11 26 min Parabola y ax bx c a 0 passes through the point A 3,0 B 2, 3 with x 1 as the axis of sym

Take x=1 as the axis of symmetry.

y=a(x-1)^2+k

Substitute ab.

0=4a+k

3=a+ka=1,k=-4

y=(x-1)^2-4

So y=x 2-2x-3

y=0,x^2-2x-3=(x-3)(x+1)=0,x=3,x=-1x=0,y=-3

Therefore, the intersection points with the x-axis (3,0), (1,0), and the y-axis intersection (0,-3)x=1 are the symmetry axes, and the vertices (1,-4).

Just trace it.

Pa=Pb, so P is on the perpendicular bisector of Ab.

Obviously, the perpendicular bisector of ab and x=1 have an intersection.

So p exists.

ab slope = (0 + 3) (3-2) = 3

So the slope of the perpendicular bisector of ab is -1 3

AB midpoint (5 2, -3 2).

So the perpendicular bisector of ab is y+3 2=-1 3*(x-5 2)x=1,y+3 2=1 2,y=-1

So p(1,-1).

The equation 9a+3b+c=0 is obtained from a(3,0), 4a+2b+c+3=0 is obtained from b(2,-3), b=-2a is obtained by x=1 as the axis of symmetry, and the solution of three unknowns is a=1, b=-2, c=-3

Draw the picture yourself.

Let p(1,y>=-4), a(3,0), b(2,-3), and the distance between the two points be obtained.

Let's assume that there is a point p that equalizes the distance between the two points.

1-3)2+(y-0)2=(1-2)2+(y+3)22 are all squared.

Get y=-1, meet the condition of y>=4, so there is a p point, the coordinates are (1, -1) I have been transformed for a long time, and the points will be given to me.

y=x^2-2x-3

You can't do this with the boss image, you just go and take a few points and connect them with smooth lines.

There is a p-coordinate (1,-1).

Let y=a(x--1) 2+m (x=1 is the vertical axis of the calendar) be substituted with (3,0) and (2,-3) to obtain: 0=a(3-1) 2+m --3=a(2--1) 2+m

a=1 m=--4 y=(x-1 2)--4 The expression hall is bad: y=x 2--2x--3

Summary. Dear, what about the <> of the full topic

It is known that the parabola y=x +bx+c passes through the point a(3,0), and the intersection point with the y-axis is b, and the symmetry axis is a straight line such as jujube stove x=1, find.

Dear, what about the <> of the full topic

Solve it for your classmates right away.

Do you know about point b in other places?

Do you know the coordinates of point B elsewhere?

The answer written by the classmate on the paper is that it is okay to stop the waiter, but you should pay attention to the fact that this question is x square, not ax square, that is, a we know is equal to 1, so the student brings a = 1 in, and you can get the same result as the teacher<>

1.Solution:

Substituting the three points a, b, and c into the parabolic equation respectively, we get:

0=a-b+c

0=9a+3b+c

3=c so we get: a=-1, b=2, c=3

The parabolic analytic formula is y=-x +2x+3

2.Solution: Exists, q has 3 coordinates.

Let the distance from Q to the straight line MB be m, and the distance from P to the straight line MB be n

s△qmb=(1/2)×|mb|×m，s△pmb=(1/2)×|mb|×n

To make s qmb=s pmb, you only need q points such that m=n.

The equation for the straight line bc is: x 3 + y 3 = 1, i.e. y = -x + 3

Point p is the intersection of the axis of symmetry and the parabola, and the coordinate of p is (1,4).

The M point is the intersection point of the axis of symmetry and BC, and the M coordinate is (1,3).

The slope of the straight line pc k1 = (4-3) (1-0) = 1, and the slope of the straight line mb is -1, then pc mb, then |pc|=n

Make p with respect to c as the point of symmetry p', then p'The coordinates of are (2 0-1, 2 3-4), i.e., (-1, 2), and p'c mb, and |p'c|=|pc|=n

Make a straight line l: parallel to bc and pass p, i.e. the slope is -1, and the straight line through p: y=-x+5

Make a straight line l': parallel to BC and via P', i.e., the slope is -1 and passes through p'The straight line: y=-x+1

l bc, the distance from any point on l to the straight line mb is equal and equal to |pc|, i.e. n

Again'//bc，∴l'The distance from any point to the straight line mb is equal and equal to |p'c|, i.e. n

l and l'The area of the triangle formed by any point above and m and b is equal to the area of pmb.

Simultaneous l and parabolic equations give (1,4),(2,3).

Lianli L'With the parabolic equation, we get ((3+ 17) 2, (-1- 17) 2), (3- 17) 2, (-1+ 17) 2).

So there are 3 coordinates at point q, which are (2,3), (3+ 17) 2, (-1-17) 2), (3-17) 2, (-1+ 17) 2).

3.Solution: Exists.

The r coordinate is (1 + 2,2).

There are two cases, and there is also a point g [5,0].

The radius of the circle is 2

When x=0 and x=2, the values of y are equal, so the axis of symmetry is x=1, and the following equation is known to be listed:

c=4a+2b+c

16a+4b+c=12-7

a+b+c=3-7 (bring the axis of symmetry x=1 in).

The first question of the joint solution a=1, b=-2, c=-3 has been solved.

2) Draw the image first, extend BM called the y-axis and the point n, know the triangle bqp similar and triangle bon from the similar triangle, write the equation of the straight line bm by the two-point formula, so that x=0, you can get the intercept of the y-axis, that is, the length of on, so that the proportion can be written from the similarity, qp on=qb bo, you can get the length of qp, s=1 2oa multiplied by oc+1 2(qp+oc)t, only t is unknown in this formula, and the other quantities can be solved, as for the range of t, The zero cut-off points b and m,1 can be taken

Solution: (1) Equation x -10x + 16 = 0

x-2）（x-8）=0

x = 2 or x = 8

Then ob=2, oc=8

The coordinates of point b are (2,0) and point c (0,8).

Let the parabola be y=a(x+2) +b

Substitution. 16a+b=0（1）

4a+b=8（2）

12a=-8

a=-2/3

b=32/3

The parabolic equation is y=-2 3(x+2) +32 3=-2 3x -8 3x+8

Answer: 1. The analytical formula can be set from the coordinates of b and c as: y=a(x--1)(x--5)=ax +bx+c comparison coefficient and c=3 generations from the coordinates of point a

a=3 5,b=--18 5 So the analytic formula is: y=3 5x 18 5x 3, axis of symmetry x (1 5) 2 3 2, point A is the symmetry point of the axis of symmetry x=3 (on the parabola), point M is the symmetry point of the x-axis, connect a m, the intersection points of the x-axis and the axis of symmetry are the points e and f, then the perimeter of the quadrilateral AMEF is the smallest, and the perimeter of the quadrilateral AMEF can be found: the coordinates of a are (6,3) m The coordinates of the point are (0, 3 2) so the linear equation of a m can be found:

y=3 4x--3 2 so e(2,0),f(3,3 4) so the perimeter = am + me + ef + fa = am + m e + ef + fa = am + m a =3 2+ [6 + 3 2) +3 ]=3 2 + ( 261) 2 3, there must be: as ac midpoint d, as a straight line de must cross the parabola at the point n, d point coordinates can be found from the midpoint formula: d (5 2, 3 2) e point coordinates (2, 0) to obtain de linear equation:

y=3x--6, a system of equations composed of straight line equations and parabolic equations can be used to find the coordinates of the intersection point.

Substituting the values of the three points of abc into the parabola, we find a, b, and c

Isn't there a formula for the axis of symmetry, which can be found by substitution.

The first question can be used with the intersection formula y=(x-x1)*(x-x2), where x1 and x2 are the abscissa of the two intersection points with the x-axis, and the symmetry axis will come out.

The second question can be answered by making point d.

The third question is to draw a picture, which is not very convenient for the claw chicken party.

Solution: (1) a(1,0) and b(3,0) pass through the parabola y ax2 bx c, and the parabola can be set to y a(x 1)(x 3).

c(0,3).

After the parabola, substitution, we get 3 a(0 1)(0 3), that is, a = 1.

The analytic formula of the parabola is y (x 1) (x 3), i.e. y x2 2x 3.

2) Connect BC, the intersection point of the straight line BC and the straight line L is P.

Then the point p at this time minimizes the circumference of the pac.

Let the analytic formula of the straight line bc be y kx b, and substitute b(3,0),c(0,3) into the following equations

Solution: The functional relation y x 3 of the straight line bc.

When x 1, y 2, is the coordinate of p (1,2).

3) Exists. The coordinates of point m are (1,,(1, ,1,1),(1,0).

Quadratic function synthesis problem, undetermined coefficient method, the relationship between the coordinates of the points on the curve and the equation, the properties of the perpendicular line in the line segment, the trilateral relationship of the triangle, the properties of the isosceles triangle.

Analysis] (1) The intersection point formula can be set, and the pending coefficient can be obtained by using the undetermined coefficient method.

2) From the diagram, it is known that points A and B are symmetrical about the symmetry axis of the parabola, then according to the symmetry of the parabola and the shortest line segment between the two points, it can be seen that if BC is connected, then the intersection point of BC and the line L is the eligible P point.

3) Since the waist and bottom of MAC are not clear, it should be discussed in three cases: MA AC, MA MC, AC MC; You can first set the coordinates of the M point, and then use the M point ordinate to represent the three sides of the MAC, and then solve it according to the above three cases

The axis of symmetry of the parabola is:

x=1, let m(1,m).

a(－1，0)、c(0，3)，∴ma2＝m2＋4，mc2＝m

2－6m＋10，ac

If ma mc, then ma

2 mc2, get: m

2 4 m2 6m 10, get: m 1.

If ma ac, then ma

2 ac2, get: m

2 4 10, get: m

If MC AC, then MC

2 ac2, get: m

2 6m 10 10, get: m 0, m 6, when m 6, m, a, c three points are collinear, do not form a triangle, do not fit the topic, so give up.

In summary, it can be seen that the eligible m point and the coordinates are (1,,(1, ,1,1),(1,0).