Math problems in junior high school, about quadratic functions finding specific processes .

1) According to Vedda's theorem and 3oa=ob, an equiquantitative relation about a and b can be obtained, and another relation of a and b can be obtained by substituting the coordinates of p point into the parabola, and the value of the undetermined coefficient can be obtained by concentrating the two formulas, and the analytic formula of the parabola is obtained; (2) As shown in the figure, take the symmetry point of point A about the y-axis, then a co= aco, if the intersection point of the line a c and the parabola is n point, then if mco a co, then the condition that must be met is that the abscissa of m is between the abscissa of a and the abscissa of n, according to which the value range of m abscissa can be found (the abscissa of m cannot be 0, otherwise the acute angle mco cannot be formed) Solution: (1) p(4,10) on the image, 16a-4(b-1)-3a=10; -3a 0, a 0,x1x2= -3a a=-3 0, x1 0,x2 0,x2=-3x1 x1+x2=x1+(-3x1)=-2x1=- b a,x1x2=-3x1 2=-3, x1 2=1,x1 0, x1=-1, x2=3, b+1=2a , simultaneous solution: a=2,b=3, y=2x 2-2x-6; (2) There is a point m, so that mco aco, point a is symmetrical point a (1,0) with respect to the y axis, let the line a c be y=kx+b, and since the line a c passes (1,0),(0,-6), then there is:

k+b=0 b=6, the solution is {k=6 b=-6 y=6x-6, the analytic formula of the simultaneous parabola is: {y=6x-6 y=2x 2-4x-6, the solution is {x=0, {x=5 y=-6 y=24 that is, the intersection point of the line a c and the parabola is (0,-6),(5,24), and the value range of x that meets the meaning of the question is -1 x 0 or 0 x 5 This question mainly examines the determination of the analytical formula of quadratic functions, the application of Veda's theorem, axisymmetric graphs, and the intersection of function images

Solution: (1) p(4,10) on the image, 16a-4(b-1)-3a=10; -3a 0, a 0,x1x2= -3a a=-3 0, x1 0,x2 0,x2=-3x1 x1+x2=x1+(-3x1)=-2x1=- b a,x1x2=-3x1 2=-3, x1 2=1,x1 0, x1=-1, x2=3, b+1=2a , simultaneous solution: a=2,b=3, y=2x 2-2x-6; (2) There is a point m, so that mco aco, point a is symmetrical point a (1,0) with respect to the y axis, let the line a c be y=kx+b, and since the line a c passes (1,0),(0,-6), then there is:

k+b=0b=6, the solution is {k=6b=-6 y=6x-6, the analytic formula of the simultaneous parabola is: {y=6x-6y=2x2-4x-6, the solution is {x=0y=-6, {x=5y=24 that is, the intersection point of the line a c and the parabola is (0,-6),(5,24), and the value range of x that meets the meaning of the question is -1 x 0 or 0 x 5

Set x people to go on a tour and the cost is y

When 40 people, the cost is 70 per person, for a total of 2800

Get a system of equations.

y=100x （x<=25）

y=x(100-2(x-25)) 2540) can be found that when there are less than 25 people, the maximum is 2500, so it is impossible to have fewer than 25 people, and the same way is that the number of people cannot exceed 40

Then the number of people should be between 25 and 40, which is the second function that is simplified to y=150x-2x 2

When y=2700

We get the equation x 2-75x+1350=0

So you get x=45 or 30

And because x<40

So x=30

Set up this tour with x people.

According to the title, x[100-2(x-25)]=2700,100-2(x-25) 70

The solution is x 40, x1 = 45, x2 = 30

x=30 is what is sought.

Let the equation of this quadratic function y=ax +bx+c satisfy the condition: -b 2a=3

4ac-b²)/4a=-1

16a+4b+c=-3

This results in a=-2, b=12, c=-19

The equation is y=-2x +12x-19

Hope it helps. Forget about it.

I don't know what you mean by "respectively" to find the corresponding functional relation, but it should be like this if you write it directly according to the conditions you give.

y=-2x2+12x-19

The 2 written after the x means the square of the x.

The axis of symmetry is a straight line x=3, the maximum value is -1 (indicating that the curve of this function is open downward), and the expression of a unary quadratic function can be obtained by passing through the three conditions of point c(4,-3).

This kind of function problem can be easily and quickly understood by drawing a picture!

y=x²-(a+2)x+9=(x-(a+2)/2)²+9-[(a+2)/2]²

The vertices are on the axes, and there are two cases.

If the vertex is on the x-axis, there is.

9-[(a+2)/2]²=0

a+2)/2|=3

a+2 = 6, a=4 or a=-8

If the vertex is on the y-axis, the axis of symmetry is x=0

a+2)/2=0，a=-2

Combining the two cases, there are three possible values of a, a=-8 or a=-2 or a=4

According to y=x-(a+2)x+9 on the coordinate axis, the vertex of the parabola is (a+2 2,0), and the point is on the parabola, (a+2 2) -a+2) (a+2) 2+9=0, and the solution is a=4 or a=-8

There are two cases for vertices to be on axes.

1. When the vertex is on the x-axis, the discriminant formula b 2-4ac=(a+2) 2-4*9=0 solves a as 4 or 8

2. When the vertex is on the y-axis, the symmetry axis is on the y-axis (a+2)=0, and the solution a is 2

Finally, it would be nice to write a summary of the above.

The most detailed....

If the vertex is on the x-axis, the maximum value is 0, and a is obtained. If the vertex is on the y-axis, then the axis of symmetry is the y-axis

y=(mx-2m-2)(x-1)

So the abscissa is (2m+2) m and 1

2m+2) m=2+2 m is a positive integer.

then 2 m is an integer.

So m= 2, 1

Then 2+2 m=3,1,0,4

where x=1 and 0 are rounded.

So m = 2 or 1

So y=2x -8x+6 or y=x -5x+4

Proof: (1) If m=0, then the equation is -2x+2=0, and there must be a real root (2) If m≠0, then = -(3m+2) 2-4m*(2m+2)=m2+4m+4=(m+2)2

Regardless of the value of m, (m+2)2 is greater than or equal to 0, i.e. 0 The equation has a real root.

synthesis (1) (2), so that no matter that m is any real number, the equation has a real root.

Quadratic function y= m x2 - (3m + 2) x + 2m + 2 = (m x - (2m + 2)) (x-1).

Therefore, the abscissa of the two intersections with the x-axis is 1, (2m+2) m=2+2 m, because the abscissa is a positive integer, so m can only be -2 (rounded, at this time, x is 1, which does not meet the two intersection points), 1, 2;

So the quadratic function is y=x2-5x+4 or y=2x2-8x+6

mx -(3m+2)x+2m+2=0, when m≠0 the value of the discriminant (3m+2) -4m(2m+2)=(m+2) so x1=(m+2) m, x2=(3m+2-m-2) 2m=1. So when the integer m=1, 2. When m=1, y=x -5x+4.

When m=2 is y=2x -8x+6.

Let the coordinates of b and f be (x1,y1) and (x2,y2)y2-y1=2x2=x1 2-x2 2, respectively

Find x2=1 12

So c = 145 144

Vertex formula, s=

So when t=20, s is the most big=600

Therefore, the Feizhou Yinqiao key machine can only stop 600m after landing.

How do you have time to find a distance? The title is written incorrectly, you.

1.Solution: From y=x -x+m, we can know that y=(x-1 2) +m-1 4).

Therefore, (1) the opening direction is upward, the axis of symmetry is x=1 2, and the vertex coordinates are (1 2, m-1 4).

2) When m>1 4, the vertex is above the x-axis;

When m=1 4, the vertex is above the x-axis;

When m<1 4, the vertex is below the x-axis.

2.Solution: According to the problem, the parabolic equation y=-1 4x +4 constructs the coordinate system, and the rectangle is also put into the coordinate system, then the coordinates of the four vertices in the square are (-4,-2),(4,-2),(4,0),(4,0) respectively

1) When the tunnel is a one-way street, the truck can drive in the middle of the tunnel, and the positions of the vertices on both sides of the truck in the coordinate system are (-1,2) and (1,2).

At x = -1 and 1, the coordinates on the parabola are (-1,4-1 4) and (1,4-1 4), respectively

The height corresponding to the change point is: 2+(4-1 4)=

That is, they are all higher than 4m, so they can pass through the tunnel.

2) If there is a double lane in the tunnel, change the truck to study the central axis, the central axis out of the tunnel is 6m high, the x-axis coordinate value at the other side is 2, and the corresponding parabolic ordinate value y=4-(1 4)*2 =3, that is, the tunnel height here (2+3)m=5m>4m

Explain that the truck can still pass through the tunnel at this time.

1.(1) Opening direction: up Axis of symmetry: x= vertex coordinates (, m-1 4).

2)m>1\4 m=1\4 m<1\4

2.(1) and (2) can pass this question, I just did it in the morning

(2008 Guigang) It is known that the roots of the two real numbers of the unary quadratic equation x2-4x-5=0 are x1 and x2, and x1 can send the graph over and the coordinates of the key points :(1) are obtained from the equation x2-4x-5=0.