Common Core Algebra I.Unit 4.Lesson 3.Nonproportional Linear Relationships

Let's begin. Now, for the last few lessons, we've been looking at what are known as proportional relationships. And two variables are proportional, let's say Y and X if when we divide them, we always get the same constant. Now, when we look at a proportional relationship like this, we can always rearrange it by using a property of equality and multiply by X on both sides, telling me that Y or whichever variable that you call it can always be found by multiplying X by some constant. Proportional relationships when we graph them. We'll always always pass through the origin. And that kind of makes sense because if I've got the equation Y equals K times X, when X is equal to zero, it doesn't really matter what the proportionality constant is. Y will always also be zero. Thus, their graph will pass through the .00. In this particular in this particular relationship, we can see that when X is one, Y is equal to three, that's all you really need for proportional relationship, right? That tells me that Y will always be three times X that's this guy. But what if it's not proportional? You know, what if it's a linear relationship? That does not pass through the origin. 

That's what we're going to be dealing with today. Okay? So I'm going to clear this out. And let's move on. So here, in the first problem, we have a nice linear function. Meaning that when we graph it, it ends up being the graph of a line. Very important, the graph of a line. All right, and let's start by having you do letter a this should be pretty easy. Evaluate F of negative two and F of one, what two coordinate points do these function values correspond to? So pause the video right now and really quickly do this. All right, let's go through it. Remember, the negative two and the one are X coordinates. So if I go to X equals negative two, Y is equal to negative one. So F of negative two is equal to negative one, that corresponds to the point. Negative two comma negative one. On the other hand, if I go to X equals one, I find that Y is one, two, three, four, 5. So F of one is equal to 5. One, comma 5. Okay? Now letter B asks us to calculate the average rate of change of F from X equals negative two to X equals one. 

So nicely enough over here. So remember the average rate of change. I'm going to abbreviate as a dot R dot C, kind of like arc, right? That's going to be change in Y value, F of one, minus F of negative two, divided by the change in the X values, one minus negative two. All right, this is literally change in Y divided by change in X well, we know that F of one is 5, and we know that F of negative two is negative one. I just put the parentheses there because it's a double negative and it kind of looks weird. And then we've got one minus negative two. 5 minus negative one is positive 6, one minus negative two is positive three. And that gives us an answer of two. And what is that known for? Or known as that's known as the slope of the line. The slope of the line is two. And you've seen that before in previous courses. Let me illustrate it. Let me actually erase this. 

So it's not sitting there. And let me illustrate it in red. If I pick a point, like this one right here, and I go to the right one, I will go up to any point that's true. Go to the right one, go up to right one, up to, et cetera. All right. Now, is there a proportional relationship between Y and X? How can you check? Well, you probably already know that there's not a proportional relationship. And because it doesn't go through the origin and on top of that, the actual title of the lesson is non proportional relationships. But here's how you check. In a proportional relationship, why divided by X will be a constant. It won't change. But look what happens when we take the point negative two negative one and the .1 comma 5. If I do Y divided by X here, I'd get negative one divided by negative two, and let me just simplify that a little bit. That'll be one half. Here on the other hand, if I do Y divided by X, I'd get 5 divided by one, and that would just be 5. These things in a proportional relationship would be the same. So now they're not the same. All right, they're just not the same. Now based on our 8th grade coursework, what relationship does exist between the two variables, write the equation and check it for points a so from your 8th grade coursework, you worked extensively with the equations of lines, extensively. 

And what you should have seen was that all lines, even ones that are proportional, can be written in the form Y equals MX plus B, where M is the slope of the line. And B is the Y intercept. Y intercept, sorry, that I kind of got connected to the N so we already know the slope, the slope is equal to two. And the Y intercept, well, one, two, three is equal to three. So this thing has the equation Y equals two X plus three. Now, kind of running out of room here. So I'm going to go back and get rid of this. And watch as I check this thing, why don't we just check one of the two points? Let's check the point negative two negative one. By checking it, what I mean is that when I put in an input of negative two, right? And I'll get negative four plus three, I get an output of negative one. Yay. Right? So this equation checked with that input output pair. Well, there's a lot of blue and a lot of lines all over the place here. All right. We are going to work extensively with equations of lines in this unit, so you'll get a lot more work on it. Still, pause the video now if you need to copy down anything you need.

 And then we're moving on. All right, here we go. Scrubbing. All right, so the slope intercept form of a linear function. F of X equals Y, F of X always equals Y and that's equal to MX plus B linear parameters. This is kind of a cool word, a parameter. A parameter is a constant, a constant, in any model. In any model, right? See, because there's constants, and there's variables. X and Y, they're variables. But our parameters, well, those are the slope, also known as the average rate of change, also known as the change in Y over the change in X and there's the Y intercept, right? And those are the two parameters for our linear functions. The slope being how fast Y changes compared to X and the Y intercept in my mind being where the function starts, right? The function being the Y values. So I'm going to clear this out. And then we're going to work more with equations of lines. Okay, let's keep working with equations of lines. 

Exercise number two, given the linear function G of X equals one half X plus one, do the following. A, create a limited table of values to help graph the function. All right. Let's do it. We could definitely use our graphing calculator here, so we've seen how to do that, but we're going to do this by hand, I think. Okay? Now, it says a limited table of values. Okay? Because we could go nuts here. We could go all over the place. Now how do we do this when we don't know anything else? Well, I look at my graph paper. And I noticed that my graph paper starts at negative 6, and it ends at positive 6. Okay? Might as well start at negative 6, and remember the rule is one half X plus one. So I'll have one half times negative 6 plus one, one half of negative 6 is negative three, and negative three plus one is negative two. So I have the pair negative 6 negative two. That's going to be right about here. I'm going to do letter B at the same time. I hope that's okay. Now, I could put negative 5 ends with my formula, but taking one half of an odd number is not going to be that nice. So let me skip over negative 5. Go to negative four. Right? Here, I'm going to get one half of negative four, which is negative two plus one, which is negative one. So the point negative four negative one lies on this line. And that's going to be right here. 

Now many of you probably remember quick ways to graph equations of lines. That's okay. But I'm going to go through the most generic way. Create a table of values one half of negative two is negative one plus one is zero. So negative two comma zero. You probably see the pattern already, right? You could probably fill the rest of it in just using pattern. When X is zero, I'll have one half of zero plus one. Which is zero plus one, which is one. That's going to be right here. Hopefully I'll have enough room. I seem to be getting better at drawing my straight lines if I draw them faster. One half of two is one, one plus one is two. Whoops. Two comma two. That's quite look right. Please forgive me. For we'll have one half of four plus one, one half of four is two plus one is three. And that's right here. And then finally 6. One half of 6 plus one, three plus one is four. All right. Man. That's good. Let's go here. Let's connect it. I should really extend it a little bit farther this way. And. I have some narrows. Because it certainly does go forever in both directions. Now, letter C says illustrate the slope of the function graphically. Well, as we know, right? Slope is one half. 

Actually, let me put that in red. All right. Slopes one half. And we know that's the change in Y over the change in X so if I grab a point right here, then I go this way, always let X increase, then what I see is that I go to the right two, and I go up one. And that's how we interpret a slope of one half, right? We interpret it as right. Two, and up, one. And letter D circle the graph's Y intercept, there it is. Please note there it is in our model, right? So everything's lining up. I'm going to clear all this out, so write down anything you need to make sure you've drawn those pictures on the graph. All right, here it goes. Okay. Now, exercise three says use information about slope and Y intercept to graph Y equals negative three fifths X plus four on the grid. Then we're going to do the rest of the problem. But let's start with that. I'm going to rewrite the equation. Down here. Maybe a little bit bigger. Negative three fifths X plus four. All right. 

Now notice our slope here is negative. Now, we should make that negative attached to either the numerator or the denominator, and I would prefer that you attach it to the numerator always. I'd like to move to the right. Since the denominator is the change in X, we have change in Y divided by change in X if we keep that denominator positive, then we'll move right 5. But this, we're going to move down three. But from where? Well, that's where the Y intercept helps, right? Here's a nice starting value. So if I go one, two, three, four, make sure it's a Y intercept, not an X intercept. And then I move to the right 5, one, two, three, four, 5, one, two, three, there's another point. Now we're talking about a line. So we really only need two points. But it doesn't hurt to then kind of go in the opposite order. One, two, three, four, 5, one, two, three. All right. We can now, of course, connect these with a nice straight line. Hopefully your line is going to be better than light. But who knows? That's actually pretty good. I like it. Get some errors. And we have a nice line graft. Now, it says pick two points off the graph and calculate the average rate of change, and verify that it's equal to the slope. 

Well, there's not a lot of points that we can really pick off that are nice. But these are two good ones, right? So that's at one, two, three, four, 5. One, two, three, four, 5, 6, 7. So we have the point negative 5 7. And one, two, three, four, 5, 5 comma one. So the average rate of change, F of B minus F of a, divided by B minus a is still just the same as the change in Y divided by the change in X so that's going to be one -7, one -7, divided by 5 minus a negative 5. One -7 is negative 6 and 5 minus negative 5 is positive ten. Now that may not look like the correct slope, but we can always reduce it in this case, at least by dividing both numerator and denominator by two and that will give me negative three fifths. 

Which is exactly the slope that I should have gone. All right? Average rate of change in slope are the same thing. Okay? So I'm going to clear this out. Write down anything you need to. Okay. Let's wrap up. So a lot of what we did today, I hope, was review. In other words, you worked extensively in both 7th grade and 8th grade common core math on equations of lines, graphing lines, and the slope and the Y intercept of a line. Today, we basically just reviewed that. Y equals MX plus B okay? So let me just say for now, thank you. Thank you for coming and watching another common core algebra one lesson by E math instruction. My name is Kirk weiler, and even though I seem to be garbling the ending to this video, but don't forget. Keep thinking. And keep solving problems.