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March 29, 2015

March 29, 2015

Posted by **Jay** on Tuesday, November 15, 2011 at 8:24am.

- Math -
**Reiny**, Tuesday, November 15, 2011 at 8:40am2 17

4 28

6 42

The points (2,17), (4,28), and (6,42) do not lie in a straight line, so the relations is not linear.

Let's assume they form a quadratic of the form

y = ax^2 + bx + c

subbing in the values, I get

4a + 2b + c = 17

16a + 4b + c = 28

36a + 6b + c = 42

start by subtracting pairs to eliminate the c, then solving for the remaining a and b

( I ended up with a=3/8, b=13/4, and c=9 )

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