Equation for an Arch

Date: 09/09/97 at 01:30:17
From: Rob Cashell
Subject: Find quadratic equation from graph

Hi,

I am trying to draw an arch that will go in the ceiling of a building.
The arch will be at a maximum height of 28 inches and will have 
reached the " x-axis " at 21 feet (252 inches).

I have played with the quadratic equation some and have been unable to 
come up with the equation of this arch. Can you help? Also, I am 
curious about the math that is used to do this.  

vertex(0,28) x-intercepts (252,0) and (-252,0) ; the parabola will 
open downward (negative 1st term).

Thank you very much, 

Rob Cashell, summertime carpenter and college student.

Date: 09/20/97 at 14:44:02
From: Doctor Chita
Subject: Re: Find quadratic equation from graph

Hi Rob:

You have come up with a classic problem in analytic geometry.  
That is, you can use algebra to solve a problem in geometry. 

One way to do this is to start with the standard form of a quadratic 
function (equation): ax^2 + bx + c = y. A point that lies on the graph 
of a function satisfies the equation of the function. You have the 
coordinates of three points on the parabola: the vertex, (0, 28), and 
the x-intercepts, (252, 0) and (-252, 0). Therefore, you need to find 
a, b, and c.

Substitute the coordinates for x and y in the general equation and 
write three equations:

        a(0)^2 + b(0) + c = 28
    a(252)^2 + b(252) + c = 0
  a(-252)^2 + b(-252) + c = 0

Solve these equations for a, b, and c, and you will have your 
equation. (The  problem is not too difficult, because you have the 
value of c immediately in the first equation. Finding a and b becomes 
relatively simple.)

Hope this helps.

-Doctor Chita,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/