Two junior high school function questions, high score reward! Urgent!

1): Let the inverse proportional function be y1 and the primary function be y2

Since the two points A and B are on the equation, the coordinates of the two points are brought into the equation.

n=k1 2,-2=k1 -1 (simplified to 2=k1) So n=2 2=1

n=k2*2+b (because n=1, so 1=k2*2+b), 2=k2*(-1)+b, the two equations can be solved.

k2=1 b=-1

So y1=2 x y2=x-1

So the two-formula relation is y2=2 y1-1 (since 2 y1-1=x-1, you can check it.) so that the two equations are equal) and x is greater than 0

2): Exists.

Because APO is similar (I'm inconvenient to play the symbol, forgive me) AOB, and AO=OB=root number 5 (calculated with coordinates), APO is an isosceles triangle with AO as the base. So angular poa = angular oap. Because the two triangles are similar, the sides are proportional.

ap/ao=ao/ab

So ap*ab=ao 2 because ab=3 times the root number 2, ao=root number 5, so ap=5 6 multiplies the root number 2 , so let p (x,y) then (1-y) 2+(2-x) 2 open the root number = 5 6 and multiply the root number 2, and then follow the equation of y=x-1 (p on the straight line y2).

The solution is x=7 6 y=1 6

2: (1) Since the triangle AOE and the triangle BOF are right-angled triangles, to prove that the area of the 2 direct triangles is equal, it is only necessary to prove that ae*oa=bf*ob.

Because e(k 3,3) f(4,k 4) (note the relationship between e and f and a,b).

then ae*oa=3*k 3=k bf*ob=4*k 4=k

So ae*oa=bf*ob

The original formula is proven. 2) After E to make the ob perpendicular line, and intersect with g, then the area of the polygon AEFBO is rectangular AOGE + trapezoidal EGBF=3*K 3+(K 4+K 3)(4-K 3) 2=(156K-K 2) 72

So when k = 13 12, the area of the polygon aefbo is the largest =

I'll do it tomorrow.

2.(1) The hyperbolic expression is y=k x, and AC bc intersects point e and f point respectively, so the coordinates of point e and f can be found according to this, and the result is expressed by k: point e (k 3, 3) point f (4, k 4) area of aoe = (1 2) * 3 * (k 3).

Area of BOF = (1 2) * 4 * (K 4).

Therefore, the area is equal.

2) Based on the above results, the area of OEF can be expressed as the area of a rectangular OACB minus OAE, OBE and ECF.

That is: 12-(1 2)*3*(k3)-(1 2)*4*(k4)-(1 2)*(4-k3)*(3-k4).

Then s oef-s efc = above - (1 2) * (4-k 3) * (3-k 4).

Simplified: -(k 2 8)+2k-6 *k 2 is the square of k* proposes -(1 8), gets: -(1 8)*(k 2-16k+48) in parentheses can be written: (k-8) 2-16

This gives us the following equation: -(1 8)*(k-8) 2+2, i.e., the maximum value is 2, and k=8

Question 1: n=k1 2 -2=-k1 gives n=1 k1=2 and n=2k2+b -2=-k2+b gives n+2=3k2 because n=1 so k2=1

Substituting n=1 k2=1 into n=2k2+b gives b=-1y=2 x y=x-1

(1,-2), with the fixed-point formula y=a(x-1) 2-2

Substitute over the origin (0,0).

0=a(0-1)^2-2

A=2, so y=2(x-1) 2-2

Or write y=2x 2-4x

First, let the quadratic equation be: y=ax*x+bx+c

Crossing the origin (0,0).c=0。vertices (-b 2a, (b*b-4ac) 4a).

b=0,a=0,(rounded) a=2,b=-4y=2x*x-4x

First of all, after the origin, then c=0, you can set y=ax +bx, so the axis of symmetry -b 2a=1, vertex (0-4b) 4a=-2, the solution is a=1 2, b=-1

So the analytic formula is y=x 2 -x

Connect OC, cross AB at point M, cross point C to make a perpendicular line to the X axis, and cross point D.

It is easy to prove that the OC is perpendicular AB (equilateral triangle AOC, bisector AMB).

ab = root number (3 + 3 2) = 2 root number (3); So the angle oab = 30 degrees. Angular coa = 60 degrees.

Equilateral triangle COA.

od = 3/2.

cd = root number (3 2 - 3 2) 2 ) = (root number 3) *3 2

Bring in y = k x to get k = xy = od*cd = (root number 3) *9 4

2)"If the ABC is rotated 180 around the AC midpoint, the PCA is obtained", if rotated in the XY plane, the AC midpoint is the diagonal intersection of the parallelogram ABCP.

Angular OAC = 60 degrees; Angle cap = 90 degrees, so angle pax = 30 degrees.

ap = bc = root number 3, so the p point coordinates are ( 3 + 3 2 , ( root number 3 ) 2).

px * py = (9 2) * root number 3) 2 = (root number 3) * 9 4 = k

The xy = k condition is satisfied, so point p is on the hyperbola.

Thirty degrees in a right triangle corresponds to half of the hypotenuse ok? And then similar haha.

The analytic formula is: y=3t where the range of t is 0, and when t=2, the area of the quadrilateral abc is half the area of the triangle abc.

1) The height of the triangular BPC is 4-t

s(abpc)=s(abc)-s(bpc)=1/2*6*(4-(4-t))

3t and 0<=t<=4

2) From the formula for the area in 1), we can see that when t=2, the area of the quadrilateral abc is half the area of the triangle abc.

a(x1,y1) and b(x2,y2) are primary functions y=kx+2y1=kx1+b

y2=kx2+b

Subtract the two formulas to obtain:

y1-y2=k(x1-x2)

Since k>0, the description (y1-y2) is the same as (x1-x2).

Then t=(x1-x2)(y1-y2)>0c

Choose c because the image of a primary function is a straight line.

So the monotonicity increases.

A and b in the two points, the ordinate of the large abscissa must be large, and the small horizontal must be small, so it is easier to understand.

Therefore, (x1-x2) and (y1-y2) must have the same name, so the product is greater than 0 and c

First you have to draw a picture, and then you have to make a hypothesis.

a(x1,y1) and b(x2,y2) are primary functions y=kx+2y1=kx1+b

y2=kx2+b

Subtract the two formulas to obtain:

y1-y2=k(x1-x2)

Since k>0, the description (y1-y2) is the same as (x1-x2).

Then t=(x1-x2)(y1-y2)>0

So choose C

Choose c k=(y2-y1) (x2-x1)>0 and t=(x1-x2)(y1-y2) so the positive and negative of t and k are consistent.

t=(x1-x2) divided by (y1-y2) is the same as t=(x1-x2)(y1-y2), which does not affect the positivity and negativity of t, so substituting a and b into the equation and subtracting them gives (y1-y2) divided by x1-x2) = k>0, and c is selected

The angle is equal to the angle formed by the diagonal line and the long side, which is similar with three angles. From cos = 4 5, we can see that there is a trilateral ratio of a right triangle, so the diagonal line and the side conform to this ratio. The length of an edge is given, and the length of any line segment can now be calculated.

2.Your image opening should be larger, twice as large as y x.