2. The image of quadratic function is parabola. When the value of x is equal to -2a/b, the function reaches the extreme value of 4ac-b 2.
-2a/b = 3，4ac-b 2 = 4，ax 2+bx+c = a4 2+B4+c =-3。 The abc value is obtained by connecting equations.
3. Both functions pass through the same point, that is to say, the expressions of the two functions at point AB are the same, with a = (x 1, 0) and b = (0, y2). Substitute the coordinate values of AB into X-3 = X 2+BX+C, and the minimum formula is 4ac-B 2.
or
1. Knowing the image passing point of a quadratic function (0. 1)(2.4)(3. 10), find the relationship of this quadratic function.
Let the relationship of quadratic function be y = ax 2+bx+c.
Substitute into three-point coordinates to get
c= 1
4a+2b+c=4
9a+3b+c= 10
a=3/2 b=-3/2 c= 1
2. The image passing point of quadratic function is known (4. -3), and there is a maximum when x=3. 4. Find out the relationship of this quadratic function.
Let the relationship of quadratic function be y = a (x+b) 2+c b) 2+c.
When x=3, there is a maximum value of 4, and we can know that b =-3 c = 4 a.
Replace (4. -3) -3=a+4 a=-7。
3. The image of the linear function y=x-3 intersects with the X and Y axes at points A and B respectively, and the image of the quadratic function y = x2 (square) +bx+c passes through points A and B. Find the relationship between the coordinates of A and B and the quadratic function and its minimum value.
The image of linear function y=x-3 intersects with X axis and Y axis at point A and point B respectively.
A(3，0) B(0，-3)
Substitute y = x 2+bx+c to get.
c=-3
b=-2
y=x^2-2x-3=(x- 1)^2-4
The minimum value is -4.