Discussion on the classification of junior high school mathematics

When -3≤x≤ 1, the corresponding y value is 1≤y≤9. There are two cases:

(1).k & gt0. At this time, x=-3, y= 1, x= 1, y=9,

-3k+b= 1，k+b=9，

The solution is k=2, b=7,

kb= 14。

(2).k & lt0. At this time, x=-3, y=9, x= 1, y= 1,

-3k+b=9，k+b= 1，

The solution is k =-2 and b = 3.

kb=-6。

If you don't understand, please take a look at this:

There are two situations about this problem.

① When k > 0, the function value increases with the increase of independent variables, so we can get two points (-3, 1) (1, 9) respectively.

Substitute these two points respectively: k=2 b=7 ∴kb= 14.

② When k < 0, the function value decreases with the increase of independent variables, so we can get two points (-3,9) (1,1) respectively.

Substitute these two points respectively: k=-2 b=3 ∴kb=-6.