Question 1060415: I need your assistance to help find the quadratic function that models the data below please.
xI -3 -2 -1 0 1 2 3 4 5 6 7 8 9
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yI 54 26 -8 0 2 14 36 68 110 162 224 296 378
I need help with this please. When I look at this all i'm seeing is a chart I don't know how I would work this out
Found 2 solutions by rothauserc, MathTherapy:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! We know that we have an x^2 term since this is a quadratic
:
the general form is
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ax^2 + bx +c = 0, where a, b, c are constants
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we need just three points to find the function
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take the first three point (-3, 54), (-2, 26), (-1, 8), then
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f(-3) = 54
f(-2) = 26
f(-1) = 8
:
9a -3b +c = 54
4a -2b +c = 26
1a -1b +c = 8
:
we use Cramer's rule calculator to find a, b and c
:
a = 5, b = -3, c = 0
:
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f(x) = 5x^2 -3x
***************
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
I need your assistance to help find the quadratic function that models the data below please.
xI -3 -2 -1 0 1 2 3 4 5 6 7 8 9
-----------------------------------------------------------------------
yI 54 26 -8 0 2 14 36 68 110 162 224 296 378
I need help with this please. When I look at this all i'm seeing is a chart I don't know how I would work this out
The EASIEST way to do this is to take the 3 EASIEST points to get the quadratic function. These points are: (0, 0), (- 1, - 8), and (1, 2)
Quadratic function: 
(0, 0)
 --------- Substituting point (0, 0) for (x, y)
0 + 0 + c = 0____ ------ eq (i)
(- 1, - 8)
 ---- Substituting point (- 1, - 8) for (x, y), and 0 for c
a - b = - 8 ----- eq (ii)
(1, 2)
 --------- Substituting point (1, 2) for (x, y), and 0 for c
a + b = 2 ------- eq (iii)
2a = - 6 -------- Adding eqs (ii) & (iii)
- 3 + b = 2 ------- Substituting - 3 for a in eq (iii)
 ------- Substituting - 3 for a, 5 for b, and 0 for c
Quadratic function: 
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