Nine Years Math Quadratic Functions Finale 15

There are no pictures, and this question is very difficult.

Summary. Hello dear, please send me the original question** to give you a better one! Thank you for your cooperation!

Hello dear, please send me the original question** to give you a better one! Thank you for your cooperation!

The original question was taken and sent to me.

I want a difficult problem about quadratic functions in the middle school entrance examination, not to help me solve it.

Hello dear! This question is the finale of the secondary function high school exam in Heilongjiang Province last year, and it is difficult to make you doubt your life!

Good. Kiss, can you try to make it?

Analysis: This point is at the intersection point of the circumscribed circle with a, b, and c as triangles and x=1 step: 1Find the analytic formula y=-2x for the straight line ac and its perpendicular bisector analytic formula y=x 2,2Find the analytic formula y=-x-1 for the straight line bc and the analytic formula for the perpendicular bisector y=x-2,3Simultaneous finding the intersection of two perpendicular bisectors is the center of the circumscribed circle, denoted as e(4,2)4, and d(1,a) can be obtained by using eb=ed

Let x=0,y=-1 then b(0,-1)a(-1,2).

The first point d=c

The second point assumes the existence of ac= 20, ab= , and using similar triangles, we can know that the equation for the da*ab=ab*bc column is solved (1,2+ 46

It should be intersected with the x and y axes at a and c points respectively.

If that's the case, here's how to solve it:

a（-3,0），c（0,6）

Substituting y=-x +bx+c to find b,c

b=-1，c=6

y=-x²-x+6

b（2,0）

The parabola y=-x -x+6, and the straight line y=1 2x+a are synthesized to get x +3 2x+a-6=0, and the m(x1,y1),n(x2,y2) coordinates are obtained.

x1+x2=3/2,x1×x2=a-6

y1y2=（1/2x1+a）（1/2x2+a）=1/4x1x2+1/2(x1+x2)+a²=1/4(a-6)+3/4+a²

If there is x1x2+y1y2=0, it exists.

That is, if 1 4 (a-6) + 3 4 + a + a-6 = 0 has a solution, then it exists.

Let the general expression of the parabolic equation be y=ax 2+bx+c.

a(0,6): y=(0)a+(0)b+c=c=6

b(-3,0): y=(9)a+(-3)b+c=9a-3+c=0

c(6,0): y=(36)a+(6)b+c=36a+6b+c=0

The joint solution is: a=-1 3, b=1, c=6

The parabolic equation is: y=-(1 3)x 2+x+6

Let p(x,0), please make your own plot according to the title: p(x,0) and pe ab are intersected with ac in e.

bc|=9, |ab|=45^.5=3(5^.5), ac|=72^.5=6(2^.5)

pe|=|ab|·|pc|/|bc|=(45^.5)(6-x)/9=(5/9)^.5(6-x)

ae|=|ac|·|bp|/|bc|=(72^.5)(x+3)/9=(8/9)^.5(x+3)

Triangle ape area = |pe|·|ae|·sin(angular AEP) = (6-x)(x+3)(40 81).5·sin (angular AEP).

Triangle ape area = |'=(-2x+3)[(40/81)^.5·sin(angular AEP)]=0=> x=

The maximum area of the triangle APE occurs at p(,0). The maximum area can be calculated from the above equation, but here it can be obtained using the peculiarities of the geometry. P is the midpoint of BC, and then E is the midpoint of AC, so the area of (APC) = (APB) area, (APE) area = (BPE) area = (ABC) area 4=(1 2)(9)(6) 4=

Let g(x, -1 3) x 2+x+6), please make your own plot according to the title: g(x, y) [on the parabola], connect ga and gc.

The equation for the straight line ac is y=6-x, i.e., x+y-6=The vertical distance to the straight line ac is:

d=|(x) +1/3)x^2+x+6) +6)| / (1+1)^.5

-1/3)x^2+2x|/(2^.5)

Thus, the (agc) area is |ac|·d/2=(9/2)|-1/3)x^2+2x|/(2^.5)

Let (AGC) area = (AEP) area, i.e.

9/2)|-1/3)x^2+2x|/(2^.5)=27/4

To solve this binary equation, we get two solutions: x=3(1+, ie.

At g(3 2, 27 4) or g(9 2, 15 4) (agc) area = (ape) = 27 4

The end

(2) Let b(a,b) pass points a and b as the perpendicular lines of the x-axis and y-axis respectively, and the perpendicular feet are m and n. Because ab is the diameter of the circle, aob=90°, so because b(a,b) is in the parabola, so . The solution is , i.e., , so a=2bBecause b(a,b) is in the parabola, so .

So b(8,4) or (-1,-half) (some solutions can't stick over).

(3) When b(8,4) is obtained from the title, the area of the quadrilateral ABCD is the largest. Let d(x,y) give the analytic formula for the straight line bc as y=x-4

The perpendicular line of the x-axis is crossed by the point d, and the straight line BC is crossed at gBecause A, B, and C are fixed points, when the area of the triangle BCD is the largest, the area of the quadrilateral ABCD is the largest. The area of the triangle BCD is equal to the area of the triangle DCG + the triangle VDG, so.

The process of solving the problem is as follows, when x one》-1, 3 x +1 one, the limit of 1 ax+1 is also , if not , the whole limit does not exist. So when x-a-1, ax+1-0, a=1b=lim(1 x+1-3 x +1)=lim(x-2) (x -x+1)=-1