Equation for an ArchDate: 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/ |
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