Common Core Algebra II.Unit 8.Lesson 1.Square Root Functions

Hello, I'm Kirk Weiler, and this is common core algebra two. By E math instruction. Today, we're going to be doing unit 8 lesson number one on square root functions. Now, unit 8 is all about roots, radicals, exponents, their properties. And what we're going to be doing today is really reviewing what the square root function looks like. And some of its limitations in terms of its domain. So let's jump right into it with a square root problem. Okay. Now whenever we investigate a new type of function, we always want to make sure that we understand exactly what the most basic type of the function looks like. So the base function, F of X equals the square root of X, we're asked in letter a to graph without the use of our calculator on the grid shown, label its equation. Well, let's do it. Let's create a little table of values. Now, what we should already know is that it's impossible to take the square root of negative numbers. But we can certainly take the square root of zero, right? The square root of zero is zero. Think about that. Zero times zero is zero. So the square root of zero, zero, no problem. Now, the square root of one, well, that happens to be one. Yes, not plus or minus one. This isn't all square roots. This is only the positive square root. If I wanted the negative square root, I would have to be graphing this. But I'm not. All right? Now, I could certainly put two there. The problem is the square root of two is an irrational number. It's about 1.4. And we could certainly use our calculator for that. But we're not going to actually include that value of X and our table. In fact, the next value of X that has a square root that's nice, at least integer value is four. And of course, the square root of four is two. The next one after that, that has a nice square root value is 9, and it's square root is three. So let's graph those points. That's where our graph paper kind of runs out. We have zero zero. One, one, four, two. And 9 three. And we end up having this classic function, right? This half of a parabola. Because that's really what a square root function tends to be is half of a parabola. Now let her be says, using your calculator to generate a table of values, graph Y equals G of X on the same grid and label its equation. Start your table at negative ten to see certain values of X, not in the domain of this function. So let's actually do this together. I know you've generated plenty of tables on your graphing calculator, but I'd like to look at this one together. So let's bring out the TI 84 plus. All right, that's pretty loud snap. Sorry about that. But at least it brought our calculator out. So let's hit whites. We're not going to put F of X in there. We're going to just put G of X N so let's clear out an equations that might be in your calculator. All right. And let's put into Y one. The square root of X plus three be careful folks. You're going to have to maybe hit the right arrow a little bit to get out of that square root symbol so that you can then put down minus two. Make sure the subtraction by two is not under the square root. But once you've done that, really take a look at your function. And let's go into the table. All right, now our graph paper starts at negative 6, but let's go into the table setup. And it says start at negative ten. So we will make our table go by ones. And that's all set up. Let's pop into the table. Okay. Now what we see is when X is negative ten, we see an error in the Y column. And when X is negative 9, we see an error in the Y column. When X is negative 8, we see an error. And the Y column. In fact, we have to go a little ways. Let me do this in red. We have to go a little ways until we finally hit X equals negative three, right, one X is negative three, we get Y is negative two. All right? Now, when we hit X equals negative two, we get a Y value of negative one. But that's basically where this stops being very, very nice, right? In fact, we have to keep going all the way till we hit Y equals, or sorry, X equals one. Before we get another nice value of Y, which ends up being zero. And we can kind of keep going and going, but notice all those kind of ugly Y values. They're not very pretty looking. But finally, if we get to an X value of 6, we end up having that Y value of one. Okay? So let's graph this thing. What do we have? One, two, three, one, two, negative two, negative one, one, zero, and two, three, four, 5, 6, one. All right, and there's that square root graph. All right. Let's take a look at letter C, a little domain and rage, right? Remember that domain are the set of all allowable inputs, all the inputs that will give us an output. And the range is all the outputs. So C, what you can do in terms of writing the range, sorry, the domain and range on each of these, okay? Pause the video now. All right. Well, when you look at the first graph, the only X values that are allowed in the first function are X values that are greater than or equal to zero. So a little set builder notation, we'd say all values of X X greater than or equal to zero. The range is actually exactly the same. All those Y values turn out to be greater than or equal to zero. As well. Right? Now, when we go over to the other one, we see that its domain now is all values of X that are greater than or equal to negative three. And its range are all values of Y, such that Y is greater than or equal to negative two. And of course, what we're really seeing here, right? Forget about the square root function. But we're just seeing those fundamental shifts again, right? By adding three to X, we took the original square root function, and we shifted it three units left. And then by subtracting two from the overall function, we dropped it down two units. All right, so our shifting formulas, which we saw in the last unit, and in another previous unit, still work just fine. Okay? So let's pause the video right now and write down anything you need to. Clear out the text. And let's take a look at exercise number two. Which of the following equations would represent the graph shown below. All right. Now, let me concede something right now. You could take each one of those four equations, enter them into your calculator and graph and see which one looks like that particular graph. And you know what? There'd be nothing wrong with that. Really, nothing wrong. It's going to kind of miss the purpose. All right? So let's talk about how we can do this without using our graphing calculator. But first, if you think you can do this problem without my help, pause the video and take a shot at it. Okay. Well, what do we really know? The original square root equation looks like this. Okay. This is the square root of X so how has this one been transformed? Well, it's been shifted for units to the left. Right? That's the guy that's in dashed there. But then that one was reflected across the X axis. And how do we do that? By negating the entire function, you remember that? So our equation must be negative square root of X plus four. And again, you could totally do that completely on your calculator. But there's really two transformations that have occurred to the basic simplest square root function. We've shifted it four units left, and we reflected it. And by the way, the order there really doesn't matter. You could definitely reflect it first, and then shift it for units to the left, and you'd get the same graph. Okay, pause the video now and write down anything you need to. Okay, I'm going to clear out the text. And let's go to the back side of the sheet. All right. Exercise three. For the remainder of the sheet, we're going to be talking a little bit about the natural domain of square root functions. Okay? So let's take a look at number three. It says which of the following values of X does not lie in the domain of the function Y equals root X -5. Explain why it does not lie there. Okay? So think about this. One of these four values of X when you substitute it into that function will not give us an output. But which one? Pause the video now and think about it. Okay, so let's talk about it. Well, let's just go there. If I took X equals 6 choice one and I substituted it in, I'd get Y equals the square root of 6 -5, which is the square root of one, which is one. All right, great. Let's say I tried X equals two. I get Y equals the square root of two -5. Which is the square root of negative three, which is, oh, I can't do that. I can't find the square root of a negative number, because a negative times a negative, let's try that again. A negative times a negative is a positive. And a positive times a positive is a positive. So yeah, that can't be in the domain. There's no output. Now you might think, but wait a second. Let's try X equals 5. And that would give me the square root of 5 -5. Which is the square root of zero, which is zero. I like zeros and output, no problem there. What about X equals 7 surely? That's not in the domain. And some students will think it's not in the domain because they'll look at the square root of two and they'll go well now. You can't find the square root of two. Well, yes, you can. You can definitely find the square root of two. It's definitely a number. It's definitely a real number. It just is irrational. It's not nice. We can't say, of course, you know? Unlike the square root of one or the square root of zero, or we kind of come up with a nice rational answer. The square root of two is irrational, you know? Accurate to a few decimal places. It's about 1.414, et cetera. All right, so it's a real number. It's just not a nice number. Where is the square root of negative three? That can't be any real number I know, because multiply two numbers together. You don't get a negative. Two of the same numbers together. I'm sorry. All right, so let's keep working with this idea, which basically boils down to one simple thing. What is under the square root for the values of X that get plugged in must be greater than or equal to zero. It's got to be greater than or equal to zero, or you can't find its square root. All right. Pause the video now and write down anything you need to. Okay, let's clear it on out. Exercise four. Determine the domain for each of the following square root functions, shown inequality that justifies your work. Well, some students will like to do these by kind of guessing in check. Well, let me put a four in there. Yeah, that works. Let me put a negative 7 in there. Now that doesn't work. But ultimately, here's how I would want you to do every one of these if you were sitting in my class. I would want you to take the quantity underneath the square root and set it greater than or equal to zero. Because that's the deal, right? Don't have the square root there. Just have the quantity under the square root. But this is a nice linear inequality. So its domain is all numbers greater than or equal to negative two. So if you pick a number greater than or equal to negative two, you throw it in that function, you're going to get an output. It may be an irrational numbers and output, but you're going to get an output. Try to solve the other two as well by doing exactly what we just did. All right. Let's go through them. Letter B is really no different. I get three X minus two is greater than or equal to zero. Add the two to both sides, three X is greater than or equal to two. Divide by three on both sides, X is greater than or equal to two thirds. Now, there was a little tricky piece in letter C here, I'll get 8 minus two X is greater than or equal to zero. Subtract 8 from both sides and I'll get negative two X is greater than or equal to negative 8. But as soon as I divide both sides by negative two, that is going to flip flop my inequality. X is less than or equal to four. And it's important because if you're left with something like if you don't flip flop that, you would be left with X is greater than or equal to four. And if you took, let's say it 5, which is certainly the case here, and you tried to figure out the Y value. Right, you'd have the square root of 8 minus ten, which would be the square root of negative two. And that, for the time being, is impossible. Bleed a little bit over into the black. That's okay. That's impossible. Okay, so finding the domain of square root functions, especially when the function underneath the square root or the expression underneath the square root is linear. It's pretty easy. But let's take a look at one more that's slightly more challenging. First, pause the video and write down anything you might need to. Okay, let's clear out the text. And tackle that last problem. Consider the function F of X equals the square root of X squared plus four X -12. It says using your calculator using your calculator to sketch the function on the axis given. Again, not the greatest grammar there, but we'll correct that by the time the worksheet gets in your hands. So what I'd like you to do is put that function into your calculator exactly like that. You know, use the sort of axes on here, one, two, three, four, 5, 6, 7, 8, so this goes from. Negative 8 to positive 8, right? And, you know, you can play around with the window as you see fit on the Y but why don't you put that in and sketch it, okay? All right. Well, what we find is a graph that looks something like this. Like that. And one, two, three. Oh, no. That's at 6. Sorry. Something like that. Let me erase that other dot. All right, and that's the graph that we see. But why do we see that graph? Why is there this sort of chunk in the middle here? This chunk in the middle here, where there's nothing, and there really isn't anything, if you made the graph much bigger, you wouldn't see anything up here or down there. So let's take a look at letter B, it says set up and solve a quadratic inequality that yields the domain of F of X you see again, I don't really care what's underneath here. All right? The plain fact is the only way I can evaluate this function is if the quantity sitting under the square root is greater than or equal to zero. But this is cool because it goes back to a topic that we had a couple units ago granted. It was a little while ago, which was solving quadratic inequality. So let's review how to do that. We've got the squadra inequality and the first thing that we're going to do with quadratic inequality is strangely enough, we're going to change it into an equation. All right? Then what we're going to do is solve that equation, oftentimes with the zero product law. All right, we get X equals negative 6. At X equals positive two. That's step one. Step two is we're going to plot those two things on a number line. And we're going to decide whether we should have open circles or filled in dots. Now, these two values of X make the equality true so they make the inequality true. Now step three, we're going to test values, for instance. If I took X equals negative 7, all right? And I tested it in this inequality. What I get is 49 plus 28 -12 is greater than or equal to zero. Of course I didn't actually work out what that all was. Let me do that really quick 49 plus 28 -12. One thing I know for certain is it is greater than zero. And I'll get 65 is greater than or equal to zero. So that's true. Let me pick a value in between, like X equals zero. If I put that into my quadratic inequality, I'd get negative 12 is greater than or equal to zero. And that's definitely false. And finally, if I chose something like X equals three, and I substituted it in and got 9 plus 12 -12 is greater than or equal to zero. That would be 9 is greater than or equal to zero. And again, that would work out just fine. Oh, actually, you know what? That's -28. And that's 9. So that's true as well. Therefore, the domain of this function. Is all these values. And we see that in the graph, right? That's the whole point of domain. You know, there should be nothing in between negative 6. And positive two, and there isn't. But then the function exists from negative two to the right, and negative 8 to the left. So we could say that our final domain is all values of X such that X is less than or equal to negative 6. Or very important, X is less a greater than or equal to positive two. That word or very important. All right, kind of cool. So let's pause the video now and write down anything you need to. Okay, clear out the text. Let's wrap up this lesson. So in today's lesson, we reviewed what the basic square root graph looked like. We saw that it got shifted just the way all other functions get shifted. And then we played around with the domain of the function. Specifically, because the domain of square root functions can be interesting. Simply due to the fact that you can't take the square root of a negative number. All right? So I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler. And until next time, keep thinking and keep solving problems.