Quadratic Question

A general quadratic can be written as  y = ax² + bx  + c  where a, b and c are constants that can be found by plugging in the x and y values of the points specified:

4 = 4a + 2b + c          (1)

0 = 4a - 2b + c           (2)

0 = 16a + 4b + c       (3)

Subtract (2) from (1):     4 = 4b    so   b = 1

Subtract (2) from (3):     0 = 12a + 6b

 or   0 = 12a + 6

or  12a = -6                so  a = -1/2

 Put a and b back into (2):    0 = 4*(-1/2) -2*1 +c

or 0 = -4 + c   so  c = 4  

The quadratic is y = -x²/2 + x + 4     (you should check that this goes through the three specified points by substituting the x values in and checking if the right y values appear).

1) (2.4)

2) (-2.0)

3) (4,0)

y = ax² + bx + c

1)  4 = 4a + 2b + c

2)  0 = 4a - 2b + c

3) 0 = 16a + 4b + c

1-2) 4 = 4b

        b = 1

2-3)  0 = -12a - 6b

        0 = -12a - 6

        a = - 1/2

2)     0 = -2 - 2 + c

        c =  4

y = - 0.5x² + x + c

: - )   !