If the image of quadratic function y = x2 + MX + n has only one intersection (- 2,0) with X axis, then M=_____ ,n=_____ .

If the image of quadratic function y = x2 + MX + n has only one intersection (- 2,0) with X axis, then M=_____ ,n=_____ .

∵ there is only one intersection (- 2,0) between quadratic function and X axis
∴（-2）^2-2m+n=0
∴x=-b/2a ,-m/2=-2
∴m=4,n=4

In the quadratic function y = x2 + MX + N, if 3m-n = 2, then its image must pass through the point

Let x = - 3, then no matter what the values of M and N are, y = 9-3m + n = 9 - (3m-n) = 9-2 = 7
So over fixed point (- 3,7)

The quadratic function y = x2 + MX + m-2 is known
(1) If the distance between the two points of intersection obtained from the parabola of x-axis is the root sign 3, write out the analytic formula of the function at this time. (2) when m takes what value, what is the minimum distance between the two points?
Great Xia, please help me. I need it urgently!!!! Waiting online!!!!

(1) When the distance between the two points of intersection obtained by cutting the parabola on the x-axis is the root sign 3, that is the equation:
The difference between the two of x2 + MX + m-2 = 0 is the root sign 3
X1-x2 = root 3,
(X1-X2)^2=3,
(x1 + x2) ^ 2-4x1 * x2,
X1 + x2 = - m, X1 * x2 = m-2
M ^ 2-4m + 5 = 0, the equation has no real solution
So the function doesn't exist
(2) Let the distance between two points be K
M^2-4M+8-K^2=0,
4K^2-16>=0,K>0.
K> The minimum distance between two points is 2
In this case, M = 2