函数y=loga(x+3)-1的图像恒过定点A,若点A在直线(1/m)x+(1/n)y+8=0上,其中m>0,n>0,

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函数y=loga(x+3)-1的图像恒过定点A,若点A在直线(1/m)x+(1/n)y+8=0上,其中m>0,n>0,
函数y=loga(x+3)-1的图像恒过定点A,若点A在直线(1/m)x+(1/n)y+8=0上,其中m>0,n>0,

函数y=loga(x+3)-1的图像恒过定点A,若点A在直线(1/m)x+(1/n)y+8=0上,其中m>0,n>0,
求什么

y= log a (x+3) -1 ===> x+3 = a ^(y+1) ;要A对任何a都成立,必然y+1 = 0, x =-2;
A为 (-2,-1)
-2/m -1/n +8 = 0 ==> n = m/(8m-2);
z = 2*m +n = 2*m + m/(8m-2);
z' = 2 + 1/(8m-2) - 8m/ (8m-2)^2 =0 ==> m = 3/8;
==> min(2*m+n ) = 9/8