When it s time to study, ask a few questions about inverse proportional functions in mathematics

1 If the image of an inverse proportional function passes through the point p(-1,4), its functional relation.

Solution: k=(-1) 4=-4

Its functional relation is y=-4 x

2. If the image of the proportional function and the image of the inverse proportional function have an intersection coordinate of (-1, -2), then what is the coordinate of the other intersection.

Proportional function: y=2x

Inverse proportional function: y=2 x

Simultaneous equations, solved: x[1]=1,y[1]=2;x[2]=-1，y[2]=-2

The coordinates of the other intersection point are (1,2).

3 The value of k is known that the point p(a,b) is on an image with an inverse proportional function y=2 x, and if the point p is symmetrical with respect to the y axis is on an image with an inverse proportional function y=k x.

Solution: ab=2

Point p is symmetrical with respect to the y-axis (-a, b).

k=（-a）b=-ab=-2

1.If you don't pass through the origin, then the other point you pass through is set to (a,b) y-b=ab (x-a) 2

Solution: Because P1OA is a right-angled triangle with an unfortunate waist, and the vertical X-axis of P1B is at the point B, then the point B bisects OA1, and P1B=OA1 2

Set the point A1 coordinates (x1,0).

So oa1=x1 p1b=x1 2

Because the point p1 is on an image with an inverse proportional function y=4 x.

So the area of the triangle op1b is 2, (s=1 2*xy=4 2=2) then s p1oa=4oa1·p1b 2=x1*x1 4=4 is solved to x1=4

In the same way, let a2 ( a1a2=x2-4

Therefore, p2(x2+4) 2 , (x2-4) 2 pass p2 to make the perpendicular line of the x-axis and hang down the foot for d

then s p2a1d=2

The uproar was sent to ((x2+4) 2 · (x2-4） 2） 2=2x2 2=32

Solve the equation as: x=4 2 or x2=-4 2 (rounded) so a2(4 2,0).

Attachment: x 2 = 4 (x + 1).

This is a quadratic equation of one element, and the root can be found using Veda's theorem:

x=4 4 root number 2 2

x1 = 2 + root number 2 x2 = 2 - root number 2

The correct solution downstairs will not waste the tongue.

Because the image of the primary function y=-x+m and y=(- 2) 3x+2 all pass the point c, and because d is the perpendicular y-axis of dc to c, so the point c is on the y-axis, so the point c is the intersection of y=(- 2) 3x+2 and the y-axis, so c(0,2), from which m=2,a(2,0)Let d(x0,2), trapezoidal area = 1 2 (cd + ae) * de=4, 1 2 (x0 absolute value + x0 absolute value + 2) * 2 = 4, 2 * x0 absolute value + 2 = absolute value = 1, according to the image x0 = -1, so according to d (x0, 2), k = x0 * 2 = -2Ha ha.

Drawing. Because the image of the primary function y=-x+m and y=(- 2) 3x+2 all pass the point c, and because d is the perpendicular y-axis of dc to c, so the point c is on the y-axis, so the point c is the intersection of y=(- 2) 3x+2 and the y-axis, so c(0,2), from which m=2,a(2,0)Let d(x0,2), trapezoidal area = 1 2 (cd + ae) * de=4, 1 2 (x0 absolute value + x0 absolute value + 2) * 2 = 4, 2 * x0 absolute value + 2 = absolute value = 1, according to the image x0 = -1, so according to d (x0, 2), k = x0 * 2 = -2

It can be solved by passing two functions through two points, a and b.

m=-8，n=2，b=-2，k=-1；

So the equation for the straight line ab is y-2=((2+4) (-4-2))(x+4)=y-2=-x-4

That is, x+y+2=0;

1.So the coordinates of its intersection with the x-axis are c(-2,0),a(-4,2),b(2,-4).

According to their graph, we can see the intersection point of the AB line segment and the y-axis d(0,-2), so the area of AOB = the area of ACO + the area of COD + the area of BOD = (1 2)*2*2 + (1 2)*2*2 + (1 2)*2*2=3*2=6

2.From the above, we can see that the equation kx+b-m x=0

That is, -x-2+8 x=0

->x^2-2x+8=0

->x^2+2x-8=0

->x+4)(x-2)=0

->x1=-4,x2=2

3.From 2, the solution of the inequality is -4

This is -8 x=-x+2

The solution is x1=4 and x2=-2

Substituting these two points into the a(4,-2)b(-2,4) primary function and the x,y axis each have an intersection point, that is, (0,2) and (2,0) points b are I-2i=2 from the y-axis, and 4 from the x-axis

The distance between the point a and the x-axis is i-2i=2

s△aob=6

o The straight-line distance is 2, which is high.

The simultaneous equation yields the coordinates ab(4, 2), and the distance (2,4) is 6 root number 2

The area is 6 roots, 2 * 2 * roots, and 2 roots

The two equations can be combined to form a(4,-2)b(-2,4) Then ab = 6 times the distance from the root number 2 o to ab = the area of the root number 2 abc = 6 times the root number 2 times the root number 2 times the half is equal to 6

s△aoc=s△aob-s△boc

Need to know the area of aob, boc. AOB is easier to find, because the coordinates of point A are known, and S AOB=

The main thing is to find the area of the boc.

The perpendicular line of the x-axis is made by d, and the perpendicular foot is e

The product of the horizontal and vertical coordinates of the two points on the hyperbola is constant, S boc=s doe and d is the midpoint of OA, and the coordinates are (-3, 2).

s△doe=

s△aoc=s△aob-s△boc=s△aob-s△doe=12-3=9.

Key points] The product (absolute value) of the horizontal and vertical coordinates of the points on the hyperbola is equal and thus equal to the area of the triangle enclosed by the coordinates.

I can only give you ideas, but I can't give you answers, otherwise you would never do this kind of question:

Knowing the coordinates of point A, you also know the area of the triangle OAB and the coordinates of point D.

You can find out by the d-coordinates and the properties of the hyperbola, which can be seen as (y-a x=b).

Point C coordinates (-6, X) and then know the area of the OCB, and then we can find the area of the AOC.

The coordinates of point A are (-6,4), then OA=2 13 [is 2 root number 13].

d is the midpoint of oa, then the coordinates of d are (-3, 13), the expression of the hyperbola can be found as y=(-3 13) x, the abscissa of point c is the same as a, both are -6, then the ordinate of c is y=(-3 13) (-6)=(13) 2, so ac=oa oc=4 (13) 2, the area of aoc is: ac ob= [4 (13) 2] 6=12 (3 13) 2.

1) When combustion, it is proportional, let y=k1x, from the meaning of the question, we can know that when x=8, y=6 so 6=8k1, the solution is k1=3 4

So y=3 4x

After combustion, it is inversely proportional, let y=k2 x

From the meaning of the title, we can see that when x=8, y=6

So 6=k2 8, and the solution is k2=48

So y=48 x.

It is divided into two phases: at burning: 3=x*(3 4), x=4 from 4 minutes to 3 mg, after combustion: 3=48 x, x=16 to 16 minutes to start below 3 mg

16-4>10 Disinfection is effective.

1) When burning, y=kx 6=8k=6 8 y=x*(6 8)After combustion: y=k1 x 6=k1 8, k1=48 y=48 x(x>=8).

2) Divided into two phases: during combustion: 3=x*(6 8), x=4 from 4 minutes to 3mg, after combustion: 3=48 x, x=16 to 16 minutes to start below 3mg

16-4>10 Disinfection is effective.

Solution: P1(1,K),P2(2,K 2),pn(n,k Wang Yuanxing n).

s1=k-k/2,s2=k/2-k/3,s3=k/3-k/4,..sn=k/n-k/(n+1)

s1+s2+s3=k-k/2+k/2-k/3+k/3-k/4=k-k/4

s1+s2+s3+s4+…+sn=k-k sleepy 2+k 2-k 3+k 3-k 4+.k cavity tease n-k (n+1)=k-k (n+1)

Proposition n: The point (n, n 2) is the intersection of the straight line y=nx and the hyperbolic y=n 3 x (n is an integer);

When x=n, Minson.

The value of the linear liquid lift function is y1=n 2

The hyperbolic function takes the value y2=n 2

y1=y2 and thus the point (n, n 2) is an intersection of the straight line y=nx and the hyperbola y=n 3 x (n is an integer).

The point (n,n2) is the intersection of the line y=nx and the hyperbolic line y=n3/x.

n 2 is the square of n and n 3 is the third power of n.

Proof: x=n, as y=nx=n 2, y=n 3 x=n 2 is the intersection of the straight line y=nx and the hyperbolic y=x n3/n.