Question 1196721: Find a quadratic model for each set of values. (-1,10), (2,4), (3,-6)
Found 3 solutions by MathLover1, greenestamps, MathTherapy:
Answer by MathLover1(20849)   (Show Source):
You can put this solution on YOUR website!
Find a quadratic model for each set of values. ( , ), ( , ), ( , )
use point (,)
 ..........eq.1
use point (,)
 ..........eq.2
use point (,)
 ..........eq.3
solve system
..........eq.1
..........eq.2
..........eq.3
--------------------------------------------------
add eq.1 and eq.3
 .........divide by
 ...........solve for 
 ............1)
add eq.2 and eq.3
 ........solve for
 ...........2)
from 1) and 2) we have
 ............solve for 
 ....................3)
substitute in 1)
 ...............1a)
go to
..........eq.1, substitute from 3) and from 1a)
 .....solve for 
go to
....................3), substitute
go to
...............1a) , substitute
 ,  ,  and your quadratic is:
Answer by greenestamps(13195)  (Show Source):
You can put this solution on YOUR website!
The response from the other tutor shows a typical formal algebraic solution.
Since your post gives no indication of your level of mathematical studies, I will demonstrate a very different method for solving the problem.
Since a quadratic model is specified, consider a sequence of consecutive terms of a quadratic sequence. We are given the values for x=-1, x=2, and x=3; use variables to represent the terms of the sequence for x=0 and x=1:
10, a, b, 4, -6
The method of finite differences (do an internet search if you are interested in learning more about it) tells us that, because the model is quadratic (a second degree polynomial), the second differences of the sequence are constant. Find the second differences of the sequence and use the fact that they are equal to find a and b.
10 a b 4 -6 <-- the consecutive terms of the sequence
a-10 b-a 4-b -10 <-- the first differences
b-2a+10 a-2b+4 b-14 <-- the second differences
The quadratic sequence is
10, 12, 10, 4, -6
The set of coordinates corresponding to the sequence are
(-1,10), (0,12), (1,10), (2,4), and (3,-6)
The identical y values for x=-1 and x=1 tell us the vertex of the graph of the function is at (0,12), so the function is of the form
Use any of the other known points to determine (by whatever method you want) that a = -2.
Then the quadratic model for the given set of 3 points is
ANSWER:
Answer by MathTherapy(10549)  (Show Source):
You can put this solution on YOUR website! Find a quadratic model for each set of values. (-1,10), (2,4), (3,-6)
The solution of the system of equations from that woman who responded, as usual, is RIDICULOUSLY time-comsuming,
INEFFICIENT, and totally UNNECESSARY. I guess she's OBSESSED with complicated fractional expressions.
(- 1, 10) = (x, y)
Equation of a parabola: 
 ------ Substituting (- 1, 10) for (x, y)
10 = a - b + c ------ eq (i)
(2, 4) = (x, y)
Equation of a parabola:
 ------ Substituting (2, 4) for (x, y)
4 = 4a + 2b + c ---- eq (ii)
(3, - 6) = (x, y)
Equation of a parabola:
 ------ Substituting (3, - 6) for (x, y)
- 6 = 9a + 3b + c ----- eq (iii)
10 = a - b + c ----- eq (i)
4 = 4a + 2b + c ----- eq (ii)
- 6 = 9a + 3b + c ----- eq (iii)
6 = - 3a - 3b____3(2) = 3(- a - b)____2 = - a - b --- Subtracting eq (ii) from eq (i) ---- eq (iv)
- 10 = 5a + b ---- Subtracting eq (ii) from eq (iii) -- eq (v)
- 8 = 4a -------- Adding eqs (iv) & (v)
2 = - (- 2) - b ----- Substituting - 2 for a in eq (iv)
2 = 2 - b
10 = - 2 - 0 + c ---- Substituting - 2 for a, and 0 for b in eq (i)
Substituting - 2 for a, 0 for b, and 12 for c, now becomes: 
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