对于正整数n,设函数fn(x)=x^n(1-x),函数fn(x)的图像在(2,fn(2))处的切线与y轴焦点

fn(x)=x^n(1-x)=x^n-x^(n+1)，

∴fn'(x)=nx^(n-1)-(n+1)x^n,

fn'(2)=n*2^(n-1)-(n+1)*2^n=2^(n-1)*[n-2(n+1)]=-(n+2)*2^(n-1),

∴函数fn(x)的图像在(2,fn(2))处的切线方程是y+2^n=-(n+2)*2^(n-1)*(x-2),

它与y轴交于点（0，（n+1)*2^n).