Writing a quadratic function to fit data, estimate solutions, and make predictions

0:00- [Instructor] We're told Amirta is a high jump finalist at the World Athletics Championships. She tracks the heights of her practice jumps to get an idea of her capability during the competition. So this is time the air in tenths of a second. This is height in feet.

0:15We have a little gap here at two-tenths of a second. Amrita observes that she can use a quadratic function f to model this data and predict her height at certain times. Write a function to model Amrita's data and use it to predict her height after two-tenths of a second.

0:32So why don't you pause this video and have a go at this before we do this together? All right, now let's work on this together. What I'm actually first going to do, which you don't have to do, but it helps you visualize what's going on, is actually plot this data to feel good with Amrita's intuition that we're talking about something that could be modeled with a quadratic function.

0:54So the x-axis here, which is time we go from, we'll really go from zero to five seconds, but let's see, we go 1, 2, 3, 4, 5. 1, 2, 3, 4, 5. And it looks like our height, it gets as high as 6 at least in the data here. So 1, 2, 3, 4, 5, and 6. And I'm just going to roughly plot these data points at one-tenth of a second we're at 1.02 feet, so that's about right over there.

1:28At 3 seconds, we are at 6 feet. So it's something like that. At 4 seconds, 4.69, which is right around here. At 5 seconds, 1.07, something like this. And so if we look at these points, we can see the path of her jump. And this makes sense that the path of her jump is going to look something like that.

1:59And that indeed looks like a quadratic function. This looks like a parabola right over here. And so what we need to do is estimate an equation or try to find an equation for this type of a parabola that seems to go through these points or at least close to these points.

2:14And then use that to figure out if I take that function, if I can define that function and then I can evaluate what f of 2 is. Then we have done what they asked, to predict her height after two-tenths of a second. So the first thing I'm gonna do is, okay, should I do this based on where this intersects the x-axis?

2:38Should I do this based on the vertex? And in this situation I think I have a pretty good sense of what the vertex is. The vertex is going to be or is at this point, (3, 6) right over here. Now you might say, how do you feel confident that that's act actually the vertex or at least close to the vertex?

2:57We're not gonna get things exact. We're trying to model, we're trying to approximate things. And it's not just because this is the peak point or if we were talking about another, a problem that faced the other way, it's not because it's just the minimum point in the data, it's also because it looks like there's some symmetry around that.

3:16What do I mean by that? Well, if we assume that this right over here is the vertex, if you go two-tenths of a second more than that, you have gone down by roughly minus 5. And if you were to go two-tenths of a second before that, you have also gone down by roughly minus 5.

3:36Now you could, you might be able to do that to kind of approximate what's going on here as well. But we wanna figure out a whole function for modeling it. So let's do that. So if we know the vertex, we can think about a quadratic in vertex form. So that would be f of x is equal to some constant a times x minus the x coordinate of the vertex.

3:58Let's call that h squared, plus the y coordinate of the vertex. We know the x coordinate, we know the y coordinate, or at least we think we do. So this is a times x minus 3 squared plus 6. So the next thing to do is how do we figure out what a is? Well, we could use one of the points we know and it tends to make it a little bit more robust if you use a point further from the vertex, they didn't write it down here.

4:27We could have maybe even thought about (0, 0) at time 0, Amrita is going to be at zero feet. But I'll just go with the data that they've given. Let's use this point right over here at 1, 1.02. So we could say that f of 1, which is equal to 1.02 is equal to a times, well f of 1 when our function is at one x is 1.

4:49So 1 minus 3 squared plus 6. And so we get 1.02 is equal to 1 minus 3 is negative 2, you square that, you get 4, so 4a plus 6. And let's see if we then subtract 6 from both sides. Subtract 6 from both sides. We get, see, 1 minus 6 would be negative 5, but then we're gonna have to have 0.02 higher than that.

5:21So this is negative 4.98 is equal to 4a. And then we just divide both sides of this by 4. We have 4.98 and it's negative divided by 4 is equal to negative 1.245. So we get negative 1.245 is equal to a. And so now we know what our function looks like. We get our model.

5:55It's f of x is equal to negative 1.245 times X minus 3 squared plus 6. And then we can now use this to predict her height after two-tenths of a second or estimate really. So f of 2 is equal to negative 1.245 times 2 minus 3 squared plus 6, 2 minus 3 is negative 1.

6:24You square that, you just get 1. So it's really going to be equal to 6 minus 1.245. Let's get the calculator back. So if we just really add this to 6, so plus 6 is equal to 4.755. So this is equal to 4.755. If we wanted to round to the nearest hundredth like everything else is here, we could round that to 4.76.

6:59I'm rounding to the nearest hundredth and it's good to just think about does this make sense with the shape of this parabola. 2, 4.76 is, it's almost exactly where I originally drew that dot. It also isn't exactly where we are at time equals 4. So we're thinking about a second before the vertex, a second after the vertex or what we think is the vertex.

7:19These numbers aren't exact, which you wouldn't expect from real data, but they are pretty close. So we feel that it also has that symmetry about it.