Question

一次函数y=kx+b的自变量的取值范围是-3≤x≤6，相应函数值的取值范围是-5≤y≤-2，求这个一次函数的解析式

Answer 1

因为一次函数本身就是一个单调的函数，要么递增，要么递减，在定义域取值的两头取得最大值和最小值，但是这题没有说明K是大于0还是＜0，所以只能分类
当k＞0时，函数是递增的，这个时候满足-3k+b=-5,6k+b=-2，解得k=1/3，b=-4
此时的函数解析式为y=1/3x-4
当k＜0时，函数是递减得，这个时候满足-3k+b=-2,6k+b=-5，解得k=-1/3，b=-3
此时的函数解析式为y=-1/3x-3

k=0的时候我没有考虑，因为k是不可能为0的，为0了函数就不存在单调性了

别看我写这么多，其实过程很简单，希望对你有帮助

Answer 2

-5<=y<=-2
所以-5<=kx+b<=-2
-5-b<=kx<=-2-b
1、当k>0时-5-b/k<=x<=-2-b/k
此时-5-b/k=-3
-2-b/k=6
k=1/3
b=-4
2、当k<0时-2-b/k<=x<=-5-b/k
-2-b/k=-3
-5-b/k=6
所以，k=-1/3
b=-3

Answer 3

1 如果k≥0
-5=-3×k+b
-2=6×k+b

解得 k=1/3 b=-4
解析式为y=x/3-4

2 如果k<0
-5=6×k+b
-2=-3×k+b

解得 k=-1 /3 b=-3
解析式为y=-x/3-3

Answer 4

将x和y的对应值代入y＝kx＋b　得到一个方程组，解之，就求得k，b的值，便得到解析式。

Answer 5

y=1/3x-4 或是y=-1/3x-3

Answer 6

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