Junior 2 Mathematics Functions Please answer in detail, thank you 22 18 0 48

Let y=kx+b

Substitute x=3, y=2, x=2, y=-3.

Large 3k+b=2

Include. No. 2k+b=-3

3k+b-2k-b=5

k=5 substitute k=5

10+b=-3

The function of b=-13 is: y=5x-13

Put x=-3 into the era.

y=-28 puts y=2 into the era.

5x-13=2

5x=15x=5

Because y is greater than 1

So 5x-13 is greater than 1

5x is greater than 14

x is greater than 14 5

Let the function be y=kx+b

So there is -2=3k+b

3=2k+b

k=1 b=-5

So the function is y=x-5

x=-3. y=- 8

x=7y>1. x>6

y=(60-40)x

Later, I will use my own brains

1.(1) Solution: Let the analytic formula of the function be y=kx+b to obtain a system of binary linear equations: [-2=3k+b] Solution: k=-5; b=13-3=2k+b】

So the analytic formula for this primary function is y=-5x+132), and when x=-3, y=28

3), when y=2, the independent variable x=

4) The independent variable x is less than and greater than 0

2.(1), the analytical formula of y about x is y=60x-40x2), when there are 1000 shirts, y=20000 yuan.

When there are 1,100 shirts, y=22,000 yuan.

Therefore, the gross profit is about 20,000 to 22,000 yuan.

1 (1) Let y=kx+b, substitute x=3, y=-2, x=2, y=-3 into k=1, b=-5∴y=x-5.

2) Substituting x=-3 to get y=-3-5=-8

3) Substituting y=2 to get x=7

4) y 1, then x-5 1, so x 6

Equations can be written.

y=a(x-k)^2-m

and known synthesis.

k=-2a b

m=-c+b^2/4a

Or - because a plus or minus is not known.

i.e. k = -2a b = 1

Namely. A and B different signs.

But c is not sure of the symbol.

So @@@1错.

For equation 3, it is obtained by bringing x = 2 into y.

And when x=2 and x=0 they have the same rent.

That is, 4a+2b+c=0*a+0*b+c=c

Fu Xixin Zheng is not sure.

None of the three is right!

Brief Banquet Method:

You can consider c to be plus or minus infinity.

Not to rule it out!

It is known that y-2 is proportional to x, and when x=3 and y=1, then the functional relationship between y and x is

y=1 when y-2=kx has x=3

Get k=-1 3

So y=(-1 3)x+2

Let y-2=kx,3,1) be substituted, 1-2=3k, k=-1 3, then the functional relationship between y and x is y=-x 3+2

Let y-2 and x function i.e., y-2=kx

By taking x=3 y=1 into the condition, we get 1-2=3kk=-1 3

Then bring the k value back to the original analytic formula to get y-2=-1 3x to sort out, and finally get y=-1 3x+2

Column equations: 0=2k+b

3=0·k+b

Solution: b = 3, k = -3 2

1) Expression: y=-3 2·x+3

2) The intersection point with the x-axis (2,0), the intersection point between the group letter and the y-axis (0,3) If you approve of mine, please return to your question page in time, click mine, and then click "Evaluation" in the upper right corner, and then you can choose or open "Satisfied, the problem has been perfectly solved".

I'm an expert in knowing, and if you have any questions, you can also ask me questions here:

1.The solution k 0 is to the right of the origin, so it passes through quadrant b 0

1-m＞0m＜1

2.Solution: y-2=kx

4-2=2k

k=1∴y-2=x

2m+7-2=mm=5

1。When x=0, i.e., y=1-m<0, then if the intersection is to the right of the origin, m>1.

2。From the meaning of the question, let y-2=kx, then 4-2=k*2, and k=1;

y=x+2;Then 2m+7=m+2, and m=-5 is obtained

a-b=1

b = a-1 so y = ax + a-1

y+1=a(x+1)

Then y+1=0 and x+1=, the equation must hold.

So y=-1, x=-1

So it's (-1, -1).

b=a-1 y=ax+a-1 x=-1 y=-a+a-1 why a is equal to -1 so it must pass -1, -1

When x=-1, y=-a+b, because a-b=1, so-(a-b)=-a+b=-1, so y=-1

So pass the dot (-1, -1).

Because a-b=1, y=ax+a-1, ie.

ax+a-1-y=0, factorized to a(x+1)-(y+1)=0, because indefinite function, a is a variable, and there is always (-1, -1) satisfying the equation.

Let's assume that k is less than 0, and the area of the triangle enclosed by the straight line y=kx+2 and the line and the two coordinate axes is 2", we know that the coordinates of the intersection of the line and the x-coordinate axis are (-2, 0), so there is kx+2=0, and the solution is k=1, and the coordinates of the three vertices of the triangle are (0,0), (2,0), and (0,2); Assuming that k is greater than 0, k=-1 can be solved in the same way, and the coordinates of the three vertices of the triangle are (0,0), (2,0), (0,2);

Since you can't enter the symbols, you can only look at it this way.

The area of the triangle enclosed by the line and the coordinate axis is 2, and it can be seen that 1 2 * (intercept of the x-axis) * (intercept of the y-axis) = 2 (1).

From y=kx+2, it can be seen that when y=0, the intercept of the x-axis = -2 k; When x=0, the intercept of the y-axis = 2

Then, substituting the intercept of the x-axis = -2 k and the intercept of the y-axis = 2 into equation (1) gives k = -1

So the three vertices of the triangle are (2,0), (0,2)(0,0).

In a straight line, the b value is 2, so the intersection of the function image and the y-axis is (0,2), the triangle is considered to be 2 in height, and the base length is the distance from the intersection of the function and the x-axis to the origin according to the area, so the base length is 2

Therefore, there is (2,0) or (-2,0) at the intersection point with the x-axis

The coordinates of the three vertices of the triangle are (2,0), (0,2), (0,0) or (-2,0), (0,2), (0,0).

Substitute x=2 and y=0.

2k+2=0，k=-1

Substitute x=2 and y=0.

2k+2=0，k=1

When x=0, y=2, the intersection of the line with the y-axis is (0,2) and when y=0, x=-k 2

Because s triangle = 2, so 1 2*2*|-k/2|=2, so when k=4 or -4k=4, vertices (0,2), (2,0), (0,0)k=-4, time, (0,2), (2,0), (0,0).