2o 1q calendar

2

+bx+c passes through A (1, 0) and B (5, 0), and the equation of functional symmetry axis can be obtained. Because the ordinate of the lowest point of the function is -4, the coordinates of the parabola vertex can be obtained, and the parabola vertex can be set and solved by the undetermined coefficient method.

(2) making auxiliary lines and crossing points O.

1

Zuoao

1

P⊥x axis is at point P, connecting point O.

1

A, the structure has an angle ∠AO

1

P equals ∠ACB, find the corresponding side length, and define the solution according to the tangent function;

(3)① According to the conclusion in (2), the straight line CF

1

CF can be obtained by c (0 0,5) and o (3 3,3).

1

The analytic formula of f is easy to get.

1

Coordinates of;

② According to the symmetry, another point f on the X axis can be found from ①.

2

(-

,0).

③④△OCF

three

With △DEC, the OF is obtained according to the properties of similar triangles.

three

Solution: Solution: (1) Because the parabola y=ax.

2

+bx+c passes through a (1, 0) and b (5, 0).

So the symmetry axis of quadratic function is x=

=3,

Because the ordinate of its lowest point is -4,

So the vertex coordinates are (3, -4).

Let the analytical formula become

y=a(x-3)

2

-4;

Substitute A( 1, 0) into the analytical formula to get a( 1-3).

2

-4=0,

That is a= 1,

The analytical formula is y=(x-3)

2

-4,

The analytic function of transforming a parabola into a general formula is: y = x.

2

-6x+5； (3 points for this small question)

(2) Tan ∠ACB=

.

Beyond o point

1

Zuoao

1

P⊥x axis is at point P, connecting point O.

1

One,

From the symmetry of parabola and circle, O

1

The straight line where p is located is the symmetry axis of parabola.

Therefore, OP=3, AP=OP-OA=2, CD=AB: CD=AB=4.

Beyond o point

1

Zuoao

1

The Q⊥y axis is located in q, which is obtained by the vertical diameter theorem: DQ=CQ=2, O.

1

P=OQ=OC-CQ=3，

So tan∠ACB=tan∠AO.

1

P=

=

; (3 points for this small question)

(3)① Let CE intersect with the X axis at F.

1

Because de ∥AB, ∠DEC=∠OFC, ∠COF.

1

=∠CDE，

So ∞△OCF

1

∩△DCE。

Straight CF

1

Exceeding c (0 0,5), o (3 3,3),

The analytical formula is y=-

x+5；

When y=0, x=

, so f

1

(

,0).

②△OCF

2

Similar to △DCE, according to symmetry, another point f on the X axis can be found from ①.

2

(-

,0).

③△OCF

three

Similar to △DEC,

=

that is

=

Smooth both sides.

three

=

.

There is a point f whose coordinates are:

F

1

(

，0)、F

2

(

，0)、F

three

(

，0)、F

four

(

,0).