Given that the quadratic function f (x) satisfies f (2-x) = f (2 + x), and the intercept of the image on the Y axis is 0, the minimum value is - 1, the analytic expression of the function f (x) is obtained

Given that the quadratic function f (x) satisfies f (2-x) = f (2 + x), and the intercept of the image on the Y axis is 0, the minimum value is - 1, the analytic expression of the function f (x) is obtained

∵ f (x) satisfies: F (2-x) = f (2 + x), the axis of symmetry of the function f (x) is x = 2. Let f (x) = a (X-2) 2 + B. and ∵ the intercept on the Y axis is 0, and the minimum value is - 1. ∵ a > 0, 4A + B = 0, B = - 1. The solution is a = 14, B = - 1. ∵ f (x) = 14x2-x

In the plane rectangular coordinate system, the minimum value of quadratic function is negative 4, the symmetry axis of image is x = - 1 and passes B (3,0)
Find the analytic expression of the quadratic function

According to: the minimum value of quadratic function is negative 4, the symmetry axis of image is x = - 1
We know that the vertex is (- 1, - 4)
So we can assume that the analytic expression of the quadratic function is y = a (x + 1) &# 178; - 4
Passing through point (3,0) ∧ a (3 + 1) ∧ 178; - 4 = 0 ∧ a = ∧ 188;
The analytic expression of the quadratic function is y = &# 188; (x + 1) &# 178; - 4

If the quadratic function y = x2-2 (K + 1) x + K + 3 has a minimum value of - 4 and the symmetry axis of the image is on the right side of the Y axis, then the value of K is______ ．

The symmetry axis of the image is on the right side of the y-axis, the symmetry axis X = K + 1 ＞ 0, the solution is k ＞ - 1, the quadratic function y = x2-2 (K + 1) x + K + 3 has the minimum value - 4, the minimum value of y = 4 (K + 3) − (2k + 2) 24 = K + 3 - (K + 1) 2 = - k2-k + 2 = - 4, the solution is K2 + K-6 = 0, the solution is k = 2 or K = - 3, ∫ k = - 3 < - 1, it is not reasonable to omit, ∫ k = 2

Let f (x) be a function defined on R and satisfy the following relations: F (10 + x) = f (10-x), f (20-x) = - f (20 = x). Try to judge the parity of F (x)

f(20-x)=f[10+(10-x)]=f[10-(10-x)]=f(x)
f(20+x)=f[10+(10+x)]=f[10-(10+x)]=f(-x)
Because f (20-x) = - f (20 + x)
So f (x) = - f (- x)
F (x) is an odd function

On what line is the image of functions y = log (2) x and y = Log1 / 2 (4x) symmetric?
Please answer quickly. There will be high delivery after the answer
The first is the logarithm of 4x with the base of 1 / 2
The second is the logarithm of X with base 2
Thank you very much

On y = - 1 symmetry
Proof (my first proof)
If 2 ^ a = x
Then (1 / 2) ^ (- a) = x, (1 / 2) ^ (- 2-A) = 4x
If log (2) x = a, then y = Log1 / 2 (4x) = - 2-A,
That is, if for the same X, one of the two function values is a and the other is - 2-A, then y = (a-2-a) / 2 = - 1 is symmetric
^Represents power

The set composed of the intersection points of the images of the linear function y = x + 3 and y = - 2x + 6 is ()
A. {4，1}B. {1，4}C. {（4，1）}D. {（1，4）}

From the meaning of the problem, the simultaneous equations can be y = x + 3Y = − 2x + 6, and the solution can be y = 4, x = 1. The intersection of the graph of the first-order function y = x + 3 and y = - 2x + 6 is (1,4). The set is {(1,4)}, so D

The set composed of the intersection points of the images of the linear function y = x + 3 and y = - 2x + 6 is ()
A. {4，1}B. {1，4}C. {（4，1）}D. {（1，4）}

From the meaning of the problem, the simultaneous equations can be y = x + 3Y = − 2x + 6, and the solution can be y = 4, x = 1. The intersection of the graph of the first-order function y = x + 3 and y = - 2x + 6 is (1,4). The set is {(1,4)}, so D

The set of intersection points of images of the linear function y = x + 3 and y = - 2x + 6
I just want the correct and standard answer format
I think all the examples in the book should be set up first. Don't you want this?

Y = x + 3 and y = - 2x + 6
x+3=-2x+6
x=1
So when x = 1, the function image intersects
In this case, y = 4
So the set of intersection points is {(1,4)}

The set composed of the intersection points of the images of the linear function y = x + 3 and y = - 2x + 6 is ()
A. {4，1}B. {1，4}C. {（4，1）}D. {（1，4）}

From the meaning of the problem, the simultaneous equations can be y = x + 3Y = − 2x + 6, and the solution can be y = 4, x = 1. The intersection of the graph of the first-order function y = x + 3 and y = - 2x + 6 is (1,4). The set is {(1,4)}, so D

The set of intersection points of the images of the first-order function Y-X + 3 and y = - 2x + 6, the idea and process to solve the problem

Y = x + 3, y = - 2x + 6, x = 1, y = 4
The intersection point of the image of the linear function y = x + 3 and y = - 2x + 6 is (1,4)
The set of intersections is {(1,4)}