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A big math problem
(1) Solve equation (x-3)(x+ 1)=0, and find x=3 or-1. According to the parabola y=(x-3)(x+ 1) and the X axis intersect at two points A and B (point A is at point B). Write the formula y=(x-3)(x+ 1) as the vertex of y=x2-2x-3=(x- 1)2-4, and then you can determine the coordinates of vertex d.

(2)① According to the parabola y=(x-3)(x+ 1), calculate the coordinates of point C and point E, connect BC, let point C be CH⊥DE in H, and get CD= by Pythagorean theorem.

2

,CB=3

2

It is proved that △BCD is a right triangle. PC and DC extend respectively and intersect with X axis at points Q and R. According to the similarity of two triangles with equal corresponding angles, it is proved that △BCD∽△QOC, then

commander

OQ

=

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Civil band

=

1

three

The coordinates of q (-9,0) are obtained, and the analytical formula of straight line CQ is y=-

1

three

X-3, the analytical formula of straight line BD is y=2x-6, and the equations are solved.

y=-

1

three

x-3

y=2x-6

, you can find the coordinates of point p;

(2) It is discussed in two cases: (1) When point M is on the right side of the symmetry axis, if point N is on the ray CD, such as alternating graph 1, the intersection point of MN's Y axis at point F is extended, so that the intersection point M is the MG⊥y axis of point G. First, it is proved that △MCN∽△DBE, and then MN=2CN is obtained. The ratio of the corresponding side of similar triangles. Let CN = If n points are on the ray DC, the coordinates of m points can also be obtained; (2) When point M is on the left side of the symmetry axis, because ∠ BDE < 45, we get ∠ CMN < 45. According to the complementarity of two acute angles of a right triangle, we get ∠ MCN > 45, and any point K on the left side of the parabola has ∠ KCN < 45, so point M.

Solution: (1)∵ Parabola y=(x-3)(x+ 1) intersects the X axis at two points A and B (point A is on the left of point B).

∴ When y=0, (x-3)(x+ 1)=0,

X=3 or-1,

The coordinate of point B is (3,0).

∫y =(x-3)(x+ 1)= x2-2x-3 =(x- 1)2-4,

∴ The coordinate of vertex D is (1,-4);

(2)① As shown on the right.

∵ parabola y=(x-3)(x+ 1)=x2-2x-3 intersects the y axis at point C,

The coordinate of point ∴C is (0, -3).

The symmetry axis is a straight line x= 1,

∴ The coordinate of point E is (1, 0).

Connect BC, and point C is CH⊥DE in H, so the coordinate of point H is (1, -3).

∴CH=DH= 1,

∴∠CDH=∠BCO=∠BCH=45,

∴CD=

2

,CB=3

2

, △BCD is a right triangle.

Extend PC and DC respectively, and intersect with X axis at points Q, R. 。

∠∠BDE =∠DCP =∠QCR,

∠CDB=∠CDE+∠BDE=45 +∠DCP,

∠QCO=∠RCO+∠QCR=45 +∠DCP,

∴∠CDB=∠QCO,

∴△BCD∽△QOC,

commander

OQ

=

laser record

Civil band

=

1

three

∴OQ=3OC=9, which means q (-9,0).

The analytical formula of straight line CQ is y=-

1

three

x-3,

The analytical formula of straight line BD is y = 2x-6.

Through the equation

y=-

1

three

x-3

y=2x-6

, solution

x=

nine

seven

y=-

24

seven

.

∴ The coordinates of point P are (

nine

seven

,-

24

seven

);

② (Ⅰ) When point M is on the right side of the symmetry axis.

If the point n is on the ray CD, as shown in the standby diagram 1, extend the Y axis of the intersection point of MN to the point f, and make the intersection point m the MG⊥y axis to the point g. 。

∠∠CMN =∠BDE,∠CNM=∠BED=90,

∴△MCN∽△DBE,

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All merchant ships

=

exist

Delaware

=

1

2

∴MN=2CN.

Let CN=a, then Mn = 2a.

∠∠CDE =∠DCF = 45,

∴△CNF and △MGF are isosceles right triangles,

∴NF=CN=a,CF=

2

One,

∴MF=MN+NF=3a,

∴MG=FG=

three

2

2

One,

∴CG=FG-FC=

2

2

One,

∴M(

three

2

2

a,-3+

2

2

a)。

Substitute parabola y=(x-3)(x+ 1) to get a=

seven

2

nine

∴M(

seven

three

,-

20

nine

);

If the point n is on the ray DC, as shown in Figure 2, MN intersects the Y axis at point F, and the passing point M is the MG⊥y axis at point G. 。

∠∠CMN =∠BDE,∠CNM=∠BED=90,

∴△MCN∽△DBE,

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All merchant ships

=

exist

Delaware

=

1

2

∴MN=2CN.

Let CN=a, then Mn = 2a.

∫∠CDE = 45,

∴△CNF and △MGF are isosceles right triangles,

∴NF=CN=a,CF=

2

One,

∴MF=MN-NF=a,

∴MG=FG=

2

2

One,

∴CG=FG+FC=

three

2

2

One,

∴M(

2

2

a,-3+

three

2

2

a)。

Substituting the parabola y=(x-3)(x+ 1), we get a=5.

2

∴m(5, 12);

(ii) When the point m is located on the left side of the symmetry axis.

∠∠CMN =∠BDE < 45,

∴∠MCN>45,

And any point k on the left side of the parabola has ∠ kcn < 45,

Point m does not exist.

To sum up, the coordinates of point M are (

seven

three

,-

20

nine

) or (5, 12)