Question

Find an equation in the form y=ax^2+bx+c for the parabola passing through the points: (-5,213) (2,38) (3,77)

Answer 1

This is a very interesting question. I am so far very stumped, but am trying different approaches.

How did they teach you to do these?

Answer 2

I figured out the equation, the old school way, I'm sure certain mentors would like my technique. I see you're offline, I won't post the answer, since it is not very mathematical.

Trying to figure out how solve it logically. Working back from the answer.

Answer 3

Plug the values of x and y into the equation y = a x^2+b x+c for each of the given points.
Solve the resulting three simultaneous equations in a, b and c for a, b and c.
Plug the numeric values for a, b and c into the equation y=ax^2+bx+c for the final result.

I got y = 8 x^2 -x +8

A solution and plot using the Mathematica computer program is attached.

Answer 4

That's the excellent logic I was hoping for. I modeled it in excel and kept guessing and got the equation. Mine was more luck than skill. Thanks!

A B C D E excel columns
a= b c
= C-D+E 8 1 8

x y X^2 x c
-10 818 800 -10 8
-9 665 648 -9 8
-8 528 512 -8 8
-7 407 392 -7 8
-6 302 288 -6 8
-5 213 200 -5 8
-4 140 128 -4 8
-3 83 72 -3 8
-2 42 32 -2 8
-1 17 8 -1 8
0 8 0 0 8
1 15 8 1 8
2 38 32 2 8
3 77 72 3 8 39
4 132 128 4 8
5 203 200 5 8
6 290 288 6 8

Answer 5

Perfect. Three equations three unknowns. Easy to solve on the TI-84 matrix function and a lot easier than guessing.

Answer 6

In the real world you would us a TI-84 or the Mathematica computer program for a solution. Then the only errors would be in typing in the problem numbers plus the enormous speed advantage.