So we have to remember that switch. Suppose I wanted to graph this function. F of x equals x minus three cubed. x minus three cubed. Well once again, I'm adding or subtracting a constant directly on the x. So that means there's going to be a horizontal transformation. Horizontal transformation. And if there's a negative here, I'm going to move it in a positive direction. So I'm going to move it over three to the right. Now of course a person might say, now I don't understand why you're moving it to the right. It's a negative three. It seems like you should move it in a negative direction. Well you remember that the original cubing function passed through the origin. So when is this going to be zero? This will be zero when x is three. Three minus three is zero. Zero cubed is zero. So when I plug in three, this function will be zero. So it goes through that point. My origin has moved over three to the right. And if I go over one and up one, I get another target point. And this is the cubing function. So if I go back one and down one, I get another target point. You remember this is an odd, this was originally an odd function. So now I draw my cubic shape, which I have to remember. It's a little flatter here than a parabola, a little steeper there. And this is the graph of that function. f of x equals x minus three cubed. Now we have a couple of graphics that help to point out the differences between vertical and horizontal translations. So let's go to the first graphic that we have coming up. Let's see now, this graphic is titled a vertical shift. It's what I called a vertical translation a moment ago. Vertical shift. So on the first line there, if y equals f of x plus c, and c is greater than zero, then you notice it takes the blue curve, y equals f of x, and it shifts it up to the red curve, y equals f of x plus c. Now if you ask me what is the blue curve, what did you graph there? That's not a function that we're familiar with. I just drew sort of a generic curve and called it y equals f of x. And then when you add c, you get the new curve, the red curve, which is higher than that, because we shifted it up c. On the bottom line, suppose we want to graph y equals f of x minus c. Again this is where c is positive. Then if you go to the right, we have the original blue curve, y equals f of x, and the red curve is below it now because we shifted it down. And the new equation is y equals f of x minus c. Okay, now there's another graphic that follows this one, let's go to. Now in the second graphic titled horizontal shifts of graphs, again I start off with the blue curve that I call y equals f of x. And if you go to the left, it says y equals f of x minus c, where c is positive. Well when you subtract a positive number, it actually moves the graph to the right. And you see that the red graph is titled y equals f of x minus c, and it moves at c units to the right. On the bottom line, if I add c to the x, y equals f of x plus c, again c is positive, then it shifts the graph to the left. And the blue graph becomes the red graph on the left. So you see that could be a little bit confusing. You have to use the opposite direction here of what you would have chosen for the vertical shift. Okay, I think we should work an example where we combine both of those. Suppose I want to graph this function. I'll call it capital G. I'll say it's a function of t, because we don't always have to have functions of x. And let's say I want to graph, let's say I want to graph the square root of t minus 2 plus 1. Well, the fundamental graph that I'm thinking of here is f of t equals the square root of t. And other than the fact that I'm using a t now, rather than an x, we know that that graph looks like this. Let's just draw a little coordinate system over here. So here's the graph. This would be the t axis. And the target points were at 00 and 11. And you remember that had the shape of half of a parabola. Let's see. It's kind of hard to draw it so small. I hope you can see this. But generally I get that shape. But my function g has two shifts or two transformations. What is the vertical transformation for my function g? Anyone? Yeah, Matt? It's going to move it up one unit. It's going to move it up one. So this plus one says move it up one, exactly. And what about this other business, Stephen? Right two units. It's going to go to the right two units. So I'm going to draw this same shape, but with those two transformations involved. So let's come over here to do that. I'll make this the t axis. And let's see, my origin is going to be moving two to the right and one up. So if I go two to the right and one up, I get a point right there. That's my new origin at 21. And my second target point, well, you know from the origin you always go over one and up one. So from this point I'll go over one and up one. I get a point right here. Those are the only two points I need to plot. And then I'll draw the shape. And let's see, remember we want to make this so it kind of turns down right there. So just do the best you can on that. And this is the function g. Let me ask you a couple of questions about that graph. You remember last time we talked about compressing the graph onto the x axis, in this case the t axis to get the domain. If I press this onto the t axis, what portion of the t axis will be covered? Sam, what would you say? I have no idea. Okay, anybody? Jenny, what would you say? It would go from positive two to infinity. Exactly. So when I press this down, it covers two to plus infinity. That's the domain of this function. And let's just write that below it here. That the domain of g would be the closed interval from two to infinity. And you know, I think that's what we would have guessed all along because we don't want the number inside here to be negative. So I have to make sure that I pick t to be two or larger. You can pick two or pick any number bigger than two, but not smaller than two. And it's reflected in the graph. What is the range of this function? Can anybody tell us? Steven? One to infinity. Yeah, because to get the range, you take the graph and you press it on to the vertical axis. And so it looks like we're going to go from here up. You might say, well, it doesn't seem to go that high, but you remember this graph keeps on going so it gets higher and higher. It'll eventually reach every number above one on the vertical axis. So the range appears to be the closed interval from one to infinity. And again, that makes sense because this is a square root plus one. In fact, this is what they call the principal square root or the positive square root. So this can't be negative. And when you add one to it, it can't be smaller than one, but it could be one or it could be larger. So the range goes from one to infinity. Okay, let's move on to another transformation. This next one is a vertical stretch. Let's go back to our basic squaring function, f of x equals x squared. But I'm going to multiply it by two so that I want to graph two x squared. So the difference is I'm not adding a two outside. I'm not adding a two directly on the x, but I'm multiplying by a two. And this has the effect of actually stretching the graph vertically. Here's why. If I begin to locate what we've been calling our target points, you notice if I plug in a zero for x, two times zero squared is zero, it still goes through the origin. But what happens if I plug in a one right here? What would be f of one? Two. It'll be two. So you see instead of going over one and up one, I go over one and up two and I get a point here. And if I plug in negative one, I also get two. So you see what's happened is the target point that was here has been stretched away from the horizontal axis. And I now get a parabola that looks thinner. But you know what? It isn't thinner, it's taller. Because all the points got further away. If there had been any points below the horizontal axis, they would have been stretched down. So if I had drawn the original quadratic function, I would have had a point there and I would have had a point there. And I would have had a parabola that would come in like this. So what I'm graphing there is the original function g of x equals x squared. So you might say it got thinner. Well I guess in a manner of speaking it, you might say it got thinner. But you know, when you get taller, you look thinner. Like you may say, Dennis, you know, you're fairly thin. Well no, it's only because I'm tall. If I were only four feet tall, still this wide, I'd look very fat. But as you get taller, then you tend to look thinner at the same time. Okay, so if I put a coefficient on here, it causes a stretch unless the coefficient's smaller than one. Let's see what happens in this next case. g of x equals one-half times the absolute value of x. My fundamental graph is the absolute value function, but I'm multiplying now by a positive factor, but smaller than one. This has the effect of compressing the graph. So let's see, if I substitute a zero for x, g at zero, the absolute value of zero, zero, so it goes through the origin. And if I substitute in a one, what is g of one going to be? g of one. It's one-half. So I only go up a half right there. And same for negative one, I only go up a half. Now this time when I draw my absolute value function, I have two rays that come together, but they don't mean it at the right angle because they've been compressed. So they come in at, well they come in in such a way that they make an angle bigger than 90 degrees now because it's been compressed. It's sort of like you took that graph and you just stomped your foot on it so you kind of pushed it together and that's what the one-half does. If you put a coefficient bigger than one, it's a stretch. If you put it in a coefficient smaller than one, it's a compression. Okay, I want to put several of these transformations together before we go to the next example. What if I want to graph this function? I'll call it h of x equals three times x plus one cubed minus two. There are actually three things that are affecting the original cubic function. What's one of the changes you see in that function? See originally this was h of x equals x cubed. What three things have affected it? Lynne? Up the y-axis two. Are we going to go up two? Yes. Well, there's a nice two. Down two. Yeah, we're going to go down two. That's fine. So we're going to go down two. What other change will we make? Go left one. We're going to go left one. Yeah, if I add a plus one there, we actually move it to the left one. Okay. And what's the three going to do? Stephen. Stretch it by a factor of three vertically. Yeah. I don't know how to represent that. Maybe I'll put in a dotted line here and that's going to be stretch three. I hope you get the drift or what that's about. So if I go to my coordinate system, x-axis, y-axis, let's label or put a scale on it. Let's see. Now my origin is going to move to a new position. I have to move it down two to the left one. Okay. Let's see. Down two to the left one. So my new origin is going to be right here. And the stretch affects the other two target points. If I go over one, I go up three. And if I go back one, I go down three. That's because everything has been stretched now. And you might say, well, Dennis, shouldn't this one be stretched also? Well, you see points that were originally on the x-axis. When you stretch, it's kind of like having rubber graph paper. If you stretch the top and the bottom away from each other, the x-axis stays rigid in the middle. It's only the points off the axis to get further away. So this stretch has occurred actually now at this level, so this point doesn't move. Now I have to remember the shape of the cubic function, and that was something like this. It comes in, levels off, turned, and goes down. Now when you look at that, you say, my, that certainly is thin. Is it really thin? No, it's just tall. See, all the points got further away. So it got taller. It looks thinner. You could say that, but from this stretching point of view, it's gotten taller. Okay, let's look at another transformation. Let's see, actually we have a graphic I think maybe I'd like to show now. So let's go to the next graphic. Okay, for vertical stretching and shrinking of graphs, imagine that the blue graph that you see on each of those coordinate planes is y equals f of x. I don't know in particular what function that would be. It's just sort of a random curve. And suppose I multiply f of x times c, and c is bigger than one, then that causes the graph to stretch. And you see on the right-hand side, in that upper graph, we have the red curve that rises up higher and goes down lower than the original f of x, because it's been stretched. You notice that the origin didn't move, as well as the ends of those curves, because it seems to end along the axis. Those points stay the same, but all the other points are stretched away. On the bottom line, I multiply by a number c, but this time c is positive, but smaller than one, and this causes the graph to shrink. And if the original graph was y equals f of x, then y equals c times f of x has been compressed or it has shrunk. And so it comes up to a lower point, to a lower maximum before it comes back down. Okay, let's go to yet one more change. This is a reflection. Let me just write this name down over here. This is a reflection. And this time, I'm going to go back to the same fundamental function, f of x equals x squared. But I'm going to put a negative in front of it. A negative x squared. Well, basically what this does is it takes every y-coordinate that I got before, and if it was positive, it makes it negative, and if it were negative, it'd make it positive. Although I don't think this one ever had a negative y-coordinate to begin with. So when I graph this, my origin is still the same, because when you make the y-coordinate negative, the negative of zero is still zero. But now, well, maybe you could tell me, when I go over one, who can guess what's going to happen? If I go over one, will I go up one? Can you go over one? We're going to go down one, exactly. We're going to go down one, because you see, if I substitute in a one in here, f of one is negative times one squared, or negative one. So my function value is a negative one there, and if I plug in a negative one, I get a negative one for my y-coordinate again. I have a question. Oh, okay. On the positive x, you're still positive x-one. How did that happen? Well, let's see. So I went over to one, and I plugged in a one. Oh, so when x equals one, your y is going to be negative one. Yeah, so it's going to be negative one squared, and so we get negative one. So when you plug in a one, you get a negative one for y. So this is the point, one, negative one. That's a good question, though. Same thing over here. If you plug in a negative one, which makes it a plus one, and then you take the negative of that, so you get a negative one, so this is the point, negative one, negative one right here. Well, to kind of summarize this, what's happened is my graph used to go up, now my graph goes down, and I'll draw my parabola, like that. And you see why it's called a reflection, because it's just the mirror image of what we would have had if I had been graphing f of x equals x squared. Any other questions about that? Yeah, Matt. I had a question. Would it be a little bit more clear to write it with the parentheses so that you know it's a negative of the squared number? Oh, yeah. Well, maybe we should. Yeah, at least for now. You know, if there are no parentheses shown, then the general agreement in mathematics is that you do the square before the x. This is not a negative x quantity squared. It's the negative of x squared. Now, if you're not aware of that convention, then I could see that that would be a little confusing. But generally, when we see an exponent and a negative in front, we have to do the exponent first. That's just a general agreement in mathematics. For example, if you saw negative three squared, that would be negative nine. But if you had parentheses, negative three squared, that would be plus nine. So if no parentheses are shown, we have to decide how we're going to interpret that. And the interpretation is to do the square before you do the negative. Okay, I think we should take another example of a reflection. And let's take another one of our fundamental curves. Let's take the cube root function. Suppose we have f of x equals negative the cube root of x plus one. Well, let's see now. Just to make sure everybody remembers, the original cube root function has this shape. I'll just draw it over here. This is before we've made any transformations on it. It had three target points, and they were at zero, zero, one, one, and negative one, negative one. That's because it has an odd index. So this is symmetric about the origin. And then my graph comes in sideways like this, turns, comes around, comes back out. We could probably draw that a little bit better. And that's the graph of the cube root of x. Now, I have to make some changes in that graph. One of the changes is due to the negative, but another change is due to the plus one. The negative is going to flip it across the x-axis. We're going to get a reflection. What does the plus one do? Move it left. Very good. It's going to move it left because you see we're adding a one directly on the x. So this is not a vertical translation. That would be if I put a plus one outside. If you add a one directly to the x, it moves it horizontally in a negative direction, or to the left. So I want to draw this graph, shift it to the left one unit, and invert it. So let's see, if I shift it to the left one unit, it means my new origin, so to speak, is at negative one, zero. And from the origin, I would normally go to the right one and up one. Now I go to the right one and down one. And if I go to the left one, originally I would have gone down one. Now I go up one. And then I draw the graph. So the target points tell me how to proceed, and right here it should go, should go vertical, right there. This is a graph of capital F. Let's just try some numbers and see if that really makes sense. Let's actually plug in some values for F and see if this agrees with it. Suppose I were to find F at negative two for this function. F at negative two. Well that means put in a negative two where the x is. So I get negative the cube root of negative two plus one, which is negative the cube root of negative one. What's the cube root of negative one? Negative one. So this is negative, negative one. So I get one. So that says that negative two, I should be up at one. By golly we are. There we are right there. So we do get that point. On the other hand, what if I were to calculate F at nine? Oh I tell you what, let me change that. Let's make it F at seven. So it kind of works out in a nice manner. F at seven. Well if I plug a seven in here, I plug in a seven there, this would be minus the cube root of seven plus one. Which is minus the cube root of eight. Now the cube root of eight is two. So this is going to be negative two. And that says if I go over to seven, well seven is going to be beyond my axis the way I've shown it here. But I would go down two and that looks reasonable because here's negative two right here. Any other questions about that? Okay, let's take another change or another transformation on a function. This last one involves a stretch and a compression, but it's going to be sideways. A sideways stretch or compression. Suppose I have a little f of x and I'm going to make this change on the squaring function because we've kind of used that as our base function. Rather than multiplying by a two on the outside, suppose I multiply by a two on the inside. Now if I go to graph this function, what we're going to find is this function gets compressed and it's going to look thinner and this time it really is going to be thinner because we're actually pushing in on the sides. It didn't get taller, this time it got thinner. So let's say this is one, two, here's three, one, two, three. Let's see, if I substitute in a zero for x, I get zero for y, so it goes to the origin. But this time you notice if I go over one-half, if I plug in a one-half there, two times a half is one, what is f at one-half? One. It's one. Okay, if I go over one-half, I go up one. Now you see the way we did this before, I'd go over one and up one, but everything's been compressed. This two is making my function think my x's are coming through twice as fast as they really are. So the function thinks it has to turn up in a hurry because we're fooling it, we're giving it one-half and it thinks we gave it a one, so it turns up faster. And if I go back to minus one-half, if I substitute in a minus one-half there, that says negative one squared or one. So these points have been indented toward the middle, so I get a graph that looks like this. And it has been compressed sideways. Now from another point of view, you're probably thinking, well, wait a minute, Dennis, why don't you just square the two x and call it four x squared? Oh, well my goodness, now you're talking about the same graph, but it's a different interpretation. This is a vertical stretch of four. And that says if you go over one, you go up four. So you see if you square this and write it that way, you will get the same graph, but now the graph didn't get thinner. It got taller. But it ends up being the same graph. And if I go back to negative one, I go up four. And it's the very same graph. It's just, are we going to look at these three target points or are we going to look at these three target points? We get the same graph either way. It's fine with me if you square this and think of it that way and make a vertical stretch. But I'm saying if you put it to inside, you could interpret this as a horizontal compression. Of course, the obvious question that I think is why would we want to know about a horizontal compression if we can make this a vertical stretch? I mean, why do we have to do this two ways? Well, if you take trigonometry, the Math 1060 course, you're going to see sign graphs that go up and down that sort of undulate and those will be compressed horizontally. So it's actually in Math 1060 is where you use this idea of compressions and stretches that are horizontal. Another possibility what if I have the function g of x equals the absolute value of the absolute value of 1 half x of a 1 half x? Well, I'm thinking I should be graphing an absolute value function. If I put a 2 inside, it's a compression. So if I put a 1 half inside, it's going to be a horizontal stretch. Let's see if that actually holds true. Let's see now. If I plug in a 0, I think I will get a 0 for the function value. And if I substitute in a 2, if I plug in a 2, I get the absolute value of 1, which is 1. And you see I have to go over 2 before I go up 1. And I have to go back 2 before I go up 1. And so my target points have been stretched out sideways and I get a graph that looks like this. Now, from a different point of view, of course, a person can say, well, why don't you just take the 1 half out, take its absolute value, put it out in front, and put an absolute value of x there. Well, then that will give you the same graph, but it's not a horizontal stretch. It's a vertical compression. It's going to be a vertical compression because you see at 1, I only went up a half. And if I go back 1, I go up a half. And you'd be using these three target points. So when you graph this in this form, it's a horizontal stretch. If you graph it in this form, it's a vertical compression or a shrink. Okay, in this next example that you see on the screen, the function is f of x equals 2 times x plus 1 cubed. What transformations are evident? Now, what I'd like for someone in the class to do is to say verbally what's going to happen and what's the basic function that's being altered. Okay, Jenny, in this next function, can you tell me what was the fundamental function that should be graphed or two changes that are being made to it? Okay, the function is f of x equals x cubed is the fundamental. And it looks like it's going to go to the left one. And it's going to be squished twice as tall. Okay, stretched. It'll be stretched twice as tall. Okay, let's just try graphing that up here. The function is f of x equals 2 times x plus 1 cubed. And, of course, you notice this all depends on us knowing what these fundamental graphs look like. We had that in the last episode. And, of course, knowing where the target points would originally be. So, Jenny said we're going to move this to the left one unit. So, the new origin is at negative 1. And the target points will be stretched to units. So, if I go over 1, I now go up 2. And if I go back 1, I go down 2. And so I get a taller cubic function. And if you want to say, no, it got thinner. That's okay with me. You can say it got thinner. Who's to say? But from this point of view, there was a vertical stretch, so it got taller. Let's go to the next function. The next graphic says we're given g of t equals the absolute value of t plus 3. Let's see, Matt, what's the fundamental function that we started with transformation or transformations are evident to you? The fundamental function is the absolute value of t. And it will be shifted up 3 units. It's going to be shifted up 3 units. Right. There's only one change to be made here. So, if I graph g of t equals the absolute value of t plus 3, then I call this the t axis first of all because I've called the variable t. And my target points are all going to be raised 3 units. So the origin moves to the new, what I call the new origin up here at 3 on the vertical axis. And the other target points say move to the right and up 1, move to the left and up 1. And now I have to know what the general shape of the absolute value function is and it makes this v shape characteristic of an absolute value graph. And that is the graph of g. It's not totally accurate, but none of us are going to draw a totally accurate graph. But it is quick. But if you don't plot the target points, then you tend to draw something too flat or too steep. So I think the target points sort of keep us on track for that. Okay, another topic I want to mention to you, a short one before we end this episode is to talk about intercepts. Now, let's say I have the function f of x equals x plus 2 squared minus 9. If I were going to graph that this is basically the squaring function with two changes being made to it. Let's see, what would you say are the two changes? What's the line? The two changes, it would go left to go to the left 2 and down 9. Okay, so when I graph this save time, I'm going to quickly go to the left 2 and go down 9, and so I get a point down here. That's my new origin. And if you imagine that I go over 1 and up 1, and if I go back 1 and up 1, I realize I'm not plotting those very carefully, then I get a graph that looks sort of like this. Now, I have a different question to ask you about this graph, and that is, where does it cross the x-axis? Looks like there are two places. And where does it cross the y-axis? Now, the places where it crosses the x-axis are called the x-intercepts. Looks like we have two x-intercepts. And where it crosses the y-axis is the y-intercept. So, I'd like to locate the x-intercepts and the y-intercept. There's only one of those. And you see, anytime you cross the x-axis, we may not know the first coordinate, but the second coordinate is zero because there's no vertical elevation. Same thing over here. We don't know the first coordinate. Looks like it's probably negative, but the second coordinate is zero. So, the characteristic of an x-intercept is that y is zero. So, I'll come over to my equation and I'll make y equal to zero. You might say, well, you don't have a y. Well, the f of x and the y are interchangeable. So, I'll put a zero right here. And I have zero equals x plus 2 squared minus 9. I'd like to solve that for x. Some people might say, multiply this out and collect like terms and factor it, but I would say, take a shortcut, move the 9 to the other side, add 9 to both sides, and then take a square root on both sides. And you notice I put down both plus and minus 3 because I'm looking for every possible solution. So, one possibility, let's just put it right here, one possibility is that x plus 2 equals 3. The other possibility is that x plus 2 equals negative 3. So, we're going to get two answers for this. One of them is x equals 1. The other one is x equals negative 5. Those are the x intercepts. This must be the point 1, 0. And this must be the point negative 5, 0. Now, to get the y intercept, I do almost the same thing. If we're talking about a point on the y axis, then it's the first coordinate that's zero, but we don't know the second coordinate because we don't move horizontally. We just move down to get to that y intercept. So, this time I'm going to let x equals 0. And if I let x equals 0, I'm just going to plug into 0 right there. f of 0 equals that'll be 2 squared minus 9. 4 minus 9 is negative 5. And so, this number is negative 5 on the y axis. It's just a coincidence that I got negative 5 on the x axis as well. So, I've now found the x and the y intercepts. Anytime you have a function and you want to find its x intercepts, you let y be 0 and solve for x. When you want to find the y intercept, you let x equals 0 and solve for y in that case. Okay, let's take one more example of intercepts. You know we have time for one more example. So let's take a function g of x equals the square root of x plus 4 plus 7. Now, I want to find the intercepts for this function, but I'm not going to draw the graph. We can draw the graph afterwards if we have time. But let's see if we can do it by strictly using algebra. To find the x intercepts, I'm going to let something be 0. Who can remind us x intercepts, what's going to be 0? Y will be 0. Y is 0. It's always the other variable is 0. So, g of x and y are the same thing. So 0 equals the square root of x plus 4 plus 7. So negative 7 is the square root of x plus 4. Now, this can't happen because this is a principal square root, meaning this is the non negative square root equal to a negative 7. That's impossible. So there's no solution. What that means is there's no x intercept and when you draw the graph, this graph will not cross the x axis. To find the y intercept, I let x be 0. And if I substitute into 0, g of 0 will be the square root of 0 plus 4, plus 7. Can anyone reduce that for us? Jenny? It'll be the square root of 4 plus 7, which is 2 plus 7, which is 9. Is 9. Okay, so this graph crosses the y axis at 9. You know, I think we're running out of time so we won't be able to draw the graph to verify that. But that'll be a good activity for you to try after this episode's over. Draw the graph and see if that checks out. We'll see you next time for episode 5.