This is a simultaneous equations problem. You will get three equations and three unknowns. Here's how:

For (0,3) => x = 0, y = 3

so y = ax^2 + bx + c becomes

3 = a(0)^2 + b(0) + c

=> c = 3

ah, so we found c from the first, yah! This means we will only need to solve 2 equations. This will turn out easier than i thought.

For (1,2) => x = 1, y = 2

so y = ax^2 + bx + c becomes

2 = a(1)^2 + b(1) + c .........remember, c=3

=> 2 = a + b + 3

=> a + b = -1 ...........(1)

For (2,3) => x = 2, y = 3

so y = ax^2 + bx + c becomes

3 = a(2)^2 + b(2) + c .........remember, c = 3

=> 3 = 4a + 2b + 3

=> 4a + 2b = 0 ..............(2)

So we found c, to find a and b we need to solve the system:

a + b = -1 ................(1)

4a + 2b = 0 ..............(2)

=> 2a + b = 0 ...............(2)/2

......a + b = -1 ..............(1)

=> a = 1

but a + b = -1

=> 1 + b = -1

=> b = -2

so our parabola is:

y = x^2 - 2x + 3