Given the quadratic function y = the square of X + (the square of MX) x-2m, find the M (1) that satisfies the condition. The known vertex is on the y-axis (2). The vertex is on the x-axis (3). The image passes through the origin If you want to learn how to do it yourself, you can bypass this problem

Given the quadratic function y = the square of X + (the square of MX) x-2m, find the M (1) that satisfies the condition. The known vertex is on the y-axis (2). The vertex is on the x-axis (3). The image passes through the origin If you want to learn how to do it yourself, you can bypass this problem

1) If the vertex is on the y-axis, then the symmetry axis is y-axis,
So the axis of symmetry x = - B / 2A = 0,
That is M & # 178 / 2 = 0,
The solution is m = 0,
(2) If the vertex is on the x-axis, there is only one intersection point between the parabola and the x-axis,
So delta = 0,
That is B & # 178; - 4ac = 0,
So m ^ 4 + 8m = 0,
The solution is m = 0, or M = - 2
(3) Because the image goes through the origin
So as long as (0,0) is substituted in, then,
-2m=0,
The solution is m = 0,