SOLUTION: what is the quadraric funtion that fits each set of data points. (1,-2) (-1,8) (2,-1) The equation I came up with is: (f(x)=-2/3x^2-5x-3) Thank you!

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: what is the quadraric funtion that fits each set of data points. (1,-2) (-1,8) (2,-1) The equation I came up with is: (f(x)=-2/3x^2-5x-3) Thank you!      Log On


   



Question 438629: what is the quadraric funtion that fits each set of data points.
(1,-2) (-1,8) (2,-1)
The equation I came up with is: (f(x)=-2/3x^2-5x-3)
Thank you!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
what is the quadraric funtion that fits each set of data points.
(1,-2) (-1,8) (2,-1)
The equation I came up with is: (f(x)=-2/3x^2-5x-3)
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Quaratic Form:
ax^2 + bx + c = y
Use each x/y pair to generate an equation:
------
(1,-2) gives: a + b + c = -2
(-1,8) gives: a - b + c = 8
(2,-1) gives: 4a +2b + c = -1
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Solve that system of three equations for a,b,c:
---
I get: a = 2 ; b = -5 ; c = 1
===================================
Equation:
y = 2x^2-5x+1
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Cheers,
Stan H.
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