# Pre-Calculus

Find a quadratic function f such that f(1)=3, f(2)=5/2, f(-1)=1

I came out with 3 a+b+c=3, 4a+2b+c=5/2, a-b+c=1. And I know that I should do synthetic division but I am stuck.

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1. y = a x^2 + b x + c

if x = 1
3 = a + b + c

if x = 2
5/2 = 4 a + 2 b + c

if x = -1
1 = a - b + c

we have 3 linear equations with 3 unknowns. There are many ways to solve but lets try to eliminate c first
1 a +b + c = 3
4 a+2b + c = 5/2
------------------ subtract
- 3 a - 1 b = 1/2
and
1 a +b + c = 3
1 a -b + c = 1
---------------- subtract
2 b = 2
so
b = 1
now go back
1 a +b + c = 3
4 a+2b + c = 5/2
or
1 a + 1 + c = 3
4 a + 2 + c = 5/2
-------------------
-3 a -1 = 1/2
-3 a = 3/2
a = -1/2
so back again with b = 1 and a = -1/2
1 a + 1 + c = 3
-1/2 + 2/2 + c = 3
c + 1/2 = 6/2
c = 5/2

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posted by Damon

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