Given that the parabola y = ax square + BX + C satisfies the following conditions, find the analytic expression of the function (1) Image passing point (- 1. - 6) (1, - 2) (2.3) (2) Vertex coordinate (- 1. - 1) and passing point (0. - 3) (3) Intersection with X-axis (- 2.0) (4,0) and passing through point (1, - 9 / 2)

Given that the parabola y = ax square + BX + C satisfies the following conditions, find the analytic expression of the function (1) Image passing point (- 1. - 6) (1, - 2) (2.3) (2) Vertex coordinate (- 1. - 1) and passing point (0. - 3) (3) Intersection with X-axis (- 2.0) (4,0) and passing through point (1, - 9 / 2)

(1) According to the meaning of the title, we get A-B + C = - 6, ① a + B-C = - 2, ② 4A + 2B + C = 9, ③ 2b-2c = 4, ④ - 2b + 5C = 17, ⑤ 3C = 21, C = 7, and ④ 2B = 4 + 2C = 4 + 2 × 7 = 18

If the square of quadratic equation of one variable * - 2 * - M = 0 has no real root, the image of the first-order function y = (M + 1) * + M-1 does not pass through? Which image limit

X & sup2; - 2x-m = 0, no real root
That is, the discriminant B & sup2; - 4ac < 0
Substitute:
(-2)²-4×1×(-m) ＜0
4+4m

If the quadratic equation nx-2x-1 = 0 with respect to X has no real root, then the image of the linear function y = (n + 1) x-n does not pass through which quadrant?

The previous equation gives n

Given the quadratic function y = ax & sup2; + BX + C of X, we can find the analytic expression of this function
（1）A（0,-1）B（1,-4）C（-1,4）
（2）A（1,0）B（-1,6）C（2,3）
（3）A（-1,2）B（-2,-3）C（1,0）

By substituting the coordinates of point ABC into the analytic expression of the function, we can get the system of equations, and then solve the system of equations. According to the meaning of the problem, we can get the system of equations: C = - 1A + B + C = - 4a-b + C = 4, the solution of which is a = 1, B = - 4, C = - 1. The analytic expression of this function is y = x & # 178; - 4x-1. (2) according to the meaning of the problem, we can get the system of equations: a + B + C = 0a-b + C = 64a + 2B + C

As shown in the figure, the image of quadratic function passes through three points a, C and B, the coordinates of point a are (- 1,0), the coordinates of point B are (4,0), and point C is on the positive half axis of Y axis,
And ab = OC
1. Find the coordinate of point C; 2. Find the analytic expression of quadratic function and find the maximum value of function

-Let y = a (x + 1) (x-4) = a (x ^ 2-3x-4)
If y (0) = - 4A > 0, a is obtained

The image of quadratic function passes through ABC three points, a coordinate of point a is (- 1.0) B coordinate of point B is (4.0) C on the positive half axis of Y axis, and ab = 5 / 2oC
Finding the coordinate of point C, finding the analytic formula and maximum value of quadratic function

∵ AB = 5 ∵ C = 20, C point coordinates are (0, 20)
Let the quadratic function y = a (x-4) (x + 1) replace c (0,20)
The solution is a = - 5
The analytic formula of quadratic function is y = - 5x & # 178; + 15x + 20
The maximum value = (4ac-b & # 178;) / 4A = 31.25

The image of a quadratic function passes through three points a, C and B. the coordinates of point a are (- 1,0), and the coordinates of point B are (4,0). Point C is on the positive half axis of Y axis, and ab = 5 / 2oC
(1) Find the coordinates of point C;
(2) Find the analytic expression of the quadratic function, and find the maximum value of the function

(1) ∵ the coordinate of point a is (- 1,0), the coordinate of point B is (4,0) ∵ AB = 4 - (- 1) = 5 ∵ OC = 2 / 5ab = 2 ∵ the coordinate of point C is (0,2) ∵ the image passes through the point (- 1,0), (4,0), so the analytic formula can be set as: y = a (x + 1) (x-4), substituting x = 0, y = 2 into: 2 = - 4A, then a = - 1 / 2 ∵ the analytic formula is: y = -

Find the intersection of the quadratic function image y = x ^ 2-3x + 2 and X axis as a and B
Finding a, B coordinates
2. Find the length of line ab

1、
Y = 0 on X-axis
So y = x & sup2; - 3x + 2 = 0
(x-1)(x-2)=0
x=1,x=2
So a (1,0), B (2,0)
2、
AB=|1-2|=1

The intersection of quadratic function image and X axis is expressed by formula

y=ax^2+bx+c=0,a≠0
△=b^2-4ac>=0
x1=(-b+√△）/(2a)
x2=(-b-√△）/(2a)
The intersection of quadratic function image and x-axis is expressed as ((- B + √ △) / (2a), 0) and ((- B - √ △) / (2a), 0)

A quadratic function, when x = - 2, obtains the minimum value of - 3, the abscissa of its image and the two intersections of X axis is equal to 3. Find the analytic expression of the quadratic function

y=a（x+2）^2-3
When y = 0, x1x2 = C / a = (4a-3) / a = 3
a=3
So the analytic formula is: y = 3 (x + 2) ^ 2-3