Common Core Algebra I.Unit 11.Lesson 2.Horizontal Stretching of Functions

Hello and welcome to another common core algebra one lesson by E math instruction. My name is Kirk weiler and today we'll be doing lesson number 11. Sorry, unit number 11. Lesson number two on horizontal stretching of functions. Before we begin, let me remind you that you can find a worksheet and a homework assignment to go along with this video by clicking on the video's description. Or by visiting our website at WWW dot E math instruction dot com. As well, don't forget about the QR codes at the top of each worksheet. If you use your smartphone or a tablet, you can scan that code and you'll be brought directly to the video that's associated with that particular lesson. All right, let's begin. In the last lesson, we looked at vertical stretching and reflecting a functions across the X axis. In this lesson, we're going to actually look at the horizontal component of that. Now this is considerably more tricky. Just like horizontal shifting is. So let's get a feeling for it in the first problem. Exercise number one, the function F of X is shown on the graph below. Selected points are shown as reference. The function G of X is defined by G of X equals F of two times X notice that the multiplication by two happens before F is even evaluated, and this is tricky. All right? But let's see how it works. Letter a says, find the values of each of the following. Carefully follow the rule for G of X and show your work. All right, so watch. G of two, here's my rule. Let me put it in big, big red here it is. There's my rule, all right? Remember, follow the function rule and you can't go wrong. Okay? So G of two is F of two times two, right? I'm just putting in two for X but that means G of two must be F of four. Well, what's F of four? One, two, three, four. Oh, it's zero. All right, so G of two is zero. Fine. Whatever. All right, G of three. Well, that's going to be F of two times three. Or F of 6. One, two, three, four, 5, 6. Ah, that's negative one. Great. All right, G of negative two, that's going to be F of two times negative two. That's going to be F of negative four. Let's see, one, two, three, four, one, two, three, four. All right. I think you have it. Pause the video and finish the last three. All right, let's go through. You might even be quick enough now to say that G of four is just F of negative 8. Because it'll be two times negative 412-345-6781, two, three, four. All right, G of zero is going to be F of two times zero. I'm going to do this one out long ways again. And that's F of zero. And that's one, two. Okay, two. And G of negative three will be F of negative 6, and that's one, two, three, four, 5, 6, and that's zero. Now letter B is pretty important. It says, given the definition for G of X, the one that I've got in a box with stars by it, why can we not find the value for G of four? Explain. We'll pause the video now and see if you can figure this out. All right. Well, again, as I've mentioned in a previous lesson, if they say why can't you find it? Try to find it. Something will get in your way. So GF four, using that formula is F of two times four. And that's F of 8. Well, okay. One, two, three, four, 5, 6, 7, oh. Huh? I can't do that. Why can't I do it? Right? I can't do it. Because 8 X equals 8 is not. In the domain. Of F of X, right? So four is in the domain of F of X, but 8 is not. All right, letter C now, letter C is a very, very important. It says state because we're going to graph G in a moment. You know, we're going to graph G we got a graph G so we want some points on the graph of G and letter C says state the points that must lie on the graph of G of X based on your work in a all right, fine. But it's very important then that you understand inputs and outputs. So for instance, this point is the point two comma zero. Then we have the point negative two, four. We have the .0. Two, we have the .3 negative one. We've got the point negative four, negative four. And we've got the point negative three zero. And I really want to emphasize that because I could see some people coming along and saying, you know, let's take, let's take this point, for example, saying, oh, I've got 6 negative one. Now you don't. The input was three, the output was negative one. That F, that was just part of the rule. It's not the input output pair. So these 6 points element into the dark part of the screen. Those 6 points have to lie on the graph of G of X we're going to graph it on the next screen. So pause the video now if you need to and copy down anything you have to. All right, here we go. Let's do some graphing. All right? So letter D says graph the function G of X based on your work from B, then state the domain and range of both the original function F of X and our new function G of X what remained the same and what changed. I'm going to go back to blue. So let's plot these points. Two zero is right here. Three negative one right there. Negative two, one, two, three, four, right there. Negative four, one, two, three, four, negative four, one, two, three, four. Right there. Zero two, just right there and negative three zero is right there. I think I'm going to connect these all with a nice blue line or a set of blue lines. One is there there. And one is from there to there. All right, take a look. Let me just make sure that we have it. Let's do, let's do this, the original F of X in red, and then since this graph is already in blue will go G of X, I'm going to go back to red. Let's do some domain and range. Domain are the X's one, two, three, four, 5, 6, 7, 8. So it's smallest X is 8. And its largest X is one, two, three, four, 5, 6. Or an interval notation, negative 8 6. Its range and smallest Y value, one, two, three, four. This negative four, and its largest Y value is one, two, three, four, positive four. Interval notation negative four to four. Right. Now let's go into blue, and let's do our new function. Well, it's smallest X value now is negative four. And its largest X value is three. Or an interval notation, negative four to three. Its smallest Y value though continues. Look at this. Smallest Y value continues to be negative four. And it's largest Y value. Continues to be four. Out of room here. Right? So this is cool. Take a look. So the domain changed domain changed. And the range. Stayed the same. Now, this is really crazy though. Again, I want to circle it in red. Look at how counterintuitive this is. When we took the input and we multiplied it by two, the domain got squished by a factor of two. That's a very technical term. It got squished. It got compressed. Every X value got divided by two. In other words, this X value that was at negative 8 became this X value at negative four, but the Y values remain the same. This X value that was at negative 6 became this X value at negative three. This one at negative four became this one at negative two. Isn't that weird? So counterintuitive. So counterintuitive because you'd think, well, multiplied the X values by two, everything should get larger or wider by a factor of two, and it doesn't. Isn't that similar to the horizontal shift, you know, we add three to X, all of a sudden we shift it left three from your perspective. Sorry. You know, or we subtract two, and we shift it right to. Exactly the same thing here. Whenever we do something to an input, as our first thing, it kind of works in the opposite of what you'd expect. And you'll investigate why that is a lot more in algebra two. For now, it's just something we want to get a feeling for and be able to use. All right, there's a lot on that screen. So pause the video now and write down whatever you need to. Okay, here goes the text. All right. Oh, letter E I think we already did this. But let's write it out formally. Sorry, this is when I get ahead of myself. I do it all the time in class by students hate it. So describe what happened to the graph of F of X when we multiplied the functions input by two. All X values. Got divided. By two to produce. The graph of G, which I just think is cool. It's really amazing to me. That it works out that way. Also mazes to me that I can get as old as I can and as I am not no dot my eyes. Anyway, whatever. So I'm going to clear this out and then what we're going to look at is another example of this. Okay, here it goes. Let's take a look. Because it's so weird that you want to make sure you see it a bunch of times. So let's take a look. An exercise two, we've got the graph F of X equals the absolute value of X minus three, shown on the graph below. The function G of X is defined by one half times F at one half times X, then the absolute value, then minus three. So what I really want you to do is compare this formula to this one. Look at how similar they are with one key difference. In this formula, we multiply X by one half before we ever do anything else. Okay? So what we're going to do is we're going to graph our new function and we're going to do that by opening up the TI 84 plus all right. We want to go into Y equals, okay? We want to clear out anything that might be in Y one, Y two, et cetera. Take a moment. All right, let's enter that formula into Y one. We want to be careful. First thing that we have to do is get those absolute value bars. And remind you how to do that. You hit the math button, all right, and the first thing you basically see is ABS absolute value. So enter that. Now again, on the newer calculators like what I have is it automatically brings up those absolute value bars, which is super duper cool. On older calculators, like the TI 83 plus, again, one of my favorites, you'll get ABS with a parentheses. Okay? Make sure to put the other parentheses in once you've closed the absolute value bars. But for us, we're going to type in one divided by two times X all right. Now what I have to do is I actually have to hit the right arrow, all right? In order to get out of the absolute value bars, again, if you're on the TI 83, it would look like this. Now, I don't want to forget to put the minus three in there. So let's do that now, minus three. And I'm going to hand out it. Now, again, as always, check it over, make sure it looks good. Absolute value, those bars, one half times X, it's all good, bar, minus three, okay. And now I want to go into my table. All right, now that's always or my table setup. I'm sorry, I want to go into my table setup, and I have to figure out where to start the table. The key here is thinking about my graph paper, one, two, three, four, 5, 6, 7, 8, the minimum value of my graph paper is 8, the maximum is positive 8, so I'm making my table start at negative 8. So let's go into that table setup. Let's do it. All right, let's make table start at negative 8. And let's make the table go by ones. Okay. Now, check it over. Sorry, I feel like I'm moving too fast on this, and I don't want to do that. Now let's go into the table itself. So second, graph. Great. Now we don't have to write down every one of these points. We just want to write down enough of them. So for instance, I'll definitely start at the negative 8 and what it shows me is that I'm getting a Y value of one. I think I'll skip over the negative 7 because it's a decimal, but I'll maybe go negative 6. It's got a Y value of zero. Negative four has got a Y value of negative one, negative two. It's got a Y value of negative two. And zero has a Y value of negative three. And then we start getting symmetry. The kind of symmetry we saw with parabolas, but now it's with absolute values at two, and that negative two. At four, I met negative one at 6. I'm at zero and at 8, I'm at one. And now I want to graph it, right? So I'm going to go out to negative 8. And then go up one. Actually, let me erase this obnoxiousness here. So I'm going to go out to negative 8 and go up one. Then I'm going to go to negative 6, and I'm at zero. One, two, three, four, negative one, one, two, one, two, zero, one, two, three. Oh, that didn't go anywhere. Two negative two, four, negative one, 6, zero, and 8, one. I think we'll just use that line utility again. And slap scenarios on. All right. What was the effect on the graph of F of X when we multiplied the input by one half? Well, this is cool, watch. Let me show you it. Here we had an X value of negative three, but then that became an X value of negative 6. Right? Isn't that kind of cool? We could track another point here. We have an X value of negative two. That thing went over here to an X value of negative four. So each X value. On F of X was multiplied. By two to get the X value, our values on G of X so again, it worked exactly counterintuitive to what you'd think. You'd think, oh, I multiply X by one half, yeah, the graph's going to compress, but it doesn't, and actually gets wider then, right? It stretches out horizontally. And this is, I think, one of the trickiest things in common core algebra won this. This counterintuitive nature of what happens when you have a horizontal transformation. It's crazy. Crazy. All right? So I'm going to clear this out, pause the video now and write down anything you need to. Okay, here we go. Let's do one more problem. Maybe the hardest thing in function transformations is to put multiple ones together. All right? So what we're going to do in this last problem of the lesson is we're going to combine the vertical stretching that we saw in the last lesson. With the horizontal stretching that we saw on this lesson. So let's take a look. We've got our favorite letter function, F of X, graphed on the grid. All right? The new function H of X is defined as two times F of three X so two times F of three times X but a race is evaluate H of one. What point must lie in the graph of H of X based on this calculation? Why don't you go ahead and do that? Pause the video now. All right. Well, follow the rule. H of one will be two times F of three times one. So that's two of F of three. Well, what's F of three? Well, one, two, three. Here's three. One, two, three, four. So F of three is equal to four. So I'll get two times four and that'll be 8. So that'll be the .1 comma 8. Right? All right. But notice, it was really based off of the point, right? It was really the .3 comma four became the .1 comma 8. Now I want you to think about that, okay? What happened? Well, the X value was divided by three. But the Y value was multiplied by two, right? So whatever happens to Y, works kind of the way you'd think it was. The way. So hey, I multiplied the overall function by two, all the Y values get stretched or multiplied by two. But I multiply that X value first by three, and all those X values get divided by three. So the Y values on F of X are multiplied by two, and the X values. Are divided by three. And it says graph G of X by plotting three major points. Now we already have one of them. Three comma four ends up going to one comma 8. But let's take the other two. Let's take these two. So what's this point? It's one, two, three, four, 5, 6, negative 6, negative two. Negative 6, negative two. So what are we doing? We're taking the X value and we're dividing by three. So negative 6 divided by three would be negative two. And then we're going to take the Y value, and we're going to multiply by two, so that will be at negative four. And then let's do this point. Wow, that's at one, two, three, four, 5, 6, 7, 8, 9, ten, 11, 12. And the Y coordinate is zero. So again, we're going to take this and we're going to divide it by three. So 12 divided by three is four. Now this is a little bit weird with zero when we multiply that by two, it's still zero. So let's take the point, and let's graph, I think I'm going to graph this point first. I know that seems weird, but one, two, one, two, three, four. So that point went down there. All right, this one becomes one, one, two, three, four, 5, 6, 7, 8, one one up there. And this thing way out there at 1201 two three four, just becomes four zero. All right? Let me graph it in. Let me graph it in red just to make it really kind of bright and shiny. And look at that. Right? And what's really happened to this graph is we've taken it and we've compressed it by a factor of three. And then we've stretched it by a factor of two. All right, that can be pretty tough to think about. Follow the functions rule. All right. So pause the video now, write down what you need to, and then we're going to finish up the lesson. All right, here it goes. So today, we saw horizontal transformations. Specifically, horizontal stretches and compressions. You get those kind of things when you take the X variable, the input. And you multiply it by some constant before you do anything else. When you multiply by a constant greater than one, ironically, it takes the graph, and it compresses it. So if I multiply it by two, then the original graph would have all its X values halved. On the other hand, if I multiplied it by one half, all the X values would be doubled. Again, this can be quite confusing. And you'll get more and more work on it as you progress in your mathematics. Not so much in this course, because quite frankly, we're in the last unit. All right. For now, I'd like to thank you for joining me for another common core algebra one lesson by E math instruction. My name is Kurt weiler, and until next time, keep thinking. Thank you solving problems.