College Algebra: Max Height in Real World

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Okay, so I thought we'd take a look at an application using parabolas, sort of, in nature and in the real world. So you can imagine, if you ever see one of these bridges that actually sort of build on underneath, sometimes the curve of that bridge actually, for stress reasons is a shape that's sort of parabolic--it looks like a parabola. In fact, here's an example of such a bridge. You can see here's the bridge going over there, and this curve here is essentially a parabola. Suppose we're given the following information. Let's take a look at this.
So suppose you're told that, in fact, the height of this arch of the bridge is given by the function h(x) -x^2 + 27. What's x? x just represents distance from the very center of the bridge out, this way and this way. And of course, we understand that this would be negative distance then, if you go to the left, and if you go to the right.
So if you look at the graph of that parabola you can see it's a sad face parabola, so that's good. In fact, it should be this parabola right here. Well, if you put the axes there, it looks something like this, so this is the exact same picture, but now just with the axes in place. And so x just represents the distance away from the very center here. And we can ask some interesting questions, or maybe some not so interesting questions depending upon your individual point of view and tastes. However, what we can ask are the following. For example, what is the maximum height of this arch? So how high is that arch? Let's take that question first and see if we can think about it. Well, that height, that maximum value, of course, is going to appear and going to happen at the vertex. So all I have to do is find the vertex. Well, what's the vertex? Watch this, folks. I'm going to do this for you without writing anything down.
How am I going to find the vertex? Think about it. It's . So what's b? Well, b is the coefficient on the x term. Now, look down here. Notice there is no x term here. So that means the coefficient must have been zero. So I have zero divided by 2a, which is some number, but zero divided by anything is zero. So, in fact, the x value where the vertex is located is going to be x = 0. And that makes sense, because you can see the arches are symmetric right around the y-axis.
So if I let x be zero, I see 27. So the height at the maximum value, the maximum height, must be 27 feet. And if you look at this graph--I don't know if you can see it or not--but in fact, it's really drawn to scale, this is 20, feet, this is 30 feet, this is 25 feet, and you can see this is a little bit above 25, so 27 is actually a reasonable thing, in fact, the correct thing. So to be able to answer that question without doing any writing at all, that's pretty cool. All I do is find the vertex and this vertex was easy because since there was no x term, we know that the axis symmetry is the y-axis, let x be zero, there's the maximum height.
Okay, now, let's take a look at another question we can ask. For example, we could say, "What is the height of the arch 10 feet to the right of the center?" So 10 feet to the right of the center--I want to know how high is that. To here. So what's that height? It should, of course, be less than 27 that we just found. But how much less? You look at this picture, what does it look like? It looks like it should be just a little bit more than 20. But I want to find out exactly, so how would I do that? Well, what I want to do is just find the height when x = 10. So all I have to do is evaluate this at 10. So that's not a big deal at all. All I'm going to do is take a look at h(10), because I want to find out what the height is at a particular x value. So, what is that? That's times, well, 10^2, so that's 100, and then plus 27. So now my job is to carefully work that out. You can do some canceling here and you can get rid of some 2's and so forth and I think this will shrink down to something like +, and then if you get a common denominator here, I think you'd have to add on another . I just got a common denominator. That's just the same thing as saying 27. And if you combine these things, I think you see , and what does that equal numerically? It's about 22.312 feet. So just a little bit over 22 feet. That's what we were sort of saying. This looks like it was a little bit more than 20 feet, so in fact, it's actually 22.312 feet. Great. So to find the height at any particular value I just plug into the function.
Now, to show you sort of a different type of question, let's take a look at this. Now we're asked how far from the center is the arch 8 feet tall? That's a different kind of question. What I want to know now is how far do I have to go off the center so that this distance is 8 feet. Well, if you look at it visually, it seems like, well, this is 10 feet; 8 feet will probably be around here. So what would that value be? It looks like it's going to be right around 20 feet. Now, how am I going to find that? That's a different type of problem. Here, I know the height and I want to find sort of this value here. So how do I do that? Well, I set the height equal to 8 feet and find out what value of x makes that height come true.
So what I have to do now is set the height to 8 feet. Let's try that. So if I say h(x) = 8, then if I set the equation equal to 8, what do I see? I see x^2 +27, and that equals 8. Now, I'm going to solve that equation. So I bring the 27 over to this side, and if I subtract it I'd see a -19. What I see is x^2 = -19. And if now multiply everything through by , and notice this will cancel away, and I'll have , and so I'll see that x^2 = . It becomes positive because a negative times a negative is positive, and I have 64 times 19, which is 1216, and then I just have a 3 downstairs. And now I take square roots of both sides, but remember, plus or minus square root, if you're going to take a square root, and so if we take the square root of both sides, I'll see x = the square root of that thing, . And where is the plus or minus coming from? Does that make sense? It sure does, because notice that if I say to you, "Eight feet above the bridge," notice there's two places where that happens. Some value and negative some value. So, in fact, we're finding both of them, and let's see what that number is. That answer equals , and if you compute this, you'll see 20.13-something feet. So it's around 20-something-odd feet. Does that look good? It sure does, because look, if this is 8, you go over here, it does look like it should be right around 20. This picture looks pretty accurate, and the answer is either +20-something-odd feet, or minus the same thing.
Anyway, there's a great opportunity to see in action the idea of evaluating a parabola at a certain point, and seeing how high it is, versus finding where the parabola takes on a particular height. If you want to find out where the parabola takes on a height, you set the equation equal to that height, and you solve for x and find out where it happens, or if you want to find out how high something is at a point, you then plug that point in for x and solve. Two different types of uses of the quadratic and looking at a parabola.
Relations and Functions
Quadratic Functions - Basics
Maximum Height in the Real World Page [2 of 2]