SOLUTION: How do you get the quadratic equation if you are given a table of values?

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Question 567835: How do you get the quadratic equation if you are given a table of values?
Answer by issacodegard(60) About Me  (Show Source):
You can put this solution on YOUR website!
You must be given at least 3 points, call them (x1,y1),(x2,y2),(x3,y3). Then we know the equation has the form f(x)=ax^2+bx+c. We get a system of 3 equations in 3 unknowns. Solving for a,b,c in,
f%28x1%29=a%28x1%29%5E2%2Bb%28x1%29%2Bc=y1
f%28x2%29=a%28x2%29%5E2%2Bb%28x2%29%2Bc=y2
f%28x3%29=a%28x3%29%5E2%2Bb%28x3%29%2Bc=y3
gives you the equation for f(x).
I'll give an example,
If we know that three points are (-1,-2),(1,4),(2,13). Then we set up three equations about a,b,c that we know:
f%28-1%29=a%28-1%29%5E2%2Bb%28-1%29%2Bc=-2
f%281%29=a%281%29%5E2%2Bb%281%29%2Bc=4
f%282%29=a%282%29%5E2%2Bb%282%29%2Bc=13
So we have,
a-b%2Bc=-2
a%2Bb%2Bc=4
4a%2B2b%2Bc=13
We need to solve for a,b,c. So, we subtract eqn1 from eqn2 to get,
a-b%2Bc=-2
2b=6
4a%2B2b%2Bc=13
Divide eqn2 by 2 to get b=3. Then, we have
a-3%2Bc=-2
4a%2B2%283%29%2Bc=13
So,
a%2Bc=1
4a%2Bc=7
Subtract the first eqn from the second to get,
a%2Bc=1
3a=6
Divide the second eqn by 3 to get a=2. Then we have
2%2Bc=1
So,
c=-1
Therefore we find that
f%28x%29=2x%5E2%2B3x-1