It is known that the two points of the equation 2x2-3x-5 = 0 are 5 / 2, - 1, then the image of the quadratic function y = 2x2-3x-5 and the two points of intersection of the X axis

It is known that the two points of the equation 2x2-3x-5 = 0 are 5 / 2, - 1, then the image of the quadratic function y = 2x2-3x-5 and the two points of intersection of the X axis

Then the two intersections of the image of quadratic function y = 2x2-3x-5 and X axis are (- 1,0) (5 / 2,0)
The distance is | 5 / 2 - (- 1) | = 7 / 2
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If the image of quadratic function y = AX2 + BX + C passes through points (4, - 3) and the function has the maximum value - 1 when x = 3, then the analytic expression of the function is______ ．

Let the analytic formula of quadratic function be y = a (x-3) 2-1, substitute (4, - 3) into a × (4-3) 2-1 = - 3, and the solution is a = - 2, so the analytic formula of quadratic function is y = - 2 (x-3) 2-1 = - 2x2 + 12x-19. So the answer is y = - 2x2 + 12x-19

Given an intersection point P (a, b) of the image of the first-order function y = x + 2 and the inverse scale function y = K / x, and the distance from P to the origin is 10, find the values of a, B and the analytic expression of the inverse scale function
thank u very muchhhh!1

Series of equations
1 according to the distance to the origin, under the root sign (a ^ 2 + B ^ 2) = 10
Substituting the coordinates of point P into the linear function B = a + 2
3 bring the P-point coordinate into the inverse scale function B = K / A
Solving three equations
Take 2 into 1 and find a with two values, corresponding to B with two inverse proportion functions

1. According to the coordinates of three points on the image of quadratic function, the analytic expression of function is obtained
(1) (-1,3) (1,3) (2,6)
(2) (-1,-1) (0,-2) (1,1)
(3) (-1,0) (3,0) (1,-5)
(4) (1,2) (3,0) (-2,20)
2. The parabola y = ax & sup2; + BX + C passes through three points (- 1, - 22), (0, - 8), (2,8). Find its opening direction, symmetry axis and vertex coordinates
Don't think about it, just do it step by step
The equations are all listed and solved. It seems that we can use some general formulas, vertex formula and intersection formula
I don't want ideas

1. Let y = ax & sup2; + BX + C, take (- 1,3) (1,3) (2,6) into A-B + C = 3A + B + C = 34a + 2B + C = 6, the solution is a = 1, B = 0, C = 2Y = x & sup2; + 2, let y = ax & sup2; + bx-2, take (- 1, - 1) (1,1) into a-b-2 = - 1A + B-2 = 1, the solution is a = 2, B = 1, let y = ax & sup2; + BX + C, take (- 1,0) (...)