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一、函数图象恒过定点问题例1求证:无论m为何值,函数y=-2mx+2(m-1)的图象恒过定点.解取m=0,m=1代入函数解析式,得y=-2y=-2x.
First, the function image constant over fixed point problem Example 1 Prove: regardless of the value of m, the function y = -2mx +2 (m-1) of the image constant over the fixed point. Solution m = 0, m = 1 into the function analytical expression , y=-2y=-2x.