As part of a lab for physics, we need to solve a series of 3 quadratic equations for a, b, and c. The equations are just $\displaystyle ax^2+bx+c=y for x1, x2, and x3$. The three x values I'm using are 31.818, 62.727, and 93.864. The three y values which correspond to the x values are 25.909, 41.364, and 46.591. I'm lost.

2. Originally Posted by davesface
As part of a lab for physics, we need to solve a series of 3 quadratic equations for a, b, and c. The equations are just $\displaystyle ax^2+bx+c=y for x1, x2, and x3$. The three x values I'm using are 31.818, 62.727, and 93.864. The three y values which correspond to the x values are 25.909, 41.364, and 46.591. I'm lost.
I'm not sure what you're trying to find either. But if you perform a quadratic regression analysis on the values you produced, the resulting quadratic equation is:

$\displaystyle -.0053532082x^2+1.006135244x-.6847028686=0$

Am I close to what you need?

3. Actually, yes. That is exactly what I need. I have to do it with a few other sets of points to check how much they vary as the points change, though, so how could you do that without a graphing calculator?

4. Originally Posted by davesface
Actually, yes. That is exactly what I need. I have to do it with a few other sets of points to check how much they vary as the points change, though, so how did you do that?
I used a TI-84+ calculator to plot the points and perform the quadratic regression analysis. I understand that this can be done using Excel, but I haven't used that tool.

Here are the steps for the TI-83/84 calculators: Quadratic Regression

5. I don't think we're technically allowed to use graphing calculators, but I'll just do it on m TI-84, then. I know how to do it with that, but was just wondering if it could be done without. I'll try to find how to do it with Excel and post it here. Thanks a ton.