Question

求函数y=(2x-1)/(x+4)^2值域

Answer 1

y=(2x-1)/(x+4)^2
yx^2+(8y-2)x+(16y+1)=0
判别式
（8y-2)^2-4y(16y+1)>=0
-32y+4-4y>=0
36y<=4
y<=1/9

追问

y=(2x-1)/(x+4)^2
yx^2+(8y-2)x+(16y+1)=0
这两步怎么弄的？

追答

y(x+4)^2=2x-1
y(x^2+8x+16)=2x-1\
yx^2+(8y-2)x+16y+1=0

追问

判别式为什么要大于零？

Answer 2

将分母移到等式的左边 化简得
yx^2+(8y-2)x+16y+1=0、
b^2-4ac>=0
求出Y值

追问

能详细点么

Answer 3

y(x^2+8x+16)=2x-1
yx^2+(8y-2)x+16y+1=0
x有解，所以△>=0
解出y<=1/9