Junior high school function problem 8T, junior high school math function problem

No change, x+minus. ——3，2)

2.Treat x as a constant and inversely solve y=(x-3) (x-1)3The distance to the x-axis is the absolute value of the ordinate (y), and the same ...... is true—1,-3）4.Y = 16-2x is obtained from 2x + y = 16.

Because it is a 3-sided shape, the waist is longer than equal to the bottom edge.

So the range of x is x>=16 3 (find the critical value by equilaterality) 5It can be set to y=k(4x-1), and x=3 and y=6 can be brought into the with:

6=k(12-1), so k=6 11.

y=24x/11-6/11

6.Let x=0 and y=0 respectively.

x(3,0); Y is given to (0,-2). Because the coefficient of x is positive. So y increases as x increases.

7.Bring in the expression:

2=-3k-1

a+1=ak-1

Solve the system of equations to obtain:

k=-1；a=-1

8.Because it is parallel, k = -5

y=-5x+b, bring (2,1) in, there is 1=-10+b, so b=11

xy-x=x-3

y(x-1)=x-3

y=(x-3)/(x-1)

6.(3,0) (0,-2) increases.

7.-3k-1=2

k=-1y=-x-1

a-1=a+1

a=-18.If the straight line y=kx+b is parallel to y=-5x+1, then k is equal and k=-52*-5+b=1

b=11

Dizzy, this kind of topic is also bt

2, y=(x-3) (x-1) (x does not equal 1)4, y=16-2x 45, y=24x 11-6 11

6. The intersection point with the x-axis (3,0), the intersection point with the y-axis (0,-2), and the y-axis increases with the increase of x.

7,k=-1,a=-1

8,k=-5,b=11

2 y = ( 3 - x )/ ( 5- x )4. y = 16-2x 9(y>0 , 2x > y)5.I don't understand.

6.Intersection at the origin, y increases as x increases.

a=1b=11

(1) Proof: The quadrilateral ABCD is rectangular.

ecd=∠ade=∠aod=90°

ado+∠edc=90°，∠oad+∠ado=90°∴∠oad=∠edc

aod∽△dce

2) Pass F as FH OC and pass OC to H, and AB to N, AB=OC=7, AO=BC=4, OD=5 AOD DCE

od:ce=ao:cd ce=,cd=2 The quadrilateral ADEF is a rectangle, de=af, dab+ baf=90° and oad+ dab=90°, oad= baf edc= baf

afn≌△dec

an=dc=2,fn=ec=, fh= The coordinates of the point f are (2, from a(0,4), f (2, (7,4), c=4 4=49a+7b+c

Solution: a= b= c=4

Solution: Point f is on the parabola sought in .

The reason is that from (2), we can see that the expression of the parabola is: y=squared + when d(k,0), then dc=7-k, in the same way, from aod dce and afn dec, we get: f(7-k, 4/4k(7-k)) substituting x=7 k, y 4/k(7-k) so the point f is on the parabola found in .

Proof (1) Let the equation of the ellipse be ....(1 point).

By eliminating y, we get (1+a2)x2-2a2bx+a2(b2-1)=0 ....(2 points).

Since the straight line l is tangent to the ellipse, it =(-2a2b)2-4a2(1+a2) (b2-1)=0, which is simplified to b2-a2=1 ....4 points).

Solution: (2) A(A+1,0), B(A+1,1), C(0,1), then the midpoint of ob is .(5 points).

Because l divides the rectangular oABC into two parts of equal area, l passes the point, i.e., f(x), i.e., 2b-a=2 ....6 points) is solved by , so the equation for the straight line l is ....(8 points).

Solution: (3) is known from (2).

Since the circle m is tangent to the line segment ea, the equation can be (x-x0)2+(y-r)2=r2(r 0)....9 points).

Because the circle m is in the rectangle and its interior, ....(10 minutes) The circle m is tangent to l, and the circle m is above l, so that ....(12 points).

Substitution is obtained ....(13 minutes).

So when the area of the circle m is the largest, at this time,

Therefore, the equation for the maximum area of the circle m is ....(15 points).

(1) The angle ADO is congruent with the angle EDC, and the angle EDC is congruent with the angle DEC, so the angle ADO = the angle DEC, and the two triangles are both right-angled triangles, so the similarity theorem is obtained: AOD DCE

2) "1" by the question, the coordinates of point b are (7, 4) obviously the two points a and b are symmetrical on the axis of parabolic symmetry, so the abscissa of the parabolic vertex is 5 3, according to the nature of the parabola, an equation can be listed, and then the coordinates of points a and b are brought into the parabolic equation respectively, and two equations are obtained, and the simultaneous equation obtains a, b, c values. From (1) question: aod dce, the coordinates of point e are (7,, and the coordinates of a and d are known, and the four points form a parallelogram, and the coordinates of point f are simply obtained.

2》Take the point d known as (,0), find the point f, if you find that f is not on the parabola, it is a counterexample. If on the parabola, the f coordinate is obtained by using the point d coordinates, and the parabolic equation is substituted, and the equation is satisfied, indicating that d is any point, and f is still on the parabola.

3) Let d be (a,0), the parabola is y=ax2+bx+c, find point e according to point c, and then obtain point f from a, e, d, after various calculations, simplify, if m, n can be used to represent the parabola, then it exists, if the parabola equation has a, it means that it does not exist.

Let y=ax bx c

The intersection point with the y-axis is (0,-3 Zhihuaxun2).

So c=-3 2

Method 1: Just bring in the coordinates of the other two points to solve a and b.

Method 2 assumes x1 and x2 are the two roots of the equation ax bx c=0.

x1＋x2=-b/a=-1＋(-3)=-4x1*x2=c/a=-1*(-3)=3

Then a=-2 takes this 9, and b=8 9

y=-2x²/9＋8x-2/3

1) The parabola intersects with the x-axis at (-1,0)(-3,0), and the parabola is y=a(x+1)(x+3).

Let x=0, then y=3a, that is, the parabola and the y-axis intersect at (0,3a)3a=-3 2,a=-1 quietly hall spike 2

y=(-1 2)(x+1)(x+3)=(1 2)(x +4x+3)=-x volt 2-2x-3 2

The parabolic analytic formula is y=-x 2-2x-3 22)y=-x 2-2x-3 2=-(1 2)(x +4x+4)+(1 2) 4-3 2=-(1 2)(x+2) +1 2

The parabolic opening is downward (-1 2<0), the axis of symmetry is x=-2, and the vertex is (-2, 1 2).