Given that the vertex of square-2x + m + 1 of the quadratic function y = x is the same as that of square + nx-1 of y = 2x, find the vertex value of M, n?

Given that the vertex of square-2x + m + 1 of the quadratic function y = x is the same as that of square + nx-1 of y = 2x, find the vertex value of M, n?

y=(x-1)²+m
The vertex is (1, m)
The vertex of the square + nx-1 of y = 2x is (1, m)
So it is
y=2(x-1)²+m
That is y = 2x & # 178; - 4x + 2 + M
So n = - 4
-1=2+m
So m = - 3, n = 4

Quadratic function y = f (x) axis of symmetry x = 2 the length of the line cut on the X axis is 2, the minimum value of F (x) = - 1 the analytic expression of F (x)

Let the analytic formula of the function be y = ax ^ 2 + BX + C. according to the meaning of the problem, we get a + B + C = 0 9A + 3B + C = 0 4A + 2B + C = - 1, and the solution is a = 1 b = - 4 C = 3, so the analytic formula is y = x ^ 2-4x + 3