Given that the quadratic function f (x) = the square of AX + BX + C satisfies the condition f (- 1) = f (3) = 0, and the minimum value is - 8, the analytic solution of the function is obtained

Given that the quadratic function f (x) = the square of AX + BX + C satisfies the condition f (- 1) = f (3) = 0, and the minimum value is - 8, the analytic solution of the function is obtained

-1+3=-b/a；
b/a=-2;
-1×3=c/a;
c/a=-3;
c-b²/4a=-8;
-3a-a=-8;
a=2;
b=-4;
c=-6;
The analytic formula is f (x) = 2x & # 178; - 4x + 6;
If you don't understand this question, you can ask,

① The quadratic function y = 1 / 2x ^ 2 + 3x + 3 is matched into the form of y = a (X-H) ^ 2 + K by using the matching method. ② the requirements of y = - 2x ^ 2-4x-6 are as follows

In this paper, we are going to find the following a = 1 / 2, here a = 1 / 2, H = -3, and here a = 1 / 2, here a = 1 / 2, H = -3, k = -3, k = -3, k = -3 / 2x-3 / 2, here here here a = 1 / 2, here here here a = 1 / 2, H = -3, k = -3, k = -3 / 2Y = -2x and we are going to find the (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\herea = - 2, H = - 1, k = - 4

If the image of quadratic function y = (a + 1) x ^ 2 + 3x + A ^ 2 + 4A + 3 passes through the origin, then a=

-3
Because it's a quadratic equation, it can't be equal to - 1

Given the square of quadratic function image y = 2x + 6x + 5, when x = what is the maximum or minimum value of Y

Y = 2 (x + 3 / 2) ^ 2 + 1 / 2, so when x = - 3 / 2, there is the minimum value of Y 1 / 2