Given that the image of a function y = 2mx + (6-3m & # 178;) passes through the origin and the second and fourth quadrants, then M=______

6-3m²=0
m=±√2
M = √ 2, y = 2 √ 2x, passing through the second and fourth quadrants
M = - √ 2 y = - 2 √ 2x passing through the first and third quadrants
m=-√2

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