The parabola y = x & # 178; + BX + C is obtained by moving the parabola y = x & # 178; + BX + C upward by 2 units and then to the left by 1 unit

The parabola y = x & # 178; + BX + C is obtained by moving the parabola y = x & # 178; + BX + C upward by 2 units and then to the left by 1 unit

y=(x+1)²+2
y=x²+2x+3
b=2 c=3

The parabola y = x & # 178; + BX + 9 + C moves upward by 2 units, and then moves left by 4 units to get the parabola y = x & # 178;, find the value of BC, fast,

The solution is because it moves two units up + BX + 9 + C + 2 of y = x and four units to the left of y = (x + 4) square + B (x + 4) + 9 + C + 2  y = xsquare + 8x + 16 + BX + 4B + C + 11y = xsquare + (8 + b) x + 27 + C ∵ after translation, the analytic formula is y = xsquare ∵ 8 + B = 0 ∵ B = - 827 + C = 0 ∵ C = - 27 ∵ BC = 8 × (-)

Point a (- 2, - C) is shifted 8 units to the right to obtain point A1. Both points of a and A1 are on the parabola y = ax square + BX + C. the ordinate of the intersection point of the parabola on the Y axis is - 6, and the vertex coordinate of the parabola is?

8 units to the right
The abscissa is - 2 + 8 = 6
That is, x = - 2, x = 6, the function values are equal
So the axis of symmetry x = (- 2 + 6) / 2 = 2
Y-axis x = 0
Then y = 0 + 0 + C = - 6
So x = - 2
y=6
6=4a-2b-6
2a-b=6
Axis of symmetry x = - B / 2A = 2
So a = 1, B = - 4
y=x²-4x-6=(x-2)²-10
Vertex (2, - 10)

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

From the ordinate of the intersection of the parabola y = AX2 + BX + C and the Y axis is - 6, we get C = - 6, a (- 2,6), and point a shifts 8 units to the right to get point a '(6,6), ∵ A and a' are on the parabola, ∵ 4A − 2B − 6 = 636a + 6B − 6 = 6