Find a quadratic function to model the values in the table?

y=ax^2 +bx +c
2=a(-1)^2 +b(-1) +c
2=a-b+c

-2=a(0)^2 +b(0) +c
-2=c so the equation above becomes
2=a-b-2
a-b=4

10=a(3)^2 +b(3) -2
10=9a+3b-2
9a+3b=12
3a+b=4

4a=8
a=2
a-b=4
2-b=4,b=-2

y=2x^2 -2x -2
y=2(6)^2 -2(6) -2=58

y=ax^2+bx+c
10=9a+3b+c
-2=c
2=a-b+c
10=9a+3b-2
6=3a-3b-6
16=12a-8
24=12a
a=2
2=2+b-2
b=2
y=2x^2+2x-2
 if x=6 then y=2*36+2*6-2=72+12-2=82

y = ax^2+bx+c
Plug in three points,
-2 = c, for (0,-2) first
2 = a-b-2 => a-b = 4......(1)
10 = 9a+3b-2 => 3a+b = 4......(2)
(1)+(2), 4a = 8
a = 2, b = a-4 = -2
y = 2x^2-2x-2
At x = 6, y = 2*36-12-2 = 58

Hi,

The easiest way is to enter these 3 points into STAT on my TI83 calculator and them press QuadReg under STAT to get the equation.

y = 2x² - 2x - 2 <==ANSWER

You can also solve by making a system of 3 equations in 3 unknowns, filling each point in for x and y in the quadratic equation ax² + bx + c = y.

1a - 1b + c = 2
0a + 0b + c = -2, so c = -2
9a + 3b + c = 10

When you substitute -2 for c, the first and third equations become:

a - b = 4
9a + 3b = 12

Now multiply the first equation by 3 and add to the third equation. Then solve.

3(a - b = 4)
9a + 3b = 12


3a - 3b = 12
9a + 3b = 12
-------------------
12a = 24
 a = 2

If a = 2 and a - b = 4, then b = -2.
Since c = -2, the equation is y = 2x² - 2x - 2.


This can also be solved on the TI83 as a 3 x 4 matrix [A], using the rref([A]) command to get the data in reduced row echelon form.

1a - 1b + c = 2
0a + 0b + c = -2
9a + 3b + c = 10

[1 .-1 . 1 . . 2]
[0 . 0 . 1. . -2] = [A]
[9 . 3 . 1 . 10]


[1 . 0 . 0 . 2]
[0 . 1 . 0. -2] = [A]
[0 . 0 . 1 .-2]

Solving this system gave a = 2, b = -2, and c = -2 for the same equation as above.

I hope that helps!! :-)

You need to find three numbers, a, b, and c, that make the equation:
y = ax^2 + bx + c for the pairs of values that you listed in your table.

Substituting the x's and y's, you get three equations:
a(-1^2) + b(-1) + c = 2
a(0^2) + b(0) + c = -2
a(3^2) + b(3) + c = 10, which simplify to

a - b + c = 2
c = -2
9a + 3b + c 10

Since the middle equation tells us what 'c' is, substitute that into the 1st and 3rd equations:
a - b + -2 = 2, or
a = b + 4
and
9a + 3b + -2 = 10, or
9a + 3b = 12,
Use the first equation to subsititute 'b+4' for a in the last:
9(b+4) + 3b = 12
9b + 36 + 3b = 12
12b = -24
b = -2
Sub back into
a = (-2) + 4 = 2

This means your function is:
y = 2x^2 - 2x - 2
(You can check by putting in each x,y pair from your table.

Finally, just put in 6,
y = 2(6^2) - 2(6) - 2 = 58