Given that the vertex coordinates of the parabola y = AX2 + BX + C are (1,4), and when x = 2, y = 1, find the values of a, B, C

Given that the vertex coordinates of the parabola y = AX2 + BX + C are (1,4), and when x = 2, y = 1, find the values of a, B, C

∵ the vertex coordinates of the parabola are (1,4)
Let y = a (x-1) ^ 2 + 4
When x = 2, y = 1,
Then 1 = a + 4 and the solution is a = - 3
Then the parabola y = - 3 (x-1) ^ 2 + 4
=-3x^2+6x+1
∴a=-3,b=6,c=1

If the axis of symmetry of the parabola y = 1 / 2x2 + MX + 3 is a straight line x = 4, then the value of M is___ .

Axis of symmetry x = - B / 2A = - M / (2 * 1 / 2) = - M = 4
So m = - 4

The equation of symmetry axis of parabola y = x Square-1 is?

The equation of symmetry axis of Square-1 of y = x is: x = 0

The axis of symmetry equation x = 2 for parabolic y = xsquare-2 (a + 1) x + 2aquare-a is known
1 find the value of real number a
If there are two different points of intersection between the parabola and the x-axis, the coordinates of the point of intersection can be obtained

1. Because the equation of symmetry axis of y = x & # 178; - 2 (a + 1) x + 2A & # 178; - A is x = - [- 2 (a + 1)] / 2 = a + 1, so a + 1 = 2, the solution is: a = 1; 2, y = x & # 178; - 2 (a + 1) x + 2A & # 178; - a = x & # 178; - 4x + 1, when y = 0, X & # 178; - 4x + 1 = 0, the solution is: x = 2 ±