The analytic expressions (1) of quadratic functions are obtained respectively. The image of quadratic functions passes through points (- 1,0), (1,2), (0,3) (1) The image of the quadratic function passes through the points (- 1,0), (1,2), (0,3); (2) the vertex coordinates of the image of the quadratic function are (- 3,6) and pass through (- 2,10); (3) the intersection coordinates of the image of the quadratic function and the X axis are (- 1,0) and (3,0), and the intersection coordinates of the image of the quadratic function and the Y axis are (0,9)

The analytic expressions (1) of quadratic functions are obtained respectively. The image of quadratic functions passes through points (- 1,0), (1,2), (0,3) (1) The image of the quadratic function passes through the points (- 1,0), (1,2), (0,3); (2) the vertex coordinates of the image of the quadratic function are (- 3,6) and pass through (- 2,10); (3) the intersection coordinates of the image of the quadratic function and the X axis are (- 1,0) and (3,0), and the intersection coordinates of the image of the quadratic function and the Y axis are (0,9)

Solution (1): let the analytic expression of quadratic function be y = ax & # 178; + BX + C (a ≠ 0)
By substituting x = - 1, y = 0; X = 1, y = 2; X = 0, y = 3 into y = ax & # 178; + BX + C, the equations of a, B, C are obtained
a-b+c=0
a+b+c=2
c=3
Solving the equations, a = - 2, B = 1, C = 3
Therefore, the analytic expression of quadratic function is y = - 2x & # 178; + X + 3
Solution (2): let the analytic formula of quadratic function be vertex formula y = a (X-H) &# + K, and the vertex coordinates of function image be (h, K)
Substituting H = - 3, k = 6 into y = a (X-H) &# 178; + K, y = a (x + 3) &# 178; + 6
Then, substituting x = - 2, y = 10 into y = a (x + 3) &# + 6, we get:
a(-2+3)²+6=10
a+6=10
a=4
Therefore, the analytic formula of quadratic function is y = 4 (x + 3) &# 178; + 6
The general formula is y = 4x & # 178; + 24x + 42
Solution (3): let the analytic expression of quadratic function be the intersection expression y = a (x-x1) (x-x2), and the intersection coordinates of function image and X axis be (x1,0), (x2,0). Substituting X1 = - 1, X2 = 3 into y = a (x-x1) (x-x2), y = a (x + 1) (x-3)
Then we substitute x = 0, y = 9 into y = a (x + 1) (x-3) to get the following result:
a(0+1)(0-3)=9
-3a=9
a=-3
Therefore, the analytic expression of quadratic function is y = - 3 (x + 1) (x-3)
The general formula is y = - 3x & # 178; + 6x + 9