SOLUTION: find the quadratic function which takes the value 41 at x=-2 and the value 20 at x=5 and is minimized at x=21 Y=Ax^2 - Bx + C

 y = Ax²+Bx+C
41 = A(-2)²+B(-2)+C
41 = A(4)-2B+C
41 = 4A-2B+C

first equation:   4A-2B+C = 41

 y = Ax²+Bx+C
20 = A(5)²+B(5)+C
20 = A(25)+5B+C
20 = 25A+5B+C

second equation:   25A-5B+C = 20

A parabola has its lowest point, that is, it is
minimized at its vertex.  Its vertex has as its 
x-coordinate the value  so we set 
that = 21

 = 21

Cross-multiplying:

      -B = 42A
-42A - B = 0
 42A + B = 0

third equation:    42A + B = 0

eq. 1    4A-2B+C = 41
eq. 2   25A+5B+C = 20
eq. 3   42A+ B   = 0

Multiply eq. 1 by -1 and add it to eq. 2 to eliminate
the C's

        25A+5B+C =  20
        -4A+2B-C = -41
      -----------------
        21A+7B   = -21

That can be easily divided through by 7

eq. 4    3A+ B   = -3

Now we have a system with only 2 equations and
 2 unknowns.


eq. 4    3A+ B = -3
eq. 3   42A+ B =  0

Multiply eq. 4 through by -1 and add to eq. 3 to ekiminate
the B's.

        -3A- B = 3
        42A+ B = 0
        ----------
        39A    = 3
             A = 
             A =  

Since that's a nasty fraction, it will be easier to start over
and eliminate A from eqs. 4 and 3

3 goes into 42 14 times so we just have to multiply eq. 4 by
-14 and add it to eq. 3:

       -42A-14B = 42
        42A+  B =  0
       --------------
           -13B = 42
              B = 
              B = 

Substitute that and A =  into eq. 1: 

eq. 1    4A-2B+C = 41
      4-2+C = 41

Clear of fractions by multiplying through by 13

        4 + 84 + 13C = 533
            88 + 13C = 533
                 13C = 445
                   C = 

Those are terrible fractions, but they are correct.  I
suspect you may have made a slight typo, such as typing
a sign or a digit wrong in one of your figures. 

But the correct function based on the numbers you gave, is:

y = x² - x + 

or if you like you can use functional notation f(x) for y:

f(x) = x² - x + 

Edwin