The image of the function f (x) = ax Λ 2 + BX + C (a ≠ 0) is symmetric with respect to the line x = - B / 2A Come on, I'm going to class The solution set of numbers a, B, C, m, N, P with respect to the equation m [f (x)] Λ2 + NF (x) + P = 0 cannot be () a. {1,2} B {1,4} C {1,2,3,4} D {1,4,16,64}

The image of the function f (x) = ax Λ 2 + BX + C (a ≠ 0) is symmetric with respect to the line x = - B / 2A Come on, I'm going to class The solution set of numbers a, B, C, m, N, P with respect to the equation m [f (x)] Λ2 + NF (x) + P = 0 cannot be () a. {1,2} B {1,4} C {1,2,3,4} D {1,4,16,64}

Let the solution of the equation m [f (x)] be Y1, then Y2 must have Y1 = ax's second power + BX + C, y2 = ax's second power + BX + C. then from the picture, y = Y1, y = Y2 is a straight line parallel to the X axis, and they have intersection with F (x)

(1) The maximum value of the function y = ax + BX + C is 2. The vertex of the image is on y = x + 2, and ABC is obtained through (3, - 6)

Let the vertex be (m, n)
The maximum is 2, n = 2
The vertex is on y = x + 2, n = m + 2, 2 = m + 2, M = 0
So the vertex is (0,2)
-b/(2a)=0 b=0
(4ac-b^2)/(4a)=2 c=2
After (3, - 6) - 6 = 9A + 3B + C = 9A + 2, a = - 8 / 9
a. The values of B and C are: - 8 / 9,0,2

The maximum value of the function y = ax + BX + C is 2. The vertex of the image is on y = x + 2, and ABC is obtained through (3,6)

The maximum value of y = ax + BX + C is 2, that is, the vertex ordinate is 2, and the vertex is y = x + 2, so the vertex is (0,2)
b=0,c=2
y=ax^2+2
After (3,6) substitution, 9A + 2 = 6
a=4/9
y=4/9*x^2+2

Is there any relationship between the square of quadratic function y = ax + BX + C and y = a (X-H) + k?
I found that there are some differences between our math book and the last math book (Orange) Looking at the answer, I was confused

Man, dry the fog. Y = ax ^ 2 + BX + C can make the formula. After the formula is made, y = a (x-2a / b) ^ 2 + (4ac-b ^ 2) / 4A, let H = 2A / B, k =) / 4A (as for why using H, K is just a habit), so: y = a (X-H) ^ 2 + K