Fitting a Parabola to Your Data
What is a Parabola?
A parabola is a U-shaped curve that Algebra tells us can be described by a quadratic function, which is a special function because mathematicians have studying them for centuries. Let's take a look at an example: f(x) = x2 - 2x - 8.
This function can be factored into (x + 2) and (x - 4). (If you multiply these two factors together, you get the original function x2 - 2x - 8 back again.) That means f(x) = x2 - 2x - 8 = (x + 2)(x - 4).
Knowing the factors of a quadratic function is important because they can help us sketch the parabola that it describes.
We know that a function crosses the X axis when its value is zero. We set f(x) = 0 by setting (x + 2)(x - 4) = 0. Now, if the product of these two quantities is zero, then either one of them or both of them is zero. If (x + 2) = 0 then (solving) x = -2, and if (x - 4) = 0 then x = 4. These are the two places where this curve crosses the X axis, at x = -2 and at x = 4.
If we want to sketch the curve, we'll also need to know where the curve crosses the Y axis, which can be found using the same logic: set x to zero and solve for f(x).
f(0) = (0)2 + 2(0) - 8 = -8, so the curve crosses the Y axis at -8. With three points known, we can sketch this curve:
Fitting a Parabola to Your Data:
Note that we began with the function and calculated three points we could use to sketch the parabola. When you want to fit a parabola to your data points, you must begin with the points: that's just the opposite problem!
We say that a parabola can be described by three points, so let's use the points we found above to confirm that we'll fit them to the same function that we had above. Use A(-2,0), B(4,0), C(0,-8).
The general form of a quadratic function is f(x) = ax2 + bx + c. If we have three points, each with an (x,y) coordinate, then we have three X and three Y values. That gives us a system of three equations in three variables:
y1 = a(x1)2 +bx1 + c
y2 = a(x2)2 +bx2 + c
y3 = a(x3)2 +bx3 + c
Note that the upper case A, B, and C are points, while lower case a, b, c used here are coefficients.
Substituting our three coordinate values gives us:
0 = a(-2)2 +b(-2) + c
0 = a(4)2 +b(4) + c
-8 = a(0)2 +b(0) + c
Now lets rearrange the equations and solve for a, b, and c, to get the equation of the fitted parabola.
4a - 2b + c = 0
16a + 4b + c = 0
c = -8
Substituting -8 for c reduces it all to a system of two equations in two variables:
4a - 2b - 8 = 0
16a + 4b - 8 = 0
Divide the second equation through by 4, then subtract it from the first:
4a - 2b - 8 = 0
4a + b - 2 = 0
-------------------
-3b - 6 = 0, or b = -2
Substitute -8 for c and -2 for b, to get a:
4a - 2(-2) - 8 = 0
4a = 4, or a = 1
Finally, substitute -8 for c and -2 for b, and 1 for a to get a quadratic equation of the parabola:
f(x) = ax2 + bx + c
f(x) = x2 - 2x - 8
As expected, this brings us back to the same function that we began with, above. And if you happen to have a graphing calculator, it will perform this process for you, giving you the values for a, b, and c in an instant! See how a TI-83 calculator, for example, can be used to solve this same problem.
Fitting a Parabola to Trajectory Data:
When you shoot an arrow, throw a ball, or fire a cannon, it goes up for a while, then it comes down again. The parabola that describes that kind of trajectory opens downward, not upward, as in the case of the parabolas presented above. A parabola that opens downward is almost the same as the ones we've been examining, except they have a negative leading coefficient. That is, the x2 term is negative, as in f(x) = -x2 + 2x + 8.
Let's begin by factoring out the negative one and then factor what's inside the brackets. We'll bring the negative one back inside the brackets as a final step:
f(x) = -x2 + 2x + 8
f(x) = -[x2 - 2x - 8]
f(x) = -[(x + 2)(x - 4)]
f(x) = (-x - 2)(-x + 4)
Now let's set the function equal to zero again and solve both factors to find out where the curve crosses the X-axis.
f(x) = (-x - 2)(-x + 4) = 0
When (-x - 2) = 0, x = -2, and when (-x + 4) = 0, x = 4. Also, f(0) = -(0)2 + 2(0) + 8 = 8, so the curve crosses the Y axis at 8. With three points known, we can sketch this curve:
To fit such a curve to three data points, use the same techniques shown above. Note that Digit couldn't measure three data points directly at the crime scene, so she had to translate from the plane of the points that she could measure to the points on the trajectory before she could begin the curve fitting process. That's a different kind of problem, and she used a CAD (computer aided design) software tool to help her. More on CAD programs can be found by search for "computer aided design" on the Internet.
Return to Digit's Home Page.