It is known that the parabola y X 6X m has two intersections with the X axis A B, with AB as the dia

1) Point C is the midpoint of a,b, because a,b is the parabola y=x -6x+m has two intersections with the x-axis, so their midpoint is the intersection of the symmetry axis of the parabola and the x-axis, so the coordinates of point c (0,-b 2a) are (0,3).

2) If the parabolic vertex is also on o, then the distance from the vertex to the x-axis is the radius of the circle, i.e., (b -4ac) 2a=|xa-xb|

Write the algebraic formula of m with the root finding formula and see if it exists.

1) What is point C?

2) Because the vertex is on o.

So x=0 y=0

So m=0

y=x^2-6x+m

The axis of symmetry is x=3

c is on the x-axis, and c is on the axis of symmetry, so the coordinates of c are (3,0) parabolic vertices.

3,9-m)

According to Veda's theorem.

x1+x2=6

x1x2=m

x1-x2)^2=6^2-4m

x1-x2 is the diameter.

If you want the parabolic vertices to be on o as well, the ordinates of the vertices must be equal to the radius.

Yes, there is. 6^2-4m/4=(9-m)^2

m^2-17m+73=0

But b 2-4ac=17 2-73*4<0 so m does not exist.

The abscissa of the vertex is -1, and the distance between the two points is 10a(-6,0),b(4,0).

s△abc=15=10*|yc|/2

yc|=3 intersects the negative half axis of the y-axis at point c

c（0,-3）

Substituting a, b, and c into the parabola y=ax2+bx+c results in a=1 8, b=1 4, c=-3

y=x²/8+x/4-3

3) (Drawing) (if similar, then ABP ACB can be found during inspection) Find C (0,3) on the Y-axis positive semi-axis

Connect AC and extend AC to cross the parabola to P, and connect Pb then PAB= BAC

Easy to get ac: y=x 2+3

Synoptic {y=x 8+x 4-3

y=x/2+3

Solution: x1=-6(point a) x2=8

y2=7 p（8,7)

ap=√245=7√5

ap/ab≠ab/ac

This conjecture is not missing.

The following is an attempt to construct an abp BCA).

Easy to get: bc:y=3x 4-3

Cross point A to make an ap bc cross parabola at p

pab=∠abc

Easy to get ap: y=3x 4+9 2

Synoptic {y=x 8+x 4-3

y=3x/4+9/2

x1=-6(point a) x2=10

y2=12p``（10,12）

P A pants ab = ab bc = 2 1

abp``∽bca

According to the symmetry, p 12,12).

p``（10,12）p```12,12）

Be what you want.

y=x -2x-3 is coupled with y=0 to obtain: a(-1, 0) b(3,0).

Let c(2,y0) be substituted with y=x -2x-3:y0=4-4-3=-3 c(2, -3).

Let g(x0,-3) be substituted by y=x -2x-3:-3=x0 2-2x0-3 x0=0 x0=2 (rounded) g(0, -3).

Equation for a straight line over ag: y=-3x-3

The linear equation of cf: y=-3(x-2)-3 i.e. y=-3x+3y=-3x+3 and y=0 are combined to obtain the coordinates of the point f: f(1,0).

(1) Find the linear relation of baic(0,-3),b(3,du0) and zhi.

Through A(-1,0) as the parallel line of BC, the intersection of the DAO axis X=1 in D(1,M), (in fact, the straight line BC is translated to the left by 4 units through A) to obtain the relation of the straight line AD, so as to obtain D(1,2) Because the triangle BCD and the triangle ACB are equal in height and bottom, so the area is equal.

Since the line x=1 (i.e., the axis of symmetry) intersects the line cb at (weight 1,-2), then we get the symmetry point (1,-6) of d(1,2) with respect to (1,-2), so d(1,2) or (1,-6).

d[1,-6] or [1,2].

To three-thirds plus the root number 17 is greater than y and greater than three-thirds minus the root number 17

Solution (1) The parabola y=x squared - 1 intersects with the x-axis with a, b two points, and the y-axis intersects with the point c so that x 0 then y 1 then c(0, 1).

Let y 0 x 1 x 1

Then a(1,0) b( 1,0) or a( 1,0) b(1,0)2) let the equation for the slash ap be y kx b The intersection of the slash ap and the y axis is eap cb, then point c and e are symmetrical along the x-axis, i.e., e(1,0)a( 1 0) y kx b x 1

y＝x²－1

Get x 2 x 1 (point a) y 3 i.e. p(2,3)ap 3 2

sacbp＝½（ap＋bc）×ac＝ ⅔2 ＋½

1. c coordinates (0,-3), then oc=3

1/2ab×oc=6

ab=12/oc=4

oa=1, then ob=4-1=3

b coordinates (3,1).

The coordinates of a and b are substituted into y=ax +bx-3

a-b-3=0

9a+3b-3=0

a=1b=-2

Parabolic analytical: y=x -2x-3

2. BCD is a right triangle.

pc=pd pb=pc=pd

Let the p coordinates (m, n).

[(0-m)²+3-n)²]=√[(3-m)²+0-n)²]

m²+(3+n)²=(3-m)²+n²

m=-n by: y=x -2x-3

Get: n=m -2m-3

m=m²-2m-3=0

m²-m-3=0

m=(1±√13)/2

m>0

m=(1+√13)/2

Then n=-(1+ 13) 2