When 3 x 1, the corresponding Y value is 1 y 8, and the function relation is found

Solution: If k<0 then when x=-3, y=8 i.e. 3k+b=-8 Eq. 1 When x=1, y=1 i.e. k+b=1 Eq. 2 is subtracted from Eq. 1 to get 3k-k=-8-1 solution k=-9 2 substituted into Eq. 2, and -9 2+b=1 solution gives b=11 2, so the analytic formula of the function is y=-9 2x+11 2, if k>0 then when x=-3, y=1 i.e., -3k+b=1 Eq. 1, when x=1 y=8 i.e., k+b=8 Eq. 2

Subtracting Eq. 2 from Eq. 1 gives -3k-k=1-8 to get k=7 4 and substituting it into Eq. 2 to get 7 4+b=8 and solves b=25 4, so the resolution of the function is y=7 4x+25 4

Summing up, the relationship of the function is y=-9 2x+11 2 or y=7 4x+25 4

If k>0, then y(-3)=-3k+b=1

y(1)=k+b=8

Subtract to get: 4k=7, k=7 4, b=8-k=25 4, y=7x, 4-25 4

If k<0, then y(-3)=-3k+b=8

y(1)=k+b=1

Subtract to obtain: 4k=-7, k=-7 4, b=1-k=11 4, y=-7x 4+11 4

Solution: According to the meaning of the question, it is possible.

1) When x=-3, y=1; When x=1, y=8; or (2) when x=-3, y=8; When x=1, y=1

i.e. (1) 1=-3k+b; 8=k+b.or (2) 8=-3k+b; 1=k+b.

Solve the equation system in (1) to obtain: k=7 4, b=25 4;

Solve the system of equations in (2) to obtain: k=-7 4, b=11 4

The function relation is: y=(7 4)x+25 4 or y=(-7 4)x+11 4.

Let k<0, the function is a subtraction function, and there is:

x=-3, y=8 i.e. 8=-3k+b

x=1, y=1, i.e. 1=k+b

Linking the above two formulas, get:

k=-7 4,b=11 4 k=-7 4<0 holds.

Therefore, the original function is y=-7 4x+11 4, which is simplified to obtain 4y=-7x+11, and the same can be obtained when k>0 is obtained.

When k>0 is early, the function is incremental.

So when x=0, 0 b=2, b 2

x=2,2k2 4,k=1;

So k+b=2 1=3

Pai hall finch when k

Summary. This function is a one-time function, single increase or single decrease, so the boundary values correspond.

Knowing that the primary function y=kx+b, when -3 x 1, the corresponding y value is -1 y 8, find the analytical expression of the primary function.

This function is a one-time function, single increase or single decrease, so the boundary values correspond.

When x=-3, y=-1, when x=1, y=8-3k+b=-1k+b=8k=9 4b=23 4

When x=1, y=-1, when x=-3, y=8-3k+b=8k+b=-1k=-9 4b=5 4

The analytical formula of this primary function is: y=9 4x+23 4 or y=-9 4x+5 4

x=-3,y=1

x=1, y=9.

1=-3k+b

9=k+bk=2,b=7

y=2x+7

Or. x=-3,y=9

x=1, y=1 when the age is pure.

9=-3k+b

1=k+bk=-2,b=3

y=-2x+3

It is known that when the primary function y=kx+b is -3 x 1, the corresponding y value is 1 y 9 k<0.

3k+b=9,k+b=1

The solution yields k = -2 and b = 3

k>0.

3k+b=1,k+b=9

The solution is k = 2 and b = 7

When -3 x 1 is used, the corresponding y value is 1 y 9

Illustrates that the primary function y=kx+b image passes through (-3,1) and (1,9) or (-3,9) and (1,1).

When the image passes through the dots (-3,1) and (1,9):

3k+b=1,k+b=9

The solution is k = 2 and b = 7

When the image passes through the dots (-3,9) and (1,1):

3k+b=9,k+b=1

The solution yields k = -2 and b = 3

k = 2, b = 7 or k = -2, b = 3

Replace -3,1 in x and 1,9 in y. Solve the system of equations.

When k>0 and the function increases monotonically at -3 x 1, then -3k+b<=y<=k+b, i.e., -3k+b=1, k+b=9 is solved to k=2, b=7, and when k<0, the function is monotonically reduced at -3 x 1, then k+b<=y<=-3k+b, i.e., k+b=1, -3k+b=9 is k=-2, b=3

When k<0, the function is a subtraction function, so when x=-3, y=-3k+b=9, when x=1, y=k+b=1, the simultaneous equation is solved k=-2, b=3; When k>0, the function is an increasing function, when x=-3, y=-3k+b=1; When x=1, y=k+b=9, l simultaneous equations are solved to k=2, b=7

When k>0 function is incremental, so -3k+b=1 k+b=9, so k=2, b=7

When k<0 function is decreasing, so -3k+b=9 k+b=1, so k=-2, b=3

When x = -3, y = 1. Then -3k+b=1;When x=1, y=9, then k+b=9, and according to the two equations, k=

It's a straight line.

So the maximum and minimum are at the endpoint.

So x=-3, y=1

x=1,y=9

or x=-3, y=9

x=1,y=1

x=-3,y=1

x=1,y=9

then 1=-3k+b

9 = k + b minus 4k = 8

k=2x=-3,y=9

x=1,y=1

then 9=-3k+b

1 = k + b subtract 4k = -8

k=-2, so k=-2, k=2

Solution: When k<0, the function y=kx+b decreases monotonically on [-3,1], then -3k+b=9

k+b=1 is solved: k=-2, b=3

When k>0 and the function y=kx+b increases monotonically on [-3,1], then -3k+b=1

k+b=9 is solved: k=2, b=7

k is 2 first bring x and y correspondence into the equation, and then use the binary equation, which is very simple, b does not need to be calculated at all, it is eliminated, and k is calculated as 2

When k > 0, y increases as x increases.

Then -3k+b=1, k+b=9, so k=2

When k < 0, y decreases as x increases.

Then -3k+b=9, k+b=1, so -2

To sum up, when k>0, k=2When k<0, k=-2

When x=1, y=-1, x=-1, and y=2 are substituted into y=kx+b, respectively. 1=k+b,2=-k+b

2b=1,b=1/2,k=-3/2

i.e. y=(-3 2)x+1 2.

When x=2.

y=-3+1/2=-5/2.

When x=2, the value of the function is.