We graph those 4 points: (0,3), (1,4), (2,5), (3,0)



Its graph cannot be a second degree quadratic function (parabola) 
because it has three points in a straight line.

Since it goes through the point (3,0) The equation must
have (x - 3) as a factor.

So perhaps its equation is  F(x) = the product of (x-3) and a
quadratic, like this:

   F(x) = y = (x - 3)(Ax² + Bx + C)

We know it has y-intercept (0,3) so we can substitute that and get

          3 = (0 - 3)(A·0² _ B·0 + C)
          3 = -3C
         -1 = C

So wesubstitute that and we have:

   F(x) = y = (x - 3)(Ax² + Bx - 1) 

We substitute (1,4)

          4 = (1 - 3)(A·1² + B·1 - 1)
          4 = -2(A + B - 1)
          4 = -2A - 2B + 2
    2A + 2B = -2

Divide through by 2

      A + B = -1

We substitute (2,5)

          5 = (2 - 3)(A·2² + B·2 - 1)
          5 = -1(4A + 2B - 1)
          5 = -4A - 2B + 1
    4A + 2B = -4

Divide through by 2

     2A + B = -2

So we solve the system of equations:

      A + B = -1
     2A + B = -2

and get A = -1 and B = 0

So the function is

   F(x) = (x - 3)(-1x² + 0x - 1)
   F(x) = (x - 3)(-x² - 1)
   F(x) = (x - 3)(-1)(x² + 1)
   F(x) = -(x-3)(x²+1)

and its graph is:



Edwin