There are two points of intersection between parabola and x-axis, which are (- 2,0) (1) the parabola and X-axis have two intersection points (- 2,0) (1 / 2,0) and the intercept on Y-axis is 1, so the analytical formula of the parabola is obtained

Substituting (- 2.0) (1 / 2,0) (0,1) into y = AX2 + BX + C
4a-2b+c=0
1/4a+1/2b+c=0
c=1
The solution is a = - 1, B = - 3 / 2, C = 1
Or substitute (- 2.0) (1 / 2,0) (0, - 1) into y = AX2 + BX + C
4a-2b+c=0
1/4a+1/2b+c=0
c=-1
The solution is a = 1, B = 3 / 2, C = - 1
The analytic formula is y = - x2-3 / 2x + 1 or y = x2 + 3 / 2x-1

Coordinates of the intersection point of parabola y = x & # 178; + X-6 and X axis

Let y = 0, then there is
X ^ 2 + X-6 = 0, i.e. (x + 3) (X-2) = 0
So X1 = 2. X2 = - 3
The intersection coordinates are (- 3,0), (2,0)

Given that a and B are abscissa of the intersection of the parabola y = (x-C) (x-c-6) - 2 and X axis, a < B, then ()
A. A < C < BB. C < a < BC. A < B < CD. C < a < B or a < B < C

Let (x-C) (x-c-6) - 2 = 0, (x-C) 2-6 (x-C) = 2, the solution is: x = C + 3 ± 11, ∵ a, B are the abscissa of the intersection of parabolic y = (x-C) (x-c-6) and X axis, a < B, ∵ a = C + 3-11, B = C + 3 + 11, ∵ a-c = 3-11 < 0, B-C = 3 + 11 > 0, ∵ a < C < B. therefore, select a

It is known that there are two intersections (- 1,0), (3,0) between the parabola and the x-axis, and the ordinate of the intersection with the y-axis is - 6, then the analytic expression of the quadratic function is -

∵ there are two intersections (- 1,0) and (3,0) between parabola and x-axis
We can write y = a (x + 1) (x-3)
Also: the ordinate of y-axis intersection is - 6
∴-6=a*(0+1)*(0-3)
∴a=2
∴y=2（x+1)(x-3) = 2x^2-4x-6

There are two points of intersection between parabola and x-axis, which are (- 2,0) (1) the parabola and X-axis have two intersection points (- 2,0) (1 / 2,0) and the intercept on Y-axis is 1, so the analytical formula of the parabola is obtained