# Find a quadratic polynomial f(x) whose graph passes by (1,2), (3,-4) and (-1,4).

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Through any three noncollinear points there is exactly one quadratic function; it can be written in the form `y=ax^2+bx+c` . Substituting the known x,y pairs we get three equations in a,b, and c:

(1,2): a+b+c=2 I

(3,-4):9a+3b+c=-4 II

(-1,4):a-b+c=4 III

We can solve using linear combinations, substitution, or matrices.

I. Linear combinations:

Add row I to row II: 2a+2c=6 or a+c=3 IV

Add row II to 3rowIII: 12a+4c=8 or 3a+c=2 V

Now take row V minus row IV to get 2a=-1 or `a=-1/2`

If `a=-1/2` then `c=7/2`

Then using I we get `-1/2+b+7/2=2==>b=-1`

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Thus the equation we seek is `y=-1/2x^2-x+7/2`

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II Using matrices we need to row reduce :

`([1,1,1,2],[9,3,1,-4],[1,-1,1,4])` ==> `([1,0,0,-1/2],[0,1,0,-1],[0,0,1,7/2])`

again yielding the coefficients.