The image is required to be a quadratic function, so a≠0 requires c=0 to pass through the origin, and b≠0 requires to be fixed in 1 or 3 quadrants. (Because to take three different elements, C must be equal to 0, so it can be judged from this condition that A and B cannot both be 0. C=0, so the problem changed from 3 out of 8 to 2 out of 7.

According to the parabola vertex coordinates (-b/2a, (4ac-b 2)/4a) and c=0, the vertex coordinates become (-b/2a, -b 2/4a). If the vertex is in 1 or three quadrants, the two coordinate symbols are the same, either positive or negative.

Located in the quadrant of 1, if it crosses the origin, a must be less than 0, and -b/2a must be greater than 0, so b must be greater than 0; It is known that -b 2/4a > 0, and there is p (3, 1) XP (4, 1) = 12.

Similarly, it is located in quadrant 3, which requires one>0, and b<0 also has 12 methods.

* * * There are 24 kinds.