The parabola y = ax square + BX + C, when x = 2, x = 0, the value of Y is equal The parabola y = ax square + BX + C, when x = 2, x = 0, the value of Y is equal, the vertex m of the parabola is on the straight line y = 3x-7, and the abscissa of the intersection point with the straight line is 4 Find the analytic formula of this parabola

The parabola y = ax square + BX + C, when x = 2, x = 0, the value of Y is equal The parabola y = ax square + BX + C, when x = 2, x = 0, the value of Y is equal, the vertex m of the parabola is on the straight line y = 3x-7, and the abscissa of the intersection point with the straight line is 4 Find the analytic formula of this parabola

When x = 4, y = 3 × 4-7 = 5
That is, the point m (4,5) is on the parabola, so there is: 5 = 16A + 4B + C
Let y = a (x-4) ^ 2 + 5 = ax ^ 2-8ax + 16A + 5 = ax ^ 2 + BX + C
The comparison coefficient is - 8A = B, 16a + 5 = C
On the other hand, when x = 2 and x = 0, the values of Y are equal, that is, the axis of symmetry of the parabola is x = 1, i.e. - B / 2A = 1, so B = - 2A
So we can get a = 0

We know the square + BX + C and X of parabola y = ax

The shape of parabola y = ax & # 178; + BX + C is the same as y = x & # 178;, a = 1 is obtained. The symmetry axis is a straight line x = 3, and - B / 2A = 3, that is, B = - 6, so y = x & # 178; - 6x + C = (x-3) &# 178; + C-9, the coordinates of the highest point are (3, C-9). Substituting the straight line y = x + 1, C-9 = 3 + 1, C = 13, the analytical formula of parabola is y = x & # 178; - 6x

The parabola y = ax square + BX + passing through point a (0,3)

Then, C = 3?

The parabola is obtained by translating the image of y = - 2 (x-1) 2 three units to the right. The analytical expression of the parabola is obtained

Let t = x-3 be left plus right minus for translation
y=-2(t-4)2
Verify the original parabola when x = 1, y = 0
The point (1,0) moves to the right by 3 units to (4,0), which satisfies the new analytical formula
Give it to my brother!