As shown in the figure, the image of quadratic function y = x ^ 2 + BX + C passes through a (- 1,0) and B (4,0), intersects with the positive half axis of Y axis at C, and ab = OC (1) Find the coordinates of point C (2) Find the expression of quadratic function, and find the maximum value of function

As shown in the figure, the image of quadratic function y = x ^ 2 + BX + C passes through a (- 1,0) and B (4,0), intersects with the positive half axis of Y axis at C, and ab = OC (1) Find the coordinates of point C (2) Find the expression of quadratic function, and find the maximum value of function

one
AB=OC=4-（-1)=5
C(0,5)
two
y=a(x+1)(x-4)
=ax^2-3ax-4a
-4a=5
a=-5/4
y=-5/4 x^2+15/4 x+5
x=3/2, y max=125/16

It is known that the quadratic function f (x) satisfies the conditions (1) f (2 + x) = f (2-x); (2) the minimum value of F (x) is - 4; the sum of two squares of F (x) is 16, so we can find f (x)

A:
If the quadratic function f (x) satisfies: F (2 + x) = f (2-x), then f (x) is symmetric with respect to x = (2 + X + 2-x) / 2 = 2
Let f (x) = a (X-2) ^ 2 + C
If the minimum value of F (x) is - 4, then a > 0, C = - 4
So: F (x) = a (X-2) ^ 2-4
The sum of squares of the zeros of F (x) is 16
f(x)=a(x-2)^2-4=0
The solution is: X1 = 2 + 2 / √ a, X2 = 2-2 / √ a
According to the meaning: X1 ^ 2 + x2 ^ 2 = 16
(x1+x2)^2-2x1x2=16
4^2-2*(4-4/a)=16
So:
4-4/a=0
The solution is: a = 1
So: F (x) = (X-2) ^ 2-4 = x ^ 2-4x
So: F (x) = x ^ 2-4x

It is known that the quadratic function f (x) satisfies the following conditions: 1. F (x + 2) = f (2-x); 2. The minimum value of F (x) is - 4; 3. The sum of two squares of F (x) = 0 is 16
The analytic expression of

F (x + 2) = f (2-x); axis of symmetry: x = 2, the minimum value of F (x) is - 4, vertex (2, - 4), a > 0, Let f (x) = a (X-2) ^ 2-4f (x) = ax ^ 2-4ax + 4a-4x1 + x2 = 4, x1x2 = (4a-4) / A. the sum of two squares of F (x) = 0 is 16,16 = X1 ^ 2 + x2 ^ 2 = (x1 + x2) ^ 2-2x1x2 = 16-2x1x22x1x2 = 0x1x2 = 0, (4a-4) / a =

It is known that the quadratic function f (x) satisfies f (0) = f (4), and the sum of two squares of F (x) = 0 is 10. The image passes through (0,3) points, and the analytic expression of F (x) is obtained
The sum of two squares is 10

Because the type of function is definite, it is quadratic function, especially when f (0) = f (4), its symmetry axis is x = 2, so let its expression be f (x) = a (X-2) ^ 2 + C (a is not equal to 0), and expand to f (x) = ax ^ 2-4ax + 4A + C, because the 4A + C of the function image (0,3) is 3; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp