It is known that quadratic functions meet the following three conditions at the same time

Solution: From (1) f(1+x)=f(1-x), the symmetry axis of the quadratic function is x=1, and the maximum value of the (2) function is 15, that is, the vertex of the quadratic function image is (1,15), so let the analytical formula of this quadratic function be f(x)=a(x-1) 2+15, where a≠0, i.e., f(x)=ax 2-2ax+a+15, and let the two roots of the equation f(x)=0 be x1, x2, then x1+x2=2, x1x2=(a+15) a, The sum of squares of the two real roots by (3)f(x)=0 is equal to 7, i.e., x1 2+x2 2=7

Because x1 2+x2 2=(x1+x2) 2-2x1x2 then 4-2(a+15) a=7

Solving this equation about a gives a = -6

Take a=-6 to get f(x)=-6x 2+12x+9, that is, what you want.

The idea upstairs is also relatively clear, but the calculation process is more cumbersome when solving the problem with the selected idea, so the calculation is wrong. The landlord can understand the solution of this problem according to two ideas, which will be helpful for you to solve the problem and choose the solution method in the future. ）

From (1) f(1+x)=f(1-x), the axis of symmetry of the quadratic function is x=1. In fact, more generally, for any x r, if the function satisfies f(a+x)=f(b-x), then the function has an axis of symmetry x=(a+b) 2, which is easy to see by drawing.

You might as well let one of the functions be t, and t 1, then the other 2-t so that the quadratic function can be set as.

f(x)=a(x-t)(x-2+t)

Moreover. t²+(2-t)²=7

Namely. t=(2- 10) 2 ((2+ 10) 2 1 rounding) by (2) function with a maximum value of 15, i.e.

f(1)=a(1-t)(1-2+t)=15. a=15/[(1-t)(1-2+t)]15/[(1-t)(t-1)]

Therefore, the analytic formula is.

f(x)=-6[x-(2-√10)/2 ][x-2+(2-√10)/2]

6x²+(6√10)x+9

y=ax2+bx+c

where a, b, and c are constants, and a≠0

The highest order of a quadratic function must be quadratic.

As can be seen by definition).

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1. Passing point (3,1): y=a(x-3) 2+b(x-3)+12When x is greater than 0, y decreases with the increase of x: the opening is downward, a<0, and the axis of symmetry is in the left half plane: 3-b (2a)<=0-->b<=6a

3. When x is 2, the value of the function is less than 2: a(2-3) 2+b(2-3)+1<2-->b>a-1

Solution: a-1 -1 5 so a=-1 6, b=

(1) Let the expression of the quadratic function be y=ax +bx+c, and substitute the three-point coordinates c=-1, a+b+c=0, a-b+c=2

a=2,b=-1.The function expression is y=2x -x-1(2), and the function expression is vertex, y=a(x-1) -3, and substituted by (0,1)a-3=1,a=4

The function expression is y=4(x-1) -3=4x -8x+1(3), and the function expression is the intersection formula, y=a(x+3)(x-5), and substituting (0,-3).

15a=-3,a=1/5

The function expression is y=1 5(x+3)(x-5)=x 5-2x 5-3(4) Let the function expression be vertex, y=a(x-3) -2, because the axis of symmetry of the function is x=3, and the distance from the two intersections to the axis of symmetry is equal, so one of the intersections is 2 units to the right of (3,0), which is (5,0).

Substitution point (5,0).

4a-2=0,a=1/2

Is it okay to have a function expression y=1 2(x-3) -2=x 2-3x+5 2?

1) Since the axis of symmetry is x=-1, the abscissa of the vertex coordinates is -1, and the ordinate of the vertex coordinates is -2, that is, the vertex coordinates are (-1, -2), and the other coordinate is (-3, 0) using the vertex formula to solve, let y=a(x+k) 2+h, (h, k) be the vertex coordinates, then there is y=a(x+1) 2-2, and then substitute another point to find a.

2) Since the coordinates of the parabola and x are informed, the two abscissa values can be regarded as the two roots of a quadratic equation, and the analytical formula is set to y=a(x-x1)(x-x2), where x1 and x2 are the abscissa values that intersect with the x-axis, and the function becomes y=a(x-3)(x+1), which can be solved by substituting another coordinate.

1. Let the parabola be y=a(x+5)(x-1) and intersect with the y-axis at the point (0,5), i.e., 5=a(5)*(1), i.e., a=-1, i.e., y=-(x+5)(x-1)=-x-4x+5

2 The parabola has only one common point (2,0) with the x-axis, then y=a(x-2) passes through the point (0,2).

i.e. 2=a*(-2).

i.e. a=1 2

That is, y=1 2(x-2) =1 2x -2x+23, when x=2, the minimum value of y = -4, and the image is set to y=a(x-2) -4 over the origin

By the image over the origin (0,0).

i.e. 0=a*(0-2) -4

i.e. a=1, i.e., y=(x-2) -4

x²-4x

From 1, the axis of symmetry is known to be x=1 with the early state

From 2, we can set f(x)=a(x-1) 2+5=ax 2-2ax+a+5, and where a<0

From 3,x1 3+x2 3=17, then x1+x2=2, x1x2=(a+5)a=(1+5 a).

The resulting chaos is 17 = (x1 + x2) 3-3 x 1 x 2 (x1 + x2) = 8-6 (1 + 5 a) = 2-30 a, and obtains: a = -2

The source of the conversation is f(x)=-2x 2+4x+3

20 is less than or equal to x less than or equal to 100) (60 is less than or equal to y less than or equal to 100) Therefore, when x = 20, y = 60, that is: 60 = a(20-h) + k, so that h = 20, then: k = 60, and when x = 100, y = 100, that is:

100=a(100-20) +60, solution: a=1 160, so, eligible finch height.

The second letter is a number.

The formula is: y=(x-20) 160+60,20 x 100).

The image of the secondary rock function passes through the point (-1,0), the hall shelters (1,2), and the problem is poor!!

The image of the quadratic function passes through the point (-2,4) and the vertex coordinates are (-3,6) solution: let the analytical formula of this quadratic function be :

ya(x+3)²

Vertex formula) then, 4a (

The analytical formula for a is: y

2(x+3)²

Incorporate this into a general form: y

2x²12x