SOLUTION: Find a quadratic model for the set of values. x -2, 0, 4 f(x) 4, 6, -38

Question 1120312: Find a quadratic model for the set of values.
x -2, 0, 4
f(x) 4, 6, -38
Found 2 solutions by ankor@dixie-net.com, solver91311:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find a quadratic model for the set of values.
x -2, 0, 4
f(x) 4, 6, -38
:
x | y
-------
-2 | 4
0 | 6
4 |-38
:
Use the form ax^2 + bx + c = y
When x=0, y=6, therefore c = 6
:
when x=-2; y=4
(-2^2)a + - 2b + 6 = 4
4a - 2b = 4 - 6
4a - 2b = -2
:
when x=4; y=-38
16a + 4b + 6 = -38
16a + 4b = -38 - 6
16a + 4b = -44
simplify, divide by 2
8a + 2b = -22
:
4a - 2b = -2
8a + 2b = -22
---------------addition eliminates b, find a
12a + 0 = -24
a = -24/12
a = -2
:
Find b
4(-2) - 2b = -2
-8 - 2b = -2
-2b = -2 + 8
-2b = 6
b = 6/-2
b = -3
:
The equation f(x) = -2x^2 - 3x + 6, is the quadratic model

green line = -38

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

A general quadratic function is given by:

If then

If then

If then

Using the fact that we can create a 2X2 linear system:

Solve the 2X2 system to get the remaining two coefficients and then write your function.

John

My calculator said it, I believe it, that settles it

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