Writing a quadratic function to fit data and estimate solutions | Algebra 1 (TX TEKS) | Khan Academy

We're told Amtha 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. And so, this is timing the air intense of a second. This is height in feet. We have a little gap here at 2.10 of a second.

Amtha 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 Amtha's data and use it to predict her height after 2.10 of a second. So why don't you pause this video, 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 Amtha's intuition that we're talking about something that could be modeled with a quadratic function.

So the x-axis here, which is time, we go from - we'll really go from 0 to 5 seconds. But let's see... go one, two, three, four, five. One, two, three, four, five. And looks like our height gets as high as six, at least in the data here. So one, two, three, four, five, and six.

I'm just going to roughly plot these data points. At 1/10th of a second, we're at 1.02 ft, so that's about right over there. At 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.

So if we look at these points, we can see the path of her jump. This makes sense that the path of her jump is going to look something like that, and that looks like a quadratic function. This looks like a parabola right over here.

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. 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 2.10 of a second.

So the first thing I'm going to do is, okay, should I do this based on where this intersects the x-axis, or should I do this based on the vertex? 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 actually the vertex, or at least close to the vertex?" We're not going to get things exact; we are trying to model, we are trying to approximate things.

It's not just because this is the peak point, or if we were talking about another parabola 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. What do I mean by that?

Well, if we assume that this right over here is the vertex, if you go 2.10 of a second more than that, you have gone down by roughly -5. And if you were to go 2.10 of a second before that, you have also gone down by roughly -5. Now you could, you might be able to do that to kind of approximate what's going on here as well, but we want to figure out a whole function for modeling it. So let's do that.

If we know the vertex, we can think about a quadratic in vertex form. So that would be f(x) = a * (x - h)^2 + k, where h is the x coordinate of the vertex. Let's call that h, 3; the y-coordinate of the vertex we know; the x-coordinate we know; we think we do.

So this is a * (x - 3)^2 + 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.

We could have maybe even thought about (0, 0) at time zero. Amtha is going to be at 0 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(1), which is equal to 1.02, is equal to a * (1 - 3)^2 + 6. And so we get 1.02 = a * (1 - 3)^2 + 6.

Let's see if we then subtract six from both sides. Subtract six from both sides, we get 1 - 6 would be -5, but then we're going to have to have 0.02 higher than that. So this is negative 4.98 = 4a.

Then we just divide both sides of this by 4. We have -4.98 / 4 = -1.24. So we get 1.24 = a.

Now we know what our function looks like. We get our model: f(x) = 1.24 * (x - 3)^2 + 6. Then we can use this to predict her height after 2.10 of a second or estimate really.

So F(2) = 1.24 * (2 - 3)^2 + 6. (2 - 3) is -1; you square that, you just get 1. So it's really going to be equal to 6 - 1.24.

Let's get the calculator back. If we just really add this to six, so + 6 = 4.76. So this is equal to 4.76.

If we wanted to round to the nearest hundredth like everything else here, we could round that to 4.76, rounding to the nearest hundredth.

It's good to just think about does this make sense with the shape of this parabola? If 2 = 4.76, 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.

These 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.