The image of quadratic function y = a (X-H) & sup2; + K passing through points (- 2,0) and (4,0) determines the value of K

The image of quadratic function y = a (X-H) & sup2; + K passing through points (- 2,0) and (4,0) determines the value of K

Take (- 2,0) and (4,0) into quadratic function y = a (X-H) &# 178; + K
a(-2-h)^2+k=0
a(4-h)^2+k=0
The solution is h = 1
That is y = a (x-1) ² + K

The image of quadratic function y = a (X-H) ^ 2 + K passes through points (- 2,0) and (4,0). Try to determine the value of H

The image of quadratic function y = a (X-H) ^ 2 + k is symmetric with respect to x = h, and the image passes through points (- 2,0) and (4,0),
So h = - 2 + (4 - (- 2)) / 2 = 1

If the value of the quadratic function y = KX & # 178; + (2k-1) x + K-2 is always negative, then the value of K is constant

The value is always negative
Let's just take the maximum of the two functions,

As shown in the figure, the image of quadratic function y = x & sup2; - 5x + 4 intersects with X axis at two points a and B (point a is on the left side of point B)
The vertex is C, and there is a moving point e moving from point B to point a in one unit per second

A(1,0)B(4,0)C（5/2,-9/4)

If u = Z, a = x, x = 2K, K belongs to Z, B = x, x = 2K + 1, K belongs to Z, then complement a =, complement B=

A is the set of all even numbers,
B is the set of all odd numbers,
Obviously, complement a = B, complement B = a

Let u = Z;. A = ι x ι x = 2K, K ∈ Z, B = {x ι x = 2K + 1, K ∈ Z}, find the complement of a in U and B in U,

A is the even of all positive integers
So its complement in Z is: 2k-1 K belongs to Z, that is set B
B is the odd number of all u = Z
The complement of B in U is 2K, and K belongs to Z, that is, set a

U = z a {X / x = 2K, K belongs to Z} finding the complement of a

The complement of a = {x | = 2K + 1, K belongs to Z}

Given x = 1 / 3K-1, y = 1 / 6K + 3, z = 1 / 2K + 3, then x ^ 2 + y ^ 2 + Z ^ 2 + 2XY + 2XZ + 2yz =?

By (x + y + Z) ² = 1, X & #178; + Y & #178; + Z & #178; = 3,
x²+y²+z²+2xy+2xz+2yz=1
∴xy+xz+yz=-1
∵x+y+z=1,∴x+y=1-z,①
Substituting XY + Z (x + y) = - 1,
xy+z（1-z）=-1
∴xy=-1-z+z² ②
According to Weida's theorem: Δ = (1-z) &# 178; - 4 (- 1-z + Z & # 178;) ≥ 0,
1-2z+z²+4+4z-4z²≥0
-3z²+2z+5≥0,
3z²-2z-5≤0,
（3z-5）（z+1)\≤0
-1≤z≤5/3
M=xyz=（-1-z+z²）z
=z³-z³-z
Substituting z = 5 / 3 into the maximum value mmax = 5 / 27
Substitute z = - 1 into: minimum value mmin = - 1

The known straight line y = (1-3k) x + 2k-1
(1) When k is the value, the line passes through the origin
(2) What is the value of K when the intersection coordinates of the line and the Y axis are (0, - 2)?
(3) What is the value of K when the line intersects the X axis at (3 / 4,0)?
(4) What is the value of K when y increases with the increase of X?
(5) What is the value of K when the line is parallel to the line y = - 3x-5?

(1) If (0,0) is substituted by y = (1-3k) x + 2k-1, 2k-1 = 0 → k = 0.5 (2) (0, - 2) is substituted by y = (1-3k) x + 2k-1, 2k-1 = - 2 → k = - 0.5 (3) (3 / 4,0) is substituted by y = (1-3k) x + 2k-1, k = - 1 (4) is greater than 0, and Y increases with the increase of X

1. If M = 2K, n = 3K + 2 and 2m-n = 3, find K.. 2. Given a + B = 2 / 3, ab = 1, find (A-2) (b-2) first, then simplify and evaluate. Experts gather here quickly~·
Add a factorization 2a(x-y)+b(y-x)

1,2m-n = 2 × 2K - (3K + 2) = K-2 = 3, k = 5
2,（a-2)(b-2)=ab-2a-2b+4=ab-2（a+b）+4=1-2×2/3+4=5-4/3=11/3
3,2a(x-y)+b(y-x)=2a(x-y)-b(x-y)=(2a-b)(x-y)
I wish you a happy study!