If the value of quadratic function f (x) = AX2 + BX + C (a ≠ 0) is a nonnegative real number for all real numbers x, then the minimum value of B + Ca is______ ．

If the value of quadratic function f (x) = AX2 + BX + C (a ≠ 0) is a nonnegative real number for all real numbers x, then the minimum value of B + Ca is______ ．

The value of quadratic function f (x) = AX2 + BX + C (a ≠ 0) is always non negative real number, that is, all function values are greater than or equal to 0, so it must be a > 0 △≤ 0 {a > 0b2 − 4ac ≤ 0 {B2 ≤ 4ac} C ≥ b24a {B + C ≥ b24a + & nbsp; B {B + Ca ≥ b24a2 + Ba, let g = b24a2 + Ba = 14 (Ba + 2) 2-1 ≥ - 1, that is, the minimum value of B + Ca is - 1, so the answer is - 1

Given the quadratic function y = AX2 + BX + C, for all real numbers x, X is less than or equal to y, less than or equal to 2x2 + 1 / 4, find the value of a, B, C

Let f (x) = (2-A) x ^ 2-bx + 1 / 4-c
Then f (x) is equal to or greater than 0
If a = 2, then B = 0. C is less than or equal to 1 / 4
If a & lt; 2, then B ^ 2-4 (2-A) (1 / 4-C) is less than or equal to 0

For the function f (x) = ax ^ 2 + (B + 1) x + B-2 (a ≠ 0), if there is a real number x0 and f (x0) = x0 holds, then x0 is called the fixed point of F (x)
(1) When a = 2, B = - 2, find the fixed point of F (x);
(2) If for any real number B, function f (x) always has two equal fixed points, find the value range of real number a
(3) Under the condition of (2), if the abscissa of two points a and B on the image of y = f (x) is the fixed point of function f (x), and the line y = KX + (1 / 2A ^ 2 + 1) is the vertical bisector of line AB, the value range of real number B is obtained
Third, why let a (x1, x1), B (X2, x2) and abscissa and ordinate be equal

First of all, two points AB are on the image, and the abscissa of two points a and B are fixed points of function f (x). Then two points AB must satisfy f (X.) = X. that is to say, abscissa and ordinate of two points AB are equal

Given the function f (x) = ax3-3x2 + 1, if f (x) has a unique zero point x0 and x0 > 0, then the value range of a is
When a = 0, f (x) = - 3x ^ 2 + 1 has two zeros
When a is not equal to 0, f '(x) = 3ax ^ 2-6x = 3x (AX-2), the extreme point x = 0,2 / A is obtained
When a > 0, f (0) = 1 is the maximum and f (2 / a) = - 4 / A ^ 2 + 1 is the minimum;
There must be a zero at (- ∞, 0). How do you know?

According to monotonicity
a> When 0, the monotone increasing interval is x2 / A, and the monotone decreasing interval is (0,2 / a)
Since f (0) = 1, f (- ∞) = - ∞, the value of X is 0

From place a to place B, Xiaoming walks 24 kilometers in 5 hours and Xiaohong walks 36 kilometers in 8 hours. Which of them is fast?

Xiao Ming: 24 △ 5 = 4.8 (km)
Xiaohong: 36 △ 8 = 4.5 (km)
4.8>4.5
A: Xiao Ming is fast

2cm
/|It's high
/_ |_ 1.6m___ \ |
|Don't paint the windows!
3m | 1.2m |
|___________ |It's the window
6m
If this wall is painted on both sides, it costs 339 yuan in total. How much does it cost per square meter on average?
2) The perimeter of a rectangular lawn is 24 meters. If its length and width are increased by 3 meters, how many square meters will the lawn area increase?

1
The top 2cm, isn't it 2m
Total wall area (6 + 2) △ 2 × 3 = 12 square meters
Window area (1.2 + 1.6) △ 2 × 1.5 = 2.1
Net area of wall 12-2.1 = 9.9
339 ÷ (2 × 9.9) = 17.12 yuan
I don't know if I'm right. This picture is really hard to guess
two
24 △ 2 × 3 + 3 × 3 = 45 square meters

1. The sum of divisor and divisor is 320, and the quotient is 7. What are the divisor and divisor?
2. Add one zero to the end of a number to get another number. The sum of the two numbers is 143. Find the original number

(1) Divisor: 320 / (7 + 1) = 40
Divisor: 40 * 7 = 280
Think: divisor is one, divisor is seven, a total of eight 320 / (7 + 1)
(2) Original 143 / 11 = 13
Think: add a zero to the end of a number, and expand it to 10 times of the original number. The original number is 1, and the current number is 10, totally 11. 143 / 11 = 13
It can also be expressed as follows:
(1) Let the divisor be x and the divisor be seven X
x+7x=320
8x=320
x=320/8
x=40
(2) Let the original number be x, then the present number is 10x
x+10x=143
11x=143
x=143/11
x=13

About 200 words of reflection after mathematics test in grade one of junior high school
Contact reality: almost full score

Review 1: today, with guilt and regret, I write this review to you to show you my deep understanding of the bad behavior of speaking in class and my determination not to gossip in class any more. I am very ashamed of my mistake this time. I really shouldn't speak in the morning when I study by myself. I shouldn't

I got 38 points on the 120 point math problem. I only had 40 minutes to do the 120 minute math problem. If I pass, I need 100 words

Teacher, I'm wrong, really. Repeat it a hundred times

For the first mathematics midterm examination reflection, more than 200 words!

This time, I didn't do well in the math test. I reflected on it for several reasons
One is that I was careless and thought that the topic was very simple. Who knows it was a little more difficult than I thought. When I saw it, I felt dizzy. I thought about the problem for a long time, but I still didn't write it right. Therefore, I delayed my time and wrote the simple topic in a hurry, but I didn't check it
Second, I was too careless. I wrote the simple topic very quickly. I thought there was no problem. I didn't check it at all. It's really a mistake I shouldn't make. Next time, I must not do it so fast. Don't worry. I have time
Third, I didn't review before. I thought there was something good to review in mathematics. If I can, I can't. It's all questions. There's nothing I don't know about questions. It must be very easy
In fact, before the teacher and parents have repeatedly reminded us, we must seriously review, seriously do the problem, do not be careless, now it seems, I did not do, no wonder I did not test well, later I have to remember these, no longer make such mistakes should not be made