Mathematics in the second year of junior high school reviews the function materials once

1.The image of a copy function is the scripture.

A straight line that crosses the origin, and the straight line passes the fourth quadrant and points (2,-3a) and points (a,-6), find the analytic formula of this function.

2 (1) When b 0, which quadrants does the image of the function y=x+b pass through (2) When b 0, the image of the function y=-x+b passes through which quadrants (3) When k 0, the image of the function y=kx+1 passes through which quadrants (4) When k 0, the image of the function y=kx+1 passes through which quadrants 1Solution: Let y=kx, substitute (2,-3a) and point (a,-6), and solve it, so that a=2 or -2 (rounded), so the analytic formula is.

y=-3x2.(1) One, two, three.

2) One, three, four.

3) One, two, three.

4) One-two-four.

Hope is not made by the child's family.

1. Solution: For straight lines, when x=0, y=-1 so c(-1,0), oc=1 ob=

b(1 2,0) slope k=1

2. The ordinate of point A is the height of AOB and the bottom of OB=

s=1/2*

3. When 2x-1=1 is as above, that is, x=1

At this point a(1,1).

Exist. p(2,0) or (1,0) or (- 2,0).

Once a function, the number of x times is 1

And the x-factor is not equal to 0

So the first item is 0 or 1 time.

m²-3=0

m²=3m=±√3

m²-3=1

m²=4m=±2

At this time, the x-coefficient = m-3 + 1 = , -2≠0

So m=-2

So there are three.

y=x-2+√3

y=x-2-√3

y=-4x+1

When m=-2, y is a primary function of x! y=-4x+1

Because mx+2>0, i.e., the solution set of mx>=-2 is less than 1, m<0 x<-2 m

2/m=1 m=-2

y=-2x+2

∵mx+2>0

mx>-2

x<1 again, [sign changed].

m<0,x<-2/m

m=-2 y=mx+2 is y=-2x+2

When y=0, x=1

The coordinates of the intersection of the line y=mx+2 and the x-axis are (1,0).