The known function y 6 3m x n 4

1. When (6+3m) < 0, y decreases with the increase of x (monotonic reduction function), i.e., m<-2 n can be any real number.

2. When x=, y<0, that is, the intersection of the image with a function and the y-axis is below the x-axis, that is, n-4<0 n<4 m can be any real number.

3. When n-4=0,6+3m≠0, the function passes through the origin. That is, when there is n=4, m≠-2

When m,n is the value, y decreases with the increase of x?

6+3m<0 m<-2

When m,n is the value, the intersection of the image of the function with the y-axis is below the x-axis?

n-4<0 n<4

When m,n is what the value, the function image passes through the origin.

n-4=0 n=4

6+3m≠0 m≠-2

y=kx+b(k≠0)

When k<0, y=kx+b decreases monotonically on r.

When k>0, y=kx+b increases monotonically on r.

When x=0, y=b>0, the intersection of the function's image with the y-axis is above the x-axis.

When x=0, y=b<0, the intersection of the function's image with the y-axis is below the x-axis.

When x=0, y=b=0, the function image passes through the origin.

6+3m<0

n-4<0

n-4=0

Solution: (1) According to the problem, 6+3m 0, that is, m -2, n is any real number;

2) According to the title, n-4 0, that is, n 4, m is any real number;

3) According to the topic, 6+3m≠0, n-4=0, m≠-2, n=4

1) Y decreases with the increase of x, so k <0, i.e., 6+3m<0So, m<-2,n arbitrary.

2) The image is focused on the y-axis below the x-axis, so y<0 when x=0, y=n-4<0, so n<4,m arbitrary.

3) Because the image passes through the origin, y=0 when x=0So. n-4=0, i.e., n=4, m arbitrary.

The primary function y=(6+3m)x+(n-4), when m=-1, n=2, y=(6+3*-1)x+(2-4).

y=3x-2

The primary function intersects a with the x-axis and b with the y-axis

x=0,y=-2.

y=0,x=2/3

The area of the triangle AOB = 2 * 2 3 * 1 2 = 2 3

The secondary function y=(6+3m)x+n-4 is known

Decreases as x increases.

6+3m＜0

Hence m -22The intersection of the function image with the y-axis, below the x-axis.

n-4＜0n＜43.

When n-4=0

That is, when n = 4, the function image passes through the origin.

Solution: m -2 from 6+3m 0

From n-4 0 to get n 4

Bring the point (0,0) to y=(6+3m)x+(n-4)0=0+n-4, n=4, if m=1, n=-2, then y=9x-6, so x=0, y=-6 y=0, x=2 3, so the coordinates of the intersection of the image of the primary function and the two axes are (0,-6) and (2 3,0).

1.When m=-1, i.e., y=3x+(n-4).

y increases as x increases.

When m=1, i.e., y=9x+(n-4).

y increases as x increases.

2.When n=2, i.e., y=(6+3m)x-2

The intersection of the image of the primary function with the y-axis is below the x-axis.

At this time, 6+3m≠0

m≠-2m≠-2

n-4=0n=4m≠-2,n=4, the function image passes through the origin.

Why does y decrease as x increases?

6+3m<0

m<-2

Why is the intersection of the function's image and y-axis below x?

n-4<0

n<4 If the image of the function passes through the origin, what values should be taken for m and n?

n-4=0n=4 If m=1, n=-2, find the coordinates of the intersection of the image of the primary function and the two axes.

y=(6+3)x+(-2-4).

y=9x-6

Intersection with y-axis (0,-6).

Intersection with x-axis (2, 3,0).

This is a question about a primary function, and the problem is set up with 4 questions, as long as you master its graph and properties, the problem will be solved.

1. If y decreases with the increase of x, it means that (6+3m) 0 is m -22, and the intersection point of the image of the function and the y axis is below x, only (n-4).0, i.e., n 43, if the image of the function passes through the origin, (6+3m)≠0 i.e., m is not equal to -2 (n-4)=0, i.e., n=4

4、.If m=1, n=-2, y=9x-6 x=0, y=-6 y=0, x=2 3

1. When m and n are valued, y decreases with the increase of x?

To make y decrease with the increase of x, it is only related to k, i.e., k 0, and k = 6 + 3 m, i.e. 6 + 3 m 0, m -2

n is a real number. 2. When m and n are valued, the intersection point of the function image and the y-axis is below the x-axis?

Related to b, b 0, i.e. b = n-4

n-4 0n 4m is a real number.

3. When m and n are valued, the image of the function passes through the origin?

k≠0, b=0, i.e., n-4=0

k≠（6+3m）

That is: n=4, km≠-2, passing through the dot at this time;

(1) Solution: from the meaning of the question.

6+3m＜0

m -2m -2, n is any number, y decreases as x increases.

2) Solution: Derived from the question.

n-4＜0n＜4

6+3m≠0

m≠-2m≠-2,n 4, the intersection of the function image with the y-axis is below the x-axis.

3) Solution: from the meaning of the question.

n-4＝0n=4

6+3m≠0

m≠-2m≠-2, n=4, the image of the function passes through the origin.

(1) When 6+3m<0, y decreases with the increase of x, so m<-2, n are arbitrary values, and the condition is satisfied.

2) The point of the function image on the y-axis satisfies x=0, when x=0, y=(6+3m)*0+(n-4)<0, obtain: n<4;

And because the problem has determined that the function is a one-time function, 6+3m≠0, obtains: m≠-2 Conclusion: when n<4 and m≠-2, the condition is satisfied.

3) The function image satisfies x=y=0 through the origin, and the (0, 0) points are substituted into the function to obtain: 0=(6+3m)*0+(n-4), and obtains: n=4;

And because the problem has determined that the function is a one-time function, 6+3m≠0, obtains: m≠-2 Conclusion: when n<4 and m≠-2, the condition is satisfied.

1) When 2m+4 0, y increases with the increase of x, and the solution of inequality is 2m+4 0, and m -2 is obtained

2) When 3-n 0, the intersection of the function image and the y-axis is below the x-axis, and the inequality 3-n 0 is solved, and n 3; is obtainedWatch carefully.

3) When 2m+4≠0,3-n=0, the function image sketch is burned past the origin, then the key is imaginarym≠-2,n=3

The primary function y=(6+3m)x+(n-4), when m=-1, n=2, y=(6+3*-1)x+(2-4).

y=3x-2

The primary function intersects a with the x-axis and b with the y-axis

x=0,y=-2.

y=0,x=2/3

The area of the triangle AOB = 2 * 2 3 * 1 2 = 2 3

When m=-1, i.e., y=3x+(n-4).

y increases as x increases.

When m=1, i.e., y=9x+(n-4).

y increases as x increases.

When the silver is n=2, that is, y=(6+3m)x-2

The intersection of the image of the primary function with the y-axis is below the reed-key of the x-axis.

This is 6+3m≠0m≠-2

6+3m≠0

m≠-2n-4=0

n=4m≠-2, n=4, the function image passes through the origin.

m=-1.

y=3x+n-4

y increases as x increases.

m=-1. y=9x+n-4

y increases as x increases.

When n=2, is the intersection of the image of the primary function with the y-axis above or below the x-axis? m What other conditions need to be met.

Underneath the x, Nakoncha.

for (0.-2）

and m is not -2

What is the value of m,n respectively when the function image passes through the origin?

m is not -2and n=4. The function image passes through the origin.

I hope it will be helpful to the cave eggplant.

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If their images are parallel to each other.

6+3m=5m

n-4≠-3n

m=3n≠1 if their images intersect at the same point on the y-axis.

n-4=-3n

6+3m≠0

5m≠0n=1 m≠-2 and m≠0

The primary functions y=(6+3m)x+(n-4) and y=5mx-3n, their images are parallel to each other.

6+3m=5m

m=3, and their images intersect at the same point on the y-axis.

n-4=-3n

n=1,﹛m=3,n=1﹜

If parallel, then the slope is the same, there is 6+3m=5m, then m=3, if the two lines do not coincide, then there is (n-4) is not equal to (-3n), that is, n is not equal to 1; If it coincides, it is equal to 1

If it is the same point as the y-axis, then n-4=-3n, n=1, as for m, it is okay if it is not parallel.

1) After the origin point i.e. x=0, y=0, 0=m-4, m=42) subtract the function by 1 degree, and the slope <0,6+3m<0,m<-23) parallel, that is, the slope is equal, 6+3m=2, m=-4 34)4=m-4, m=8

5) That is, the slope is <0 and the intercept is <0

m<-2 and m<4 are obtained, and m<-2 is obtained

m=4 when the straight line crosses the origin When y increases with x6 3m>0 then m>-2 When m is parallel to y=2x 6 3m=2 then m=-4 3 When the image passes (0,4) into the equation m=8 When the image passes the two, three, four quadrants, 6 3m<0 m-4<0 gets m<-2