Are all unary quadratic functions symmetrical

No. The function image is symmetrical only if the defined domain is also symmetrical with respect to the symmetry axis.

For example, y=x 2 (x belongs to the set r) symmetry.

y=x 2 (x<1) This one is asymmetrical.

No, to see whether a function is symmetrical, mainly depends on its definition domain, only under the premise of satisfying the definition domain, and then see whether it is symmetrical, 1. to see whether the definition domain is symmetric; 2；If you define domain symmetry, then see if the function value f(x) is symmetrical!

Not necessarily, to see whether the quadratic function is symmetric, mainly depends on its definition domain, and only under the premise that the definition domain is symmetrical with respect to the symmetry axis, the quadratic function is symmetrical.

The so-called symmetry must consider the symmetry interval and the symmetry axis, as long as the symmetry interval is symmetrical with respect to the symmetry axis, then the one-dimensional quadratic function image is symmetrical.

As long as the domain is symmetrical with respect to the symmetry axis, then the unary quadratic functions are all symmetric.

Be. All images of unary quadratic functions can be translated with respect to y-symmetry.

Let the unary quadratic function be y=ax 2+bx+c

axis of symmetry x=-b 2a

Vertex (-b 2a, (4ac-b 2) 4a) vertex y=a(x+b 2a) 2+(4ac-b 2) 4a

As long as it's not limited to a certain range, you can draw a simple diagram.

The law of symmetry of the quadratic function 1, y1 = ax2 + bx + c The function with respect to the symmetry of the x-axis is y2 = -ax2-bx-c. Because the shape of the parabola does not change, but the opening is in opposite directions, a becomes -a; The axis of symmetry remains unchanged, and the axis of symmetry of y1 is.

x=−\frac

x=− 2a

The axis of symmetry of by2 should also be.

x=−\frac=−\frac

x=− 2ab

The coordinates of the intersection of 2aby1 and the y-axis are 0,c, and when the x-axis is symmetrical, it is 0,-c. 2. Y1 = ax2 + bx + c The function of y-axis symmetry is y2 = ax2-bx+c. Because the shape of the parabola does not change, and the direction of the opening does not change, a does not change; The axis of symmetry changes, and the axis of symmetry of y1 is.

x=−\frac

x=− 2a

The axis of symmetry of by2 should be.

x=−\frac=\frac

x=− 2ab

The coordinates of the intersection of 2aby1 and the y-axis are 0,c, and the coordinates of the intersection of y2 and y-axis are also 0,c, so c does not change. 3. Y1=a x-h 2 k The function of origin symmetry is y2=-a x+h 2-k. At this point, the parabola must be turned into a vertex study.

Because the vertex of y1=a x-h 2 k is (h,k) and the vertex after symmetry with respect to the origin is -h,-k, the parabolic shape does not change, and the opening direction is opposite, so a becomes -a. [The content of this document can be freely copied or modified freely, looking forward to your praise and attention, we will do better].

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Get it now. The law of symmetry of quadratic functions.

The law of symmetry of quadratic functions.

1. Y1=ax2+bx+cThe function of x-axis symmetry is y2= -ax2-bx-c.

Because the shape of the parabola does not change, but the opening is in opposite directions, a becomes -a; The axis of symmetry remains unchanged, and the axis of symmetry of y1 is.

x=−\frac

x=− 2a

The axis of symmetry of by2 should also be.

x=−\frac=−\frac

x=− 2ab

The coordinates of the intersection of 2ABY1 and the y-axis are 0,c, and Guan Chenxiang is 0,-c after the x-axis symmetry.

2. Y1 = ax2 + bx + c The function of y-axis symmetry is y2 = ax2-bx+c.

Take a look at the following two sample questions:

Example 1: The number y=ax 2+bx+c is symmetrical with g(x) with respect to (2,1).

Find the analytic expression of the g(x) function.

Solution] Let (x0,y0) be the image of a known function.

On the previous point, the point is about (2,1) symmetrical selling scatter (x,y), then according to the midpoint coordinate formula.

x0+x=4, y0+y=2, i.e., x0=4-x, y0=2-y

x0,y0) on the image of a known function, substituting the coordinates into it.

2-y=a(4-x) 2+b(4-x)+c, i.e., y=-ax 2+(8a+b)x-16a-4b-c+2

This is the analytic expression of the function g(x).

Example 2: y=x 2+x+1 analytic expression of the symmetry function with respect to x=y.

Solution] Let a(x,y) be on the function y=x 2+x+1, then a is about the symmetry point b(x) of y=x',y') on a symmetrical function on the function y=x 2+x+1.

The midpoint c of ab is on y=x.

The coordinates of c are [(x+x')/2,(y+y'2], c in the group y=x on y+y')/2=(x+x')/2 （1）

The slope of the straight line ab is perpendicular to y=x, i.e., kab*k=-1kab=-1

kab=（y-y'）/x-x') 2)

From (1) and (2) can be solved:

x'=y,y'=x

Let a(x,y) be on the function y=x 2+x+1.

Replace x'=yy'=x is substituted into the equation y=x 2+x+1 so x'=y'^2+y'+1

This is the analytic formula that is sought, i.e., x=y 2+y+1

Solution: (1)y=x 2-2x-1=(x-1) 2-2, the coordinates of a are (1,-2).

The image of the quadratic function y=ax2+bx passes through (0,0). The vertices are on the axis of symmetry of the quadratic function y=x2-2x-1 image.

The symmetry axis of the image of the quadratic function y=x2-2x-1 is symmetric c(2,0) with respect to the point c and point o

2) The quadrilateral AOBC is a diamond shape.

Point B and point A are symmetrical with respect to the straight line OC.

b（1，2）．

Substitute b(1,2),c(2,0),(0,0) into ax 2+bx+c a=-2,b=4

y=-2x^2+4x

Symmetry problem in the image: a straight line perpendicular to x or y pumping over the vertex is the axis of symmetry, and any straight line perpendicular to this axis of symmetry has an equal distance from the intersection point of the function f(x) to this line, then the function f(x) is a symmetry function with an axis of symmetry.

Quadratic function: f(x) = ax +bx+c (a≠0) symmetry problem:

Take any constant k

such that f(x-k)=f(x+k), then the axis of symmetry is x;

such that f(k-x)=f (k+x), then the axis of symmetry is k.

Conclusion: The values in parentheses are added and divided by 2, and the result is the axis of symmetry.

For the quadratic function y=ax 2+bx+c

The axis of symmetry is the straight line x = -b 2a

My original question is not much different from yours, you can just change the number yourself.

Solution: Proof: Take any point (x0, y0) on y=f(x). - Set up a point.

then y0=f(x0).

Because (x0,y0) with respect to x=m symmetry point is (2m-x0,y0) -- find the symmetry point.

Since (x+x0) 2=m, x=m-x0 and because f(m+x)=f(m-x), f(x)=f(2m-x)--proves that the symmetry point is on the function f(x).

So f(x0)=f(2m-x0), so y0=f(2m-x0) check y0=f(2m-x0).

So (also on y=f(x).)

So f(x) is symmetrical with respect to the straight line x=m.

When x=m, f(2m)=f(0) so f(x) about the symmetry of the straight line x=m I don't understand what I am asking ?

Let the original function be y=f(x).

The function about origin symmetry is.

y=-f(-x)

This is to substitute -x into the equation first, and then add a negative sign to the whole equation.

The quadratic function is not a centrally symmetric graph because it cannot be perfectly coincident after selecting 180° along a certain point. But it is axisymmetric and the axis of symmetry is x=-b 2b

A primary function is both a centrally symmetric and an axisymmetric graph. Because it's a line segment. A line segment is both a center-symmetric and axisymmetric figure.

Since y=12a is parallel to the x-axis, the symmetry point of (x,y) with respect to y=12a is: (x,12a-(y-12a))=(x,24a-y).

So, the quadratic function of y=3a(x-2a) 2+12a with respect to the symmetry of the straight line y=12a is: 24a-y=3a(x-2a) 2+12a

i.e.: y=12a-3a(x-2a) 2