 Welcome. Today we're going to be talking about some quadratic function applications. So these are big reasons why you actually need to study quadratics. And as I put up here, there are many applications that follow the quadratic model because of gravity. If you throw an item up, it's going to come back down, and then that often looks like a parabola. Okay, so here's a perfect example. A ball is thrown downward. Okay, so I guess this one didn't really go up too much. From a height of 512 feet, whose position at time t in seconds is given by this particular function. Okay, so if we were to graph the whole function, it would certainly look like a parabola. What is the height of the ball after two seconds? So is this to your input for your time, or is your to the output for your height or position? So in this case, definitely it's time because it's measured in seconds. So we want to do s of 2. So you can go ahead and plug a 2 in for t, maybe negative 16 times 2 squared minus 64 times 2 plus 512. Okay, but instead of doing all that arithmetic out, I'm going to pull out my calculator. Let's get that started. I'm going to go to the y equals. We're probably going to need a graph anyway. So let's go ahead and graph negative 16. I'm just going to use my x key. It's a variable. We'll just have to remember x is actually time. Squared minus 64x plus 512. Okay, beautiful. So I'm going to take a look at my table, second table. I want my value in x is 2, or time is 2. So that's going to give me 320 feet. And you notice how I put units on that one. So this is 320 feet. That would be my answer there. Okay, when does the ball reach its maximum height? So I see that word maximum. That means I need to find the vertex. Okay, so let's go back to the calculator. I already have it in my y equals. Let's go ahead and hit graph. That doesn't look like much of parabola to me. So we have problems with our window again. Let's go to the window. We'll leave the x's alone for now. Remember the last time we changed our y min to negative 15. But negative 15 doesn't make any sense. So let's start that at zero, since we're talking about height. And it started at a height of 512 feet. So let's go ahead and just do 525 to be safe. I picked that out of, totally out of the air. Anything bigger than 512 will certainly work. And then my scale, because if it goes by once, it's going to be really small. So I'm going to make it go by 25s, just to make it look a little bit better. So when I hit graph, oh, it's a little bit better. But, yeah, it's still not exactly perfect. So let's go ahead and adjust that window a little bit more. So let's make my y max, how about 550? That should be enough. And if you made it that to begin with, good for you. Oh, maybe not. Okay, we're just off of that. So let me go a little bit more. This is the thing about these windows sometimes. A calculator can be helpful, but not always. All right, I'm going for 600 this time. Let's hit graph, and there we go. Okay, so when does the ball reach its maximum height? Oh, but this doesn't necessarily. So here would be our maximum, right up here. And we could find that on the calculator. But notice here it's on the negative side of our x's. So having a negative time doesn't really make a lot of sense. But we'll go ahead and find it just to get practice finding it. And then we'll interpret our answer properly. So let's go to second calc. And we want the maximum this time, since our parabola opens down and it asks for the maximum. Okay, let's go ahead and toggle around. Let's get over on the other side of that maximum. Because I'm on the right side right now. Get on the left side. Hit enter. Toggle around. Let's go ahead and get the right bound. Hit enter. And my guess is somewhere in here, although I can't see it anymore because of my function. That's fine, hit enter. And you notice my maximum's at negative 2, 576. So this would be 2 seconds before it was thrown. Which again doesn't exactly make a whole lot of sense. But remember we were throwing this object down. So that's why that would be the case. So when does the ball reach its maximum? So let's write down our vertex is at negative 2, 576. And that's a negative. So this would be 2 seconds before it's thrown. Which again doesn't make any sense, but that's okay. It's thrown. Again, we want to practice this stuff. So having a negative in there, like I said, doesn't make sense in terms of the problem exactly, but that's okay. All right, what is that maximum height? So that maximum height would be 576 feet. Because that's our y value. So by finding our vertex, we actually answered 2 questions with that one. Okay, great. When does the ball hit the ground? And that was feet up there. Let's go back to the calculator. So let me clear that part of it out. So it looks like the ball is hitting the ground about right here. So that would be 1, 2, 3, about 4 seconds. Remember that's also called our x intercept, right? As far as that goes. Because that's where it crosses the x-axis at. So with this one, we want a height of 0. So this makes s of t equal to 0. So we want the x intercept. Just giving you all the different interpretations for what I'm asking here with this one. Okay, so when I go back to the calculator, let's take a look at our table and see what the value of 4 looks like. Oh, look at that, 4, 0. So after 4 seconds, this ball is going to hit the ground. So t is going to equal 4. Perfect. So again, like I said, this whole two seconds before it's thrown thing doesn't really make a lot of sense. But you've got to do some practice with this stuff. Okay, next one. Now we're talking about the number of applicants for asylum. Okay, and this is in the United Kingdom. We've got some data off a website here. And our function, this one is now in for text form. So our last function that they gave us was in standard form. This one's in for text form. So we get to practice both of them. Okay, so a of t represents the number of people in thousands. And then t, notice right here, this is an important line, is years since 1990. Okay, so we're going to have to remember that 1990 is kind of our base year if you want to think about it that way. Okay, so how many applicants filed for asylum in the UK in 2003? So are they giving us an input or are they giving us an output? Well, 2003 is definitely a year. So this is going to be a time and then you have to think about how many years was it between 1990 and 2003. So our t is going to be 2003 minus 1990. So I believe that gives us t would be 13. Okay, so we can plug 13 into the formula. Or again, we can go ahead and graph it and take a look at the calculator. Let the calculator do the work for us. Or we can just do it by hand either way. So what this would look like then would be a of 13, because that's our t, would be negative 3.5 times 13 minus 11 squared plus 81. And this is just some number crunching. But like I said, I'm going to go ahead and put into my y equals, let's clear out the last one we just did. So I've got negative 3.5 parentheses. Now I'm going to put my function in here because we're probably going to be doing other stuff with it. X minus 11 parentheses square plus 81. Now remember we've messed with our window. So let's see what happens here. And before I get too far, let's go back and refresh then what's happening. So the negative in front here for my, what we called our a value. That means our parabola is going to open down. This minus 11, that means our parabola has been moved to the right 11 and plus 81 means it's going to be up 81. So my vertex is at 1181 just again. So we have an idea, but that means our window is probably going to be quite a bit off. We said our x was 11. So let's go down and let's fix that x. Let's make it 15. Oh no, let's make it 20. Just to give us something to look at here. My y max of 600 is now a little bit obnoxious. So let's go ahead and do 90 just to be on the safe side. And let's go by 10. So when I graph this, there it goes. Good. I can see my whole parabola. So this is opening down as we thought. It got moved left 11, right 11 and up 81. So this makes sense. Okay, but back to what we wanted to do. We wanted a value of 13. So let me go ahead and look at my table. Scroll down a wee bit here until we get to 13. And that's going to give me a value of 67. And remember our value is in thousands. So this is going to be 67,000 bookends. Okay, sketch a graph of the model. Well, we already did that. So let me just go ahead and draw that in. So it's going to look something like that. And you could do yours to scale a little bit better than I do. But as long as we have the general shape, that's all that matters. All right, what is the vertex? Well, I already gave that away earlier. I said the vertex was at 1181 because those are my values for H and K. And if you don't buy that, you can always, again, go to the calculator. Look at the graph. You can eyeball 1181 or you can do that second calc and go through the steps of finding the maximum again. Okay, but I know from the formula that that is going to be my vertex. What does it mean in terms of this problem? So remember the vertex is at the maximum since this parabola opens down. So that's going to be up here, the tippy tippy top. And then, so my X value, remember, represents time. And my Y value represents applicants or A. So that means 11 years after 1990, so that would be in 2001. The, well, let's see, the number, let's see, there were 81,000 applicants, the maximum number or something like that. You could probably come up with a better sentence than I did. You can always say the maximum number of applicants was in 2011 when there were this many as well. Yeah, that wasn't worded very well. That's why I do math not English. All right, in what year did the number of asylum applicants reach its highest? Oh, we already answered that in 2001. Use the graph to estimate in what year or years the number of applicants was 25,000. Okay, so let's go back to the graph. I think I did these by 10, so 10, 20, so somewhere around in here, so we want to eyeball that. So I'm going to go back to my Y equals and go ahead and put in another equation down here in my Y2. I'm going to type 25. Now, I'm not going to type 25,000, because remember, we're assuming these are in thousands. So if I click on graph, bam, there it is. So you can eyeball this point here and this point here if you would like. We can also look at our table. Let's see, does 25 ever show up? Oh, it does. So here it's 7. Okay, so 7 years after and does it show up again? It does at 15. Okay, so use the graph to estimate. We actually could find it exactly off of our table, which is nice. It doesn't happen every time. So if it didn't happen, you can always then go to intersect and figure out where those two graphs intersect since I put the two equations on there. So let's see, so T was, what did I say, 7 and 15? Let me go back and look at the table just to make sure. 15 for sure. And 7 was the other one? Yes, so that means then in 1997. And so 15 years after 1990, that would be in 2005. Okay, give a reasonable domain and range for this model. Now, reasonable is the keyword here. So if we go back and look at our graph, oh, so domain, it looks like the parabola starts here and ends here. So let's take a look at our table again. And let's see where these values, see, if we were to pick 6, 6 doesn't make any sense because that gives us something negative. So we could maybe say 7-ish until, let's see, toggle down until this gets negative, until about 15. So really these same answers. Now there are a couple of numbers before that. So we might want to say, so I guess 6 to 16 might make sense too, even though they do give out some negatives. But, you know, that's fine. So reasonable domain, I'm going to put a parenthesis sub, oops, I said 6, 6 comma 16 for my domain. And then my range, well, 0, definitely is the lowest I can go with that one because you can't have a negative number of applicants. And then my maximum up here would be my top Y value. And that happened at 81. So for my range, it would be 0 to 81 for my range. Okay. Fantastic. Practice these some more and we're going to do some more applications next. Thank you very much.