Urgent!! High score to find a junior three dimensional quadratic function problem, know the progress

One, s=?(v1+v2), v1=20m s, v2=0m s known....Find t=two, and get a=8m (s 2) from a=(), i.e. less 8m per second....3. Set the time tx, then v0 tx ?a (tx 2) = 15 to get tx = both ten square excluding the answer greater than, get the time tx seconds, four, the same method as three, get the time is about seconds ( 5 I don't know if you can understand it?

This is the correct solution).

The solution is beyond the textbook, and I can't solve it at the junior high school level, so I have to learn the laws of Newton's kinematics in the first year of physics! There are a few equations that are OK with the first question. Later, you only have to make a table and analyze qualitatively according to the nature of the function....

1) Condition 1: s=1 2*a*t*t=25; Condition 2: a*t=20A=8, T=

2) Because a=8, the speed of the vehicle decreases by 8m s

3) vt*vt-v0*v0=2*a*s, s=15,a=8, solution: v0=, t=(vt-v0) a=

Have you changed the question, this question is not written in the third year of junior high school.

（1）5n+2x=70,5n+2x=50；

2）s=50n（70-5n）=3500n-250n^2;

s=3500n-350n^2;

3) It is clear that the first type of building has the largest floor area, and there is a maximum value when n = 7.

From x -6x+5=0 we get x 1 x 5 =0x=1 or x=5

m=1 n=5

The image of y=-x +bx+c passes through the points a(1,0) and b(0,5) then 0=-1 b c

5=cb=-4

The analytic formula of this parabola y=-x -4x+5=- x 2 9 makes y=0 x=1 or x=-5

Coordinates of c -5,0 and coordinates of d -2,9 Area of BCD = 5 2 9 5 5 2 5 = 14 When -5 x 1, y 0;When x -5 or x 1, y 0 note When x is what is the value, y 0 is the image ordinate 0 when the abscissa of the image is found.

Solution: (1) Factorized by x -6x+5=0: x 1 x 5 =0 so two roots: x=1 or x=5

i.e. m=1 n=5

From the diagram and the inscription, it can be seen that the image of y=-x +bx+c passes through the points a(1,0) and b(0,5).

Bring in two points: 0=-1 b c

5=c, so b=-4

Then the analytic formula of this parabola: y=-x -4x+5=- x 2 9 factorization: (-x+1)(x+5)=0

So x=1 or -5

The coordinates of c are -5,0, and the coordinates of d are -2,9 when x=-2 The area of BCD = 2 1 *BC*BC The height of the edge. Point to Straight Line Distance Formula...

From the image, we can see that when -5 x 1, y 0;When x -5 or x 1, y 0 answers,。。

The value of the function y decreases with the increase of x, indicating that the coefficient of x is negative, and a cross-origin : can be written

Let the equation of the straight line through the origin be y=kx, and substitute (-1,3) into k=-3

So the equation is: y=-3x

There are many more, such as: y=-2x+1 and so on.

The primary function y=kx+b, the subtractive function k<0, over (-1,3), 3=-k+b, b=k+3, y=kx+k+3

For example, let k = -1

y=-x-1+3=-x+2

Well, first of all, you can set this function to y=ax+b

From the conditions known by the question, there are an infinite number of such functions.

y decreases with the increase of x, which means a>0, now let a=1, then b=4 so, y=x+4

Answer; Since the value of the function y increases with the increase of x, then the coefficient of x can be set as long as it is greater than zero, so the relationship of the function can be set to be y=x+b

and because the image is dotted (-1, 3).

So 3=-1+b

b=4 is y=x+4

Intersect with the y-axis at the point (0,2).

Obviously, c=2 and two 00s, so a>0 x1+x2= -b a >0, so b<0 because the quadratic function opening is upwards and 0 x1 1, 1 x2 2, so x=1 y=a+b+2<0

When x=2, y=4a+2b+2>0

Get a+b<-2

2a+b>-1 ①

x1+x2 = -b/a

0＜x1＜1

1＜x2＜2

1＜-b/a＜3

Get a+b<0

3a+b>0 ②

The axis of symmetry = -b 2a = (x1 + x2) 2 belongs to the range (1 2, 3 2), so 1 2<-b 2a<3 2 is -30, thus -3-3

b<-a, b>-3a, shifts a+b>0 and 3a+b>0

It's easier to understand with a sketch.

From the meaning of the problem, we can construct a function f(x)=ax 2+bx+c, which consists of the intersection of the image and the x-axis at two points (x1,0)(x2,0), and 00From this, the answer can be obtained by processing the three equations. That's all for the answer, and the readers in the future should know!

Solution: 1. The parabola y=8(x+3+m) +7+n is based on the y-axis as the axis of symmetry.

The axis of symmetry x=-3-m=0, i.e., m=-3

(2,3) again, so 3=8*2 2+7+n, solution, n=-36, so m=-3, n=-36

2. y=2x is shifted to the right by 1 unit, that is, the original x becomes x-1, and the original parabolic analytic formula becomes y=2(x-1) 2

After 3 units downwards, it becomes y=2(x-1) The intersection point of 2-3 and the straight line y=2x-5 is a,b

Solve equation 2(x-1) 2-3=2x-5 to get x=1, or x=2, and substitute the linear equation to get y=-3, or y=-1

Therefore, the coordinates of a,b are (1,-3) and (2,-1) or (2,-1) and (1,-3).

solution, (1) Since the parabola passes the point c(0,-3), we can know that c=-3,a(-3,0), b(1,0) are all on the x-axis, that is, -3 and 1 are the solutions of the equation ax +bx-3=0, then -3+1=-b a, -3*1=-3 a, can be quickly determined, a=1, b=2

Therefore, the parabolic equation is y=x +2x-3

2) y=x +2x-3=(x+1) -4, therefore, vertex d=(-1,-4).

Therefore, |cd|=√2，|be|= 2, so |cd|=|be|=√2

That is, to make s pdc=s pbe, just make the distance from the p point to the straight line be equal to the distance from p to the straight line cd.

Let p(m,n), then the distance from p to the straight line be is |m+n-1|/√2

The distance from p to straight line cd is |m-n-3|/√2

So |m+n-1|/√2=|m-n-3|2, and m +2m-3=n

To solve the equation, there are three sets of solutions, m=2, n=5 or m=-1+3, n=-1 or m=-1-3, n=-1

Therefore, the coordinates of p are (2,5), or the coordinates of p are (-1 + 3,-1), or the coordinates of point p are (-1-3,-1).

When the coordinates of p are (2,5), the area of the triangle is s=6 2*2 2=3

When the coordinates of p are (-1 + 3, -1), the area of the triangle is = (3- 3) 2

When the coordinates of point p are (-1- 3, -1), the area of the triangle is s=(3+3)2

1. Bring in the ab c three-point coordinates.

9a-3b+c=0

a+b+c=0

c=-3 gives a=1 b=2 c=-3

y=x2+2b-3