 This video will talk about quadratic function applications. So we have a profit for a manufacturer of collectible gram-plot of clocks as given by this function here. P of x is the profit in dollars and x is the number of clocks that are made and sold. So find the y-intercept and explain what it means in this context. Well the y-intercept is just the constant and for us the constant is negative 375. So what does that really mean? Well the point is zero negative 375 and zero would be the x which is the number of clocks. So zero clocks made results in how much profit. Well the profit is the y value and that is negative three hundred seventy five dollars. So you don't make a profit if you don't make any clocks. Kind of makes sense. Find the x-intercepts. Well I'm going to graph to find my x-intercepts. So I'm going to clear out the equation that I had. I'm negative 1.6x squared plus 240x minus 375 and I'm going to go and make a window that I think will work. And from my window I'm going to go from zero clocks to and I'm going to say 200 clocks. I suppose they're going to make quite a few and we'll go every 10. And then the number of profit they're either going to make zero dollars or they're going to go up to I'll say maybe 10,000. And if I go to 10,000 I'm going to go every thousand in my scale and I look at my graph. Do I see the whole thing? I see it up. The x-intercepts that's going to be then something over here. So second trace five and enter, enter, enter. We'll find this one over here. X is equal to 148.4 and I'm going to have to draw my graph. If I want to verify that that's part of my stuff here. This is 10,000. This is 200 and then I want to find the other one. Let's go look at our table. It's somewhere between one and two. So I better go do the second trace five and I need to go to the other side. So I'm going to go to Y2 because that's a straight line which will get me over to the other side of my vertex quicker. And it's hard to see because I have all my numbers in the way. But I think I'm about over there. This looks good. So we'll try it. Enter, enter, enter. And I find that that one is x is equal to 1.6 approximately. So what did those two things mean? That means that if we, those are x intercepts. So it's actually this point zero and this point zero. So that if we make 148 or two clocks, profit will be $0. And these are rounded up. So how many clocks would be made sold to maximize the profit? Well, if you think about the graph, we've got the number of clocks going here and it goes up and up and up and up and up. And then it all of a sudden starts falling. So the maximum number would be up here at the top. It'd be the vertex. So we could do the negative b over 2a if we wanted to. But since I'm using the calculator to find other things, let's just find the vertex. Vertex is the maximum of this. So we want second trace four. Have to go to the left bound of it. And I'm on the left side, but I want to get up there closer. So enter and then go to the other side. And I just have to be definitely on the other side of that vertex. But that's good enough. Enter. Press enter again. And I find out that that's 75 and 8,625. We want to know how many clocks. So the maximum number of clocks would be 75. And the maximum profit is going to be 75 clocks earn $8,625. This next problem says an object is thrown upward into the air within an initial velocity of about 176 feet per second. After t seconds and h of t is above the ground, t seconds when h of t is above the ground is given by this function. So how high was the release point of the object? The release point would be when t is equal to zero. Well, if you put zero in here and you put zero in here for t, you're going to find out that the release point is zero feet. How far from the release point did the object reach its maximum altitude? Well, we could do negative b over 2a if we wanted to, or we could use the graph. And we could just do negative b over 2a. Let's try that. Negative b over 2a is going to be equal to the t. And negative b is negative 176 divided by 2 times negative 16, or negative 32. So coming over to my calculator to do that for me real quick, negative 176 divided by, and I could even put the whole thing in it, parenthesis 2 times negative 16. And we get 5.5. So how far from the release point did the object reach its maximum point? This is a t. That's how long. So we have to plug and check. So we have negative 16 times 5.5 squared plus 176 times that 5.5. But I have my trusted calculator over here, so that shouldn't take me too long. And we find out that it is 484 feet. It started at zero feet and it went to 484 feet. So that's how far from the release point it is. It's 484 feet. What is the maximum altitude by this object? It went instead of what? And when it does that is at 5.5 seconds. And how far from the release point did the object land on the ground? So we want to know the x-intercept. So here, if I'm going to do it all by hand, I want to know when the height is zero. That's when it lands. So negative 16t squared plus 176t. Well, I could use my quadratic formula since it's a quadratic set equal to zero. So I'm going to come over to my calculator and use the quadratic formula. A is negative 16. B is 176. C is going to be zero because we don't have a constant. And we find out that t, we use the quadratic formula, and t is equal to zero and 11. So for our case, it's going to be object lands at 11 seconds. All right, our final problem here. M and Ethan are fencing off a large rectangular area for a temporary pin for their animals. What is the maximum rectangular area they can enclose with 384 feet of fencing? And then what are the dimensions? So we have, we know that they have a fence. Let's kind of draw a picture. Four sides to this rectangle. And we know that all the way around is going to be 384 feet, which is the perimeter. And perimeter is 2l plus 2w. So that's one nice little fact. And then we don't know what the area is. That's what we're trying to find. But we do know the area is equal to length times width. So what if I do substitution and solve one of these equations for l or for w? I'm going to solve it for l. Don't ask me why, I just ask. So 384 minus 2w is equal to 2l. Divide everything in half. 384 divided by 2 is 192 minus w is equal to l. And I know that w is equal to just w. I don't know what it is, but it doesn't have any multiples or anything being added or subtracted. So now when I find area, I know that it's going to be the length times the width. But length happens to be 192 minus w. And width happens to be w. And that gives me 192w minus w squared. And I need to know what the maximum of that graph is or the maximum of that. So negative b over 2a would be okay. And I can plug in chug or I can plug it in. And I am a calculator girl. So I'm going to do it as 192w or x in our case in the calculator, minus x squared. I'm going to set my window to be 0 to 100 would probably be okay for me. And I'm going to go 0 to, well, I'll just keep it at 10,000. Let's see if we need all that. Oh, I need more than 100, but I can see my vertex. And that's all I really care about right here. So second trace 4, it's a maximum. And I think I can get to the other side of that vertex, but I may have to extend my window. So enter for the left. And I'm going to get to go right to the edge of my graph to see if I can get past that vertex. Press enter and then enter again and I might get an error. Oh, I didn't. So the vertex is 96 and 9,216. And this is a and w. So x is w. My y is my a. Draw my little graph up here to verify what I did. And up here we had 96 and 9,216. So we know that the width is 96 and the length is 192 minus 96, which is 96. And then it asks us for what is the maximum area. Area is equal to 9,216 and all of these are in feet. So what is the maximum area if they use their barn on one side of the rectangular area and the fence on three sides gives us a 384. So now we have 384 is equal to two times a length plus a width. So now we have, we can solve for w here very easily. So 384 is minus 2L is going to be equal to w and L will be equal to L. So area is equal to length times width. And length happens to be L times the width, which is 384 minus 2L, which gives us 384L minus 2L squared. That changes our equation a little bit. So I have 384x minus 2x squared. And I would like for my window to be nice for me, I wanted to go, let's say 200 this time. And I want it to go up to instead of 10,000. I'm hoping that maybe I have a little bit more area since I have one side that I can use my house for. So I'm going to say 20,000. And let's look at our graph and see what I got. Oh, I have the whole thing. So here's my graph. This is 200. This is 20,000. And my graph looks something like the length and area is equal to 96 and 18,432. So the area, maximum area, and let's put max here, is equal to 18,432 feet. That maximum length is equal to 96. And the maximum width is going to be that, what did we say width was 384 minus 2L, 192. So it's 192 feet by 96 feet.