The image of the quadratic function y = 1 / 2x & # 178; + bx-3 / 2 intersects the x-axis at point a (- 3,0) and point B, with ab as the edge above the x-axis Make a square ABCD, point P is a moving point on the x-axis, connecting DP, passing through point P, make the vertical line of DP intersect with the y-axis at point E. please write the coordinates of point d directly. When point P moves to where on line Ao (point P does not coincide with a, O), the length of line OE has the maximum value, and find out the maximum value

Substituting a, we get the equation y = 1 / 2x & # 178; + x-3 / 2, B point is (1.0), D is (- 3,4), when e is the midpoint of Ao, OE has the maximum value, we can get the maximum value of 9 / 16
Let even function f (x) be a decreasing function in [0, + ∞), then the solution set of inequality f (x) > F (2x + 1) is______ .
If the even function f (x) is a decreasing function in [0, + ∞), then f (x) > F (2x + 1) can be changed to | x | 2x + 1 | and the solution is x | 13 or X | 1, so the answer is: X ﹥ 13 or x | 1
It is known that the image of quadratic function y = xsquare - (A-2) x + a-5 intersects X axis at points a and B, and intersects Y axis at point C. when line AB is the shortest, the length of line OC is calculated
Let two intersection coordinates be: (x1,0); (x2,0)
x1+x2=a-2
x1x2=a-5
|AB|=|x2-x1|
AB^2=(x2-x1)^2=(x1+x2)^2-4x1x2=(a-2)^2-4(a-5)=a^2-8a+24=(a-4)^2+8
So when a = 4, the square of AB has a minimum, that is, AB has a minimum
So the function is: y = x ^ 2-2x-1
The coordinates of the intersection point with y axis are: (0, - 1)
So: | OC | = 1
Let two intersection coordinates be: (x1,0); (x2,0)
x1+x2=a-2
x1x2=a-5
|AB|=|x2-x1|
AB^2=(x2-x1)^2=(x1+x2)^2-4x1x2=(a-2)^2-4(a-5)=a^2-8a+24=(a-4)^2+8
So when a = 4, the square of AB has a minimum, that is, AB has a minimum.
So the function is: y = x ^ 2-2x-1
The coordinates of the intersection point with y axis are: (0, - 1)
So: | OC | = 1
F (x) is defined on [- 1,1], even function is a decreasing function on [0,1]. The solution to the inequality f (x-1) > F (2x-1)
F (x) is defined in [- 1,1]
-1≤x-1≤1
-1≤2x-1≤1
0≤x≤1
Even function is a decreasing function on [0,1]
Then [- 1,1] is an increasing function
|x-1|>|2x-1|
(x-1)²>(2x-1)²
3x²-2x
The question of increasing on {- 1,0}: is the even function on - 1,1 a decreasing function on [0,1]... Solve the inequality
Let the quadratic function f (x) satisfy f (3 + x) = f (1-x) and the sum of two squares of F (x) = 0 be 10. If the image passes (0,3), find the analytic expression of F (x)
Let f (x) = ax & # 178; + BX + C, f (3 + x) = f (1-x) get a (3 + x) &# 178; + B (3 + x) + C = a (1-x) &# 178; + B (1-x) + C 4A + B = 0. ① if the image passes (0,3), C = 3. ② if the sum of two squares of F (x) = 0 is 10, we get {[- B + √ (B & # 178; - 4ac)] / (2a)} & # 178; + {[- B - √ (B &...)
F（3+x）=F（1-x）
X = - 1 is the square of F (x) = a * (x + 1) + C
Since the two roots are symmetric with respect to x = - 1 and the sum of squares is 10, let the root on the right side be x0 and the root on the left side be - 2-x0
X0 square + (- 2-x0) square = 10
The two roots are - 3,1
So f (- 3) = f (1) = 0
And because the image is too (0, 3), so f (0) = 3
The solution is a = - 1, C = 3
So... Unfold
F（3+x）=F（1-x）
X = - 1 is the square of F (x) = a * (x + 1) + C
Since the two roots are symmetric with respect to x = - 1 and the sum of squares is 10, let the root on the right side be x0 and the root on the left side be - 2-x0
X0 square + (- 2-x0) square = 10
The two roots are - 3,1
So f (- 3) = f (1) = 0
And because the image is too (0, 3), so f (0) = 3
The solution is a = - 1, C = 3
So f (x) = - (x + 1) square + 1 = - x square - 2x + 3?
Let F X be an even function on R and an increasing function on [0, + ∞), and solve the inequality f (2x-1) about X
Because the function f (x) is an even function on R and an increasing function on [0, + ∞)
So the function f (x) is a decreasing function on (- ∞, 0]
So the inequality f (2x-1)
x＜-2
each
Let the quadratic function f (x) satisfy f (x + 2) = f (2-x) and the sum of the squares of the two real roots of the equation f (x) = 0 be 10. F (x) can be obtained by (0.3)
From F (2 + x) = f (2-x), we know that f (x) is symmetric with respect to x = 2, so we can assume that the quadratic function is f (x) = a (X-2) ^ 2 + B = ax ^ 2-4ax + 4A + B
And X1 + x2 = - B / a = 4
x1*x2=(4a+b)/a
(x1+x2)^2-2x1*x2=x1^2+x2^2=10
Substituting the point (0,3) into the function, we get 4A + B = 3
The solution is a = 1, B = - 1
f(x)=x^2-4x+3
Let f (x) = ax ^ 2 + BX + C
F (x) image over (0, 3) → C = 3
The quadratic function f (x) satisfies that the symmetry axis of F (x + 2) = f (2-x) → f (x) is x = - B / 2A = 2
The sum of squares of the two real roots of F (x) = 0 is 10, that is, (x1) ^ 2 + (x2) ^ 2 = 10 ② (Note: X1 and X2 are the two real roots of the equation)
Another X1 + x2... Expansion
Let f (x) = ax ^ 2 + BX + C
F (x) image over (0, 3) → C = 3
The quadratic function f (x) satisfies that the symmetry axis of F (x + 2) = f (2-x) → f (x) is x = - B / 2A = 2
The sum of squares of the two real roots of F (x) = 0 is 10, that is, (x1) ^ 2 + (x2) ^ 2 = 10 ② (Note: X1 and X2 are the two real roots of the equation)
And X1 + x2 = - B / a X1 * x2 = C / a ③
② The deformation is (x1 + x2) ^ - 2 (x1 * x2) = 10 ④
A = 1 and B = - 4 are obtained from simultaneous solutions
f(x)= x^2- 4x + 3
That's all. Put it away
If the even function f (x) is an increasing function on [0, + ∞), what is the solution set of the inequality f (1 / x) > F (2)?
If the even function f (x) is an increasing function on [0, + ∞), what is the solution set of the inequality f (1 / x) > F (2)?
f(1/x)＞f(2 )
Satisfy 1 / x > = 0 and 1 / x > 2
X > 0 and x < 1 / 2
The solution set is (0,1 / 2)
It is known that the quadratic function f (x) satisfies f (3-x) = f (x-1), the minimum value of F (x) is 1, and the square sum of the two real roots of the equation f (x) = x is 10
F (x) = 1 / 2x square - x + 3 / 2
Or F (x) = - 1 / 4x square + 1 / 2x + 3 / 4
Give me a point!
Let
f(x) = ax^2+bx +c
f(3-x)=a(3-x)^2+b(3-x)+ c
= ax^2-(6a+b)x + (9a+3b+c)
f(x-1)= a(x-1)^2+b(x-1)+c
=ax^2-（2a-b)x + (a-b+c)
f(3-x)=f(x-1)
compare coef of x
It is known that f (x) = 2x can be expressed as the sum of an odd function g (x) and an even function H (x). If the inequality Ag (x) + H (2x) ≥ 0 holds for X ∈ [1,2], then the minimum value of real number a is______ .
F (x) = 2x can be expressed as the sum of an odd function g (x) and an even function H (x) {g (x) + H (x) = 2x, G (- x) + H (- x) = - G (x) + H (x) = 2-x, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②, ②

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