quadratic function to model height of fountain water stream

S

sillymoiksta

New member
Joined
Oct 5, 2008
Messages
14
Hi this is the problem I need help with:

The height of a fountain's water stream can be modeled by a quadratic function. Suppose the water from a jet reaches a maximum height of 8 feet at a distance 1 foot away from the jet. If the water lands 3 feet away from the jet, find a quadratic function that models the height h(d) of the water at any given distance d feet from the jet.

There is a picture that illustrates this but I honestly have no clue where to even start. I'd appreciate help on starting the problem and what formula I'm supposed to plug the numbers into. Thank you.
 
Re: quadratic function

I would probably start by taking the picture and drawing a set of axes (x-y or d-h in this case) over the top of it. Make it so that the maximum of the water stream's shape occurs at (1, 8) = (d, h(d)) and the stream hits the ground at a distance of three feet from the fountain (d=3, h(d)=0).

Now, what shape does that look like, and what kind of equation looks like that?

You also know a few points on that graph. In fact, you know the maximum value, one of the x-intercepts (and you can figure out the other one, since this type of graph is symmetrical), ...

I hope this helps. Keep going from here if you need us to say anything more.
 
Re: quadratic function

sillymoiksta said:
Hi this is the problem I need help with:

The height of a fountain's water stream can be modeled by a quadratic function. Suppose the water from a jet reaches a maximum height of 8 feet at a distance 1 foot away from the jet. If the water lands 3 feet away from the jet, find a quadratic function that models the height h(d) of the water at any given distance d feet from the jet.

There is a picture that illustrates this but I honestly have no clue where to even start. I'd appreciate help on starting the problem and

what formula I'm supposed to plug the numbers into. <<< That's what Robots do. You need to derive the formula to use - that is what a Human-probsolver" is required to do.


Thank you.
Click to expand...

Make absolutely sure that the problem has been posted correctly.

First - tell us please:

What is a general quadratic equation of a parabola?

From the description of the problem, the symmetry axis of the parabola is not parallel to the y-axis. If assume that the jet is at the origin - then the vertex is not at the center of the two roots. That complicates the problem. That is why I wanted make sure the problem has been posted correctly (before I chase it around). If possible, scan the attached picture and post it here.

By the way, what level of mathematics is this?
 
Re: quadratic function

Thanks both of you for the help. The problem is in fact posted correctly(I checked it 3 times), Subhotosh Khan, and I would post the picture but I don't know how to on this site. Maybe you could tell me how?

To Subhotosh Khan: This is a problem from my Algebra 2 class and it's due tomorrow but since it's the last problem in the assignment I didn't get to it before the bell rang so I couldn't ask the teacher. There aren't any examples like this in the book either so I thought I'd ask here.
The general quadratic equation of a parabola is "ax^2 + bx + c" I believe. But in this section we have been using another formula a lot "y= a(x-h)^2 + k" so I wasn't sure if I was supposed to apply that equation to that problem, and that's why I asked about the formula.

To chivox: I graphed (1,8) and (3,0) on a graph as you suggested. I wasn't quite sure how you came up with the points so do you think you could explain that to me? All the rest of the problems in this section have given me the problem in "y= a(x-h)^2 + k" format so I knew how to graph it. Now I don't think I have the "a" where I could figure out the pattern of the parabola.

I'm sorry for not understanding this quicker but thank you for being patient with me. :)
 
Re: quadratic function

Actually about the image thing I see a help topic about it so I think I'll be able to figure how to upload a pic.
 
Re: quadratic function

Thanks again but I actually figured it out now thanks to additional help from my brother. Thanks for getting me started though. :D
 
Re: quadratic function

Way to go! Post what you got sometime for us to take a look. I sure am glad I don't have to go into work tomorrow and figure out how to post this silly picture of a parabola I made. It's better, I think, if you figure it out with your brother than with us, since that makes you figure out more about parabolas in general. And the trick of drawing a coordinate axes on top of a picture is a trick as old as the hills.
 
Re: quadratic function

I can sum up what I explained to my sister quickly:

The equation y = a (x - h)^2 + k looks pretty intimidating, but it's easier if you break it down into smaller components. The problem uses h for height and d for distance, but for this explanation I am going to use the equation that sillymoiksta provided that uses h and k. These are arbitrary variable names, so I can do that ;)

First off, let's focus on the basic parabolic equation: y = x^2 . This is a parabola because it's a power of 2.

Now let's see what the "a" variable means. We'll set h and k to 0 so we can ignore them. When a = 3, the equation is y = 3 x^2. This produces a skinny, steep parabola with vertex (0, 0). When a = 1/3, the equation looks like y = (1/3) x^2, and it is wider. The vertex is still (0, 0). When a = -1, the graph of y = - (x^2) is upside-down, and the vertex is at (0, 0). Therefore "a" controls the steepness and direction of the parabola and doesn't affect the vertex.

Once again let's isolate a variable and see how it changes the graph. For k = 5, we get y = x^2 + 5. If you graph this, you'll see that the shape of the parabola is exactly the same as in y = x^2, only now the vertex is moved up 5 units. For k = -5, we get y = x^2 - 5, and the vertex moves down 5 units. "k" is the y-position of the vertex.

The same thing happens for h. If h = 5, then y = (x - 5)^2 produces a graph with the vertex at (5, 0). "h" is the x-position of the vertex.

You now know how to make any upward or downward-facing parabola there is if you can write it out in this y = a (x - h)^2 + k format.

In this problem we are given a maximum point (i.e. the vertex) of (1, 8). The water level is at 0, so another point is (3, 0). Since we know how to use the vertex to make a parabolic equation, we can take the vertex point (h = 1, k = 8) and the above info to come up with this equation: y = a (x-1)^2 + 8.

We still need to find "a". Luckily we have another point, otherwise we'd be stuck. You can enter (x = 3, y = 0) into the equation and solve for "a" (check: a = -2). The water in this fountain follows a parabola of y = -2(x - 1)^2 + 8.
 
Re: quadratic function

Haha thanks chivox. Haha wow yes my brother wrote right above me basically what he went over with me and we did what he posted. My brother is a goof, but he is very good at helping me with math. Anyways haha. Thanks for your help too chivox and nice meeting you! :)
 
You must log in or register to reply here.
Top