Okay, Unit 2, Lesson 5. So we're going to look here at writing functions of a single variable, really from word problems. And so what we're after here is the output of the function is something we're interested in, and we know we can manipulate some property of this thing we're trying to model, and that's going to be the input to our function. A lot of times we're going to have multiple equations.
multiple things that describe that system. Okay, so we really only have one learning objective in this whole lesson, but we're going to spend a lot of time on it because my experience is that you need help, that students struggle with this. So let's look at our first example.
And we're going to go here with sort of what I would call a pure mathematics example. And then we'll look at an actual word problem. So here the words that we're going to use are going to describe a situation from math that you should be fairly familiar with.
So let p be a point on the graph of y equals x squared plus 3. Express the distance from the origin to the point p as a function of the variable x. So what? What I'm going to do is, and there's always a good strategy in these problems, if at all possible, draw yourself a picture, because it often can be very helpful in orienting yourself.
And let's see, I've got y equals x squared plus 3. So that's a parabola. Its vertex is at 0, 3. It doesn't really matter too much. what exactly that looks like there's my parabola come over here I'm going to say this is the graph of y equals x squared plus 3 Now, let P is going to be a point on the graph.
So let's come on here and let's just pick point P here. All right? Put this side over to the side.
There's point P. Now, don't put that at a conspicuous location. So that's kind of why I slid it out to the side.
Now we want to find the distance from the origin to the point P. Well, let's get a different color. down here and we'll do our origin, right?
So that's zero, zero. You can call that capital O for our origin if you want. And then, so it is this distance that I'm interested in, right?
So now I'm going to come up here. I'm going to start trying to do the things that we know. All right.
So distance. So if I have two points, let's call that point A. And we'll do X sub A, Y sub A, and point B. X sub B, Y sub B. Then the distance between A and B is what?
The square root of XB minus XA, quantity squared, plus. yb minus ya quantity squared. So that's just our standard distance formula for the distance between two points.
If you, you know, if that's something you really should know. If you are having trouble thinking about it, if this is point A and this is point B, then... And the base of that right triangle is XB minus XA. And the height of the other leg of that right triangle is YB minus YA.
And so those are the two sides of the right triangle. triangle, sum of the squares is equal to the square of the distance of the hypotenuse there, and so we've just solved the Pythagorean there. Okay, now, so now I would like to find the distance.
Well, I need two points, so I'm going to say, well, we'll let the origin, which has coordinate zero, zero. Be one set of points and then my other point is point P Point P could be anywhere on the curve, right? So point P is X Y so I'm just gonna do it like that.
That's any point that's on the curve. So we'll just do X comma Y Now, so the distance is going to be equal to the square root of... x minus 0 quantity squared plus y minus 0 quantity squared, which is just equal to the square root of x squared plus y squared. So that's my distance.
But notice that I want it as a function of x only. Well, what do we know? We know that if I'm on the curve, y is equal to x squared plus 3. Therefore... Distance is equal to the square root of x squared plus x squared plus 3 squared.
Now, you know, I'm going to go ahead and simplify. I'm going to back out just a little bit. And we've pretty much got it. I'm going to clean that up for us.
Right, so the x squared, that's going to be what? Foil that out. I'm going to get x squared plus x to the fourth plus 6x squared plus 9. And when I put that together, I get x to the fourth plus 7x squared plus 9. So that leaves me with distance. equals f of x, which is equal to the square root of x to the fourth plus 7x squared plus 9. That's really what we're after there, okay?
Is that function, give me an x value, and I can tell you the distance that the point on the curve is from the origin. So this is a nice kind of math problem. Let's look at a little more applied problem as our second example here. And that second example is going to be, we have a rectangular field and it's to be enclosed by a fence. The area of the field is 500 square feet.
Express the feet of fencing needed to enclose the field as a function of the length of one side of the field. So again, first thing we're going to do is draw a picture. My rectangular field, right? So what? That field has a length and a width, and we know that the area is 500 feet squared.
Okay? So we wanna find that feet of fencing needed, so that's gonna be perimeter. So I'm going to go to my perimeter.
And what is the perimeter? The perimeter is 2 times L plus 2 times W. And if you want to keep your units on there, I think that's a good thing to do. Now, I've got L's and I've got W's. I need to get this down to a single row.
I want to get it into the function of just the length, right? It's a function of just the length information. Well, I've got length and width in there. I could factor a 2 out, but that's not really going to help me.
What I want to take advantage of the fact is I know that area is equal to length times width, and in this case that must be 500 square feet. So what I can do is I can say, oh, well, that's the key. If length times width is 500 square feet, then what do I want? I want length, so I'm going to solve for w is equal to 500. And I'm going to go ahead and put my units on there because I think it's a nice piece to see how when I divide by L, which has units feet, then the width is actually 500 over L feet.
That is, the units actually make sense there, and that's always a good way to kind of help check yourself when you're working a problem like this. Now I can get to... What I'm going to say is, F and I, you know, this would be fine. So what do we, when we want length of fencing? Yeah, feet of fencing.
So total fencing. is going to be what? I'm going to call that a function and it's going to be a function of length L and that is going to be equal to 2L plus 2 times W which is 500 over L.
So what did I clean that? That's going to be 2L plus 1000 over L. All right. Now, that is fine. That is a function.
X is a function. The total fencing needed is a function of the length. If you wanted to say let X be equal to length, then you could have F of X is equal to 2X plus 1,000 over X. And I would encourage you to keep trying to think about what your units are on the output there.
Okay? All right. Two good examples there.
For the lesson practice problems, you've got some more examples to work.