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## 8.F.B.4 Transcript

Core State Standards support video, Grade 8 Math, 8.F.4. This standard says: construct a function to model a linear relationship between two quantities; determine the rate of change and initial value of the function from a description of a relationship or from two (x y) values, including reading these from a table or from a graph; interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.

This is a whole lot to chew on, so let’s stop and think. The key to success for this standard really is for the students to have a solid foundation in all of the involved concepts, including the associated symbolic representations. So, they need that first in order to really be good at constructing functions. So, here’s the critical ideas. First is this idea rate of change as mentioned in the standard, and, of course, this is the critical idea of slope. Then you have the idea of a linear relationship between two quantities. What’s important here, and make sure the students realize this, is this whole idea of linear. The whole reason that it is a linear relationship connects back to the slope, and the idea here is that the slope is always constant. And that’s how come you have a straight line.

Then you have the idea of initial value, which is your y intercept, and in a real-life type of scenario, this is what you start off with. That’s why it’s called the initial value, which in real contexts, it makes sense to call it the initial value instead of the y-intercept. But it’s important that students make the connection.

The critical ideas for constructing a linear function; really it revolves around two things, a point and the slope. The problem with just having a point to work with is that I can actually have an infinite number of lines that go through that one point. So, that’s not enough information. By the same token, if I have a slope to work with, and let’s say, for example, here’s my given line. Now, the problem is that I can have an infinite number of lines with that same slope. So, then, what is essential to construct one function, which would be one line if it’s a linear function, is that I’ve got to have both. I need a certain point, and I have to have the slope for it. And so, that determines just one unique line. So, the key foundation is that I must have a point on a line and the slope of that line.

Now, let’s go back and review some of the basics. The most critical is slope. Now, slope involves the amount of change in your y value compared to the amount of change in your x value. To find the change, we simply take the difference of the y’s over the difference of the x’s, and that gives us our slope. Now, if we do a little bit of adjusting and we multiply both sides by the difference of the x’s, then this is what we’ll end up with. And if we do just a little bit of manipulation, and let’s let one of the coordinates just be the generic x and y, then we’ll end up with your point-slope form of the equation, which would be the unknown y coordinate minus your known y coordinate. And let’s switch the order around, put the slope first, times your unknown x coordinate minus your known x coordinate, and there’s your point-slope form.

Now, for your y-intercept, we know that that’s going to be a point that’s somewhere on your y-axis. And if you stop and think about it, that point is going to have the coordinates 0, y. Now the coordinates are 0, y for my y-intercept, but in the slope-intercept formula, that y intercept, they use the letter b. If we take the point-slope form and combine that with a y intercept of 0, b, and we substitute in its place, well, here, we will put b and our x coordinate is zero. So, then, we have y minus b is equal to mx. And then we do a bit of manipulation, and viola, we have our slope-intercept form of the equation. So, notice, starting with slope and working our way over to point-slope form and the slope-intercept form that, really, it’s all connected and it’s important that students see those connections.

Now, what students need to realize is that the only difference among any of these contexts, scenarios, or applications, is how the information about the slope and the point on the line are provided. And they also need to realize that really, the main difference with the information is whether it is provided directly or indirectly. They might tell me that the slope, for example, is two thirds. But, they might be sneaky and make me work at it and have to do some things to figure out that the slope is two thirds. So, when a student is going to construct a function, they need information about a point or points and information about the slope. Then, once they have that information, they can substitute that into either the slope-intercept form or the point-slope form of the equation.

Now, in the standard, it talks about reading these values from a table or from a graph. So, they might have some situations where there really isn’t a real-life scenario. They just might be given some points in a table. And so, they’d have two or three or four points to work from there. Or, they might be given a graph, and in that, from that graph, they would have to figure out the basic information. For example, they might be able to read the graph and tell where the y-intercept is, and then by counting, they might be able to figure out what the slope is. Maybe it’s 3 over 2 or something like that. And from that, they would have the information that they need. But, again, what they’re looking for is, they’ve got to have some information that would enable them to find the slope, and they need some information that will give them at least one point that’s on the line.

What will probably happen is that students are going to be expected to construct a function based on some kind of real-life scenario. So, let’s just look at some. Let’s say there’s a print shop and they charge 3 dollars a booklet, and they have a setup fee of 50 dollars. Now, here is where the students have to make some connections between the mathematics and real life. Three dollars a booklet, that sounds like the slope to me because you’re comparing the money to the number of booklets. So, actually, my slope will be 3 over 1 where 3 is the number of booklets and 1 is the amount of money. And then, for the points, well, let’s see. That $50, that’s the initial value . . . that’s how much I have to pay even without having any booklets printed yet. So, basically, that would be my y-intercept. In essence, it’s 0, 50.

So, I have a point 0, 50, but that’s also my y-intercept. So, in this situation, I’m probably going to be better off going with the slope-intercept form. This problem is one of the simpler ones that students will run into because they were given both. They were given both pieces of information kind of directly because what they were given was that the slope was $3 a booklet. So, there’s my slope. And then, my initial value, I have to pay $50 setup fee. So, there’s my equation: y is equal to 3x plus 50, with x being the number of booklets and y being the cost to have those booklets printed.

Let’s take another example, lawn mowing. Let’s say, for example, that you have a company that comes out and does some work for you, and they charge you $40. And they’re usually out there about 1 hour. And let’s say you run into a neighbor, and they have a much bigger yard than you do, and they use the same company. And you find out that they have to pay $115 and that they’re usually out there about 4 hours. And you stop and think to yourself, well, wait a minute, $40 for an hour; seems like they should have charged my neighbors about $160. Why are they getting a discount, or maybe it’s not as simple as it seems? Maybe it’s not $40 an hour that they’re charging.

Well, stop and think about it here. It’s almost like, yes, I’m given two points here. What students need to figure out is, now, which is the dependent variable, which is the independent variable? Which is my x, and which is my y? Well, it seems like, well, the amount of money that I have to pay is going to depend somewhat on the amount of hours. So then, the money should be my dependent, and the number of hours should be my independent.

So, I have 1 hour, they charged me $40. And for the other person, my neighbor, they charged $115, and they’re usually out there for 4 hours. So, I have a point already. I can pick either one that I want. The slope, I’m not given the slope directly. I’m going to have to figure that out. To get the slope, I need the difference of the ys over the difference of x’s. So, one way to do this then, would be, if I decide to subtract in this direction, then that would be a little bit too difficult because I can tell that I’m going to end up with some negatives. So, no, I think that I would rather subtract in this direction.

So, for my y values, I’ve got 115 and I’m going to subtract 40, over 4 minus 1. Let’s see, 115 minus 40, let’s see, that is 60, that’s 75 over 3. That simplifies to 25. Okay, so, now I can tell that my slope is 25. So, that’s $25 an hour, and then I can pick either one of the two points. Let’s see, I think I’d rather go with 1 and 40. So now, it’s a matter of figuring out, well, let me see, I think I’d rather go with the point-slope form. So, what I need to do is substitute. The slope is 25. My x value is 1. My y value is 40. So, I’m going to have y minus 40 is equal to 25 x minus 25. Add 40 to both sides. That would be 15. I know it’s a little hard to see some of this, but if we do the arithmetic, it all checks out. So, we have 25x, and then we add 40 to both sides, and that gives us a plus 15. So, it wasn’t $40 an hour. What happens here is that, apparently, they charge $15 just to go out there, and then they’re going to charge me $25 for every hour that it takes them to do the job.

One more scenario; let’s take a weight loss situation. Let’s say you have a friend and they’ve gone on a diet, and after 3 months, their weight is 218, well, his weight is 218. And after 10 months on this diet, the weight has dropped to 190. So, what do we have here? That looks like that would be two points, and it looks like his weight would depend on how long he’s been on the diet. So, we almost have these points set up directly where, if my x value is the time in months, and then my y value is the weight, here’s my two points to work with.

And then I can do the work, much like I did with the previous problem, where again, I can pick either one of the two points. I think I’d rather work with 10, 190, especially with the 10. That will make things a lot easier. Problem is, just like before, I was not given the slope directly. I’m going to have to work for it in this case. Okay, the difference of the y’s would be 218 minus 190, and the difference of the x’s would be 3 minus 10. Let’s see, that’s 28 over a negative 7. And that simplifies to a minus 4, which makes sense for this slope to be a negative because, apparently, he’s losing 4 pounds a month. So, now, I pretty much have my information. I can go through and find what the equation would be, and then I could answer whatever questions that they might ask. So, for example, they might ask, well, what was his weight at the beginning of the diet, those kinds of things.

Again, I won’t go through and finish this equation out. Again, the students have what they need to work with. Again, they have a slope of minus 4, which would be losing 4 pounds a month, and I do have one point of 10, 190. And again, now it’s just a matter of substituting into the point slope-form because you do not have the y-intercept.