Given that the point (x0, Y0 + 1) is a point on the image of the quadratic function y = ax ^ 2 + BX + C (a > 0, B and C are constants), and x0 is the solution of the equation 2aX + B = 0 about X, why is there Y > yo for any real number x?

Because x0 is the abscissa of the function vertex and a > 0, the quadratic function has the minimum value, so it is easy to know that no matter what x takes, there is Y "Y0!

If an independent variable x0 is equal to its corresponding function value, x0 is called the fixed point of the function. Let f (x) = x ^ 3-2x + 2, what is the fixed point of F (x) If an independent variable x0 is equal to its corresponding function value, x0 is called the fixed point of the function. Let f (x) = x ^ 3-2x + 2, what is the fixed point of F (x)~ Chasing sub boy

An argument x0 is equal to its corresponding function value
That is, f (x0) = x0
For f (x) = x ^ 3-2x + 2
With x = x ^ 3-2x + 2
That is, x ^ 3-3x + 2 = 0
The solution is x = 1 or x = - 2
So the fixed points of F (x) are 1, - 2

The function of the independent variable x is usually recorded as f (x). F (x0) represents the function value of the function f (x) when the independent variable x = x0. The known function f (x) = x2-ax + 2, Where a is a real number 1. If a = 2, find the value of F (3) 2. If there is a real number T, 1 ≤ t ≤ 4, so that f (- T2-3) = f (4T), find the value range of real number a 3. If the inequality f (x) ≥ 2x + a holds for any 0 ≤ x ≤ 4, find the value range of A

Substitute a = 2 and get f (3) = 3
Substituting, we get a = - (T-3) * (t-1). We get a parabola with an opening downward about t, G (T) = - T ^ 2 + 4t-3, and the axis of symmetry t = 2. Therefore, on 1 ≤ t ≤ 4, G (T) max = g (2) = 1, G (T) min = g (4) = 3, that is - 3 ≤ a ≤ 1
If the parabola Y1 = x2-ax + 2 is tangent to the straight line y2 = 2x + A, then there is x2-ax + 2 = 2x + A, or there is no intersection. Because there is only one intersection or no intersection, the discriminant of the solid equation is less than or equal to 0, that is, (a + 2) ^ 2-4 (2-A) < = 0, that is, when the parabola (a + 2) ^ 2-4 (2-A) is below the X axis, the range of a is - 2 * 5 ^ 1 / 2-4 < = a < = 2 (5 ^ 1 / 2-2), and when it is tangent, x = 1 / 2 (a + 2). According to the definition of X, there is - 2 ≤ a ≤ 6, so based on the above, we can get - 2 ≤ a < = 2 (5 ^ 1 / 2-2)

In the function y = x + 1, the value range of the independent variable x is __

The value range of the function y = x + 1 independent variable x is all real numbers
So the answer is: all real numbers

Function y = - x + B when the value range of argument x is - 3

x=-3,y=3+b
x=-1,y=1+b
So 1 + B

Given the function y = (A-2) x-3a-1, when the value range of the independent variable x is 3 ≤ x ≤ 5, y can reach both a value greater than 5 and a value less than 3, then the value range of the real number a is () A. a＜3 B. a＞5 C. a＞8 D. Arbitrary real number

If A-2 > 0, that is, a > 2, the function is an increasing function,
From the meaning of the question, when x = 5, Y > 5, that is (A-2) × 5-3a-1 > 5, the solution is a > 8;
When x = 3, y < 3, i.e. (A-2) × 3-3a-1 < 3, at this time a, no matter what the real number inequality is, it is always true;
Therefore, a > 8;
If a = 2, y = - 7, not acceptable;
If A-2 < 0, that is, when a < 2, this function is a subtractive function,
When x = 3, Y > 5, i.e. (A-2) × 3-3a-1 > 5, this inequality does not hold
Therefore, this situation does not exist
Therefore, C

Given that the point (x0, Y0 + 1) is a point on the image of the quadratic function y = ax ^ 2 + BX + C (a > 0, B and C are constants), and x0 is the solution of the equation 2aX + B = 0 about X, why is there Y > yo for any real number x?