## Translate the parabola y = ax ^ 2 + BX + C down 2 units Then translate 6 units to the left, and the vertices of the parabola are - 3, - 1. And a + B + C = 9. Find the values of a, B and C

Translate the parabola y = ax ^ 2 + BX + C downward by 2 units and then to the left by 6 units, that is, y + 2 = a (x + 6) ^ 2 + B (x + 6) + CY = ax ^ 2 + 12ax + 36a + BX + 6B + C-2 = ax ^ 2 + (12a + b) x + (36a + 6B + C-2). The vertex is - 3 and - 1, so it is y = a (x + 3) ^ 2-1 = ax ^ 2 + 6AX + 9a-1. The corresponding coefficients are equal 12a + B = 6A (1) 36a +

### Translate the parabola y = ax ^ 2 + BX + C by 2 units to the left, then translate it down by 1 unit, and then coincide with the parabola y = - 3x ^ 2 + 6x to find the values of a, B and C

Left translation: y = a (x + 2) ²+ b(x+2)+c

Down translation: y = a (x + 2) ²+ b(x+2)+c-1

y=ax ²+ 4ax+4a+bx+2b+c-1

=ax ²+ (4a+b)x+4a+2b+c-1

a=-3 4a+b=6 4a+2b+c-1=0

a=-3 b=18 c=-23

### It is known that the parabola y = ax ^ 2 + BX + C translates 2 units to the left and 3 units to the down, which can be compared with the parabola y = 2x ^ 2-4x + 1 Coincidence, find the values of a, B and C

In order to get a, b, c, Set y = 2x ²- 4X + 1 first translate up 3 units, The new parabola is y - three = 2x ²- 4x+1 (i.e. y becomes Y-3), Y = 2x ²- 4x+4; Then put y&n

### It is known that point a (- 2, - C) is translated 8 units to the right to obtain that both points a ', a and a' are on the parabola y = AX2 + BX + C, and the ordinate of the intersection of this parabola and Y axis is - 6, then the vertex coordinate of this parabola is () A. （2，-10） B. （2，-6） C. （4，-10） D. （4，-6）

From the ordinate of the intersection of parabola y = AX2 + BX + C and Y axis is - 6, C = - 6 is obtained,

‡ a (- 2,6), point a shifts 8 units to the right to obtain point a '(6,6),

∵ both a and a 'are on a parabola,

∴

4a−2b−6＝6

36a+6b−6＝6 ， By solving this set of equations, we get

a＝1

b＝−4 ，

Therefore, the analytical formula of parabola is y = x2-4x-6 = (X-2) 2-10,

The coordinates of parabola vertex are (2, - 10)

So choose a

### Known quadratic function y = ax ²+ The image of BX + C passes through points a (0, a), B (1, - 2), and the axis of symmetry is a straight line x = 2

If you pass through point a (0, a), you can substitute y (0) = C = a

Go through point B (1, - 2) and substitute: Y (1) = a + B + C = - 2 to get: 2A + B = - 2

The axis of symmetry is a straight line x = 2, i.e. - B / (2a) = 2, resulting in: B = - 4A

The solution is: a = 1, B = - 4, C = 1

So y = x ^ 2-4x + 1

### If the image of the quadratic function f (x) is symmetric about the Y axis, and 1 ≤ f (1) ≤ 2, 3 ≤ f (2) ≤ 4, find the value range of F (3)

∵ the image of the quadratic function f (x) is symmetric about the Y axis, ∵ let the quadratic function f (x) = AX2 + C, which can be obtained from the meaning of the question, 1 ≤ a + C ≤ 23 ≤ 4A + C ≤ 4, f (3) = 9A + C, and make its plane area as shown in the figure below: F (3) = 9A + C obtains the maximum value at points a and B, which is obtained from a + C = 24a + C = 3, a (13, 53), the same as B (1