It is known that the intersection of image and x-axis of quadratic function has two points a (- 2,0) and B (3,0), and the function has a maximum value of 2 1) Find the analytic expression of quadratic function; 2) Let the image vertex of the quadratic function be p, and find the area of the triangle ABP It's urgent. I just don't know how to determine the a of intersection. As long as I know the first question, thank you

It is known that the intersection of image and x-axis of quadratic function has two points a (- 2,0) and B (3,0), and the function has a maximum value of 2 1) Find the analytic expression of quadratic function; 2) Let the image vertex of the quadratic function be p, and find the area of the triangle ABP It's urgent. I just don't know how to determine the a of intersection. As long as I know the first question, thank you

1) Let y = ax ^ 2 + BX + C, and substitute a (- 2,0) and B (3,0) points
4a-2b+c=0;9a+3b+c=0
We can get a = - B
The function has a maximum value of 2, that is, the ordinate of the vertex is 2
And the abscissa of the vertex is - 2A / b = 2, that is, the vertex is (2,2) substituted
We get 4A + 4B + C = 2
4a-2b+c=0
9a+3b+c=0
4A + 4B + C = 2, a = - 1 / 3, B = 1 / 3, C = 2
That is y = - x ^ 2 / 3 + X / 3 + 2
2) The vertex P (2,2), AB is 5
So the area of triangle ABP is 2 * 5 / 2 = 5