All analytical expressions of quadratic function of one variable. The requirements include: 1. Opening direction. 2. Vertex. 3. Equation of symmetry axis. 4. Maximum value 5. When the coefficient a is greater than or less than 0, the change of Y What about vertex style?

All analytical expressions of quadratic function of one variable. The requirements include: 1. Opening direction. 2. Vertex. 3. Equation of symmetry axis. 4. Maximum value 5. When the coefficient a is greater than or less than 0, the change of Y What about vertex style?

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① The vertex formula of quadratic function: y = a (X-H) ^ 2 + k
1. Direction of opening: when a > 0, the opening is upward; when A0, y has the minimum value K; when A0, y decreases with the increase of X in the left half of the symmetry axis; when A0, y increases with the increase of X in the right half of the symmetry axis; when A0, y decreases with the increase of X in the right half of the symmetry axis;
When A0, the opening is upward; when A0, y has the minimum value (4ac-b ^ 2) / 4A; when A0, y decreases with the increase of X in the left half of the axis of symmetry; when A0, y increases with the increase of X in the right half of the axis of symmetry;
When a
y=ax^2+bx+c（a≠0）
Vertex (- B / 2a, (4ac-b ^ 2) / 4A)
Axis of symmetry x = - B / 2A
A > 0, opening upward, there is a minimum, the minimum is the vertex ordinate. Axis of symmetry, left, decreasing; right, increasing.
A ＜ 0, the opening is downward, and there is a maximum value, which is the vertex ordinate. Axis of symmetry, left, increasing; right, decreasing.
The vertex formula, y = a (X-H) ^ 2 + C, where C has nothing to do with C in the general formula and is not necessarily equal. The fixed point coordinates are (h, c)... Expansion
y=ax^2+bx+c（a≠0）
Vertex (- B / 2a, (4ac-b ^ 2) / 4A)
Axis of symmetry x = - B / 2A
A > 0, opening upward, there is a minimum, the minimum is the vertex ordinate. Axis of symmetry, left, decreasing; right, increasing.
A ＜ 0, the opening is downward, and there is a maximum value, which is the vertex ordinate. Axis of symmetry, left, increasing; right, decreasing.
The vertex formula, y = a (X-H) ^ 2 + C, where C has nothing to do with C in the general formula and is not necessarily equal. The fixed point coordinates are (h, c) folded
The general formula of quadratic function of one variable y = ax ^ 2 + BX + C (a is not equal to 0)
Axis of symmetry x = - B / 2A
A > 0, opening upward, there is a minimum, the minimum is the vertex ordinate. On the left side, y decreases with the increase of X, and on the right side, y increases with the increase of X.
A ＜ 0, the opening is downward, and there is a maximum value, which is the vertex ordinate. On the left side of the symmetry axis, y increases with the increase of X; on the right side, y decreases with the increase of X.
y=a(x-h)^2+k
Axis of symmetry x = h
A > 0, opening upward, with... Unfolding
The general formula of quadratic function of one variable y = ax ^ 2 + BX + C (a is not equal to 0)
Axis of symmetry x = - B / 2A
A > 0, opening upward, there is a minimum, the minimum is the vertex ordinate. On the left side, y decreases with the increase of X, and on the right side, y increases with the increase of X.
A ＜ 0, the opening is downward, and there is a maximum value, which is the vertex ordinate. On the left side of the symmetry axis, y increases with the increase of X; on the right side, y decreases with the increase of X.
y=a(x-h)^2+k
Axis of symmetry x = h
A > 0, the opening is upward, and there is a minimum value, which is K. On the left side, y decreases with the increase of X, and on the right side, y increases with the increase of X.
A < 0, the opening is downward, and there is a maximum value, which is K. On the left side of the symmetry axis, y increases with the increase of X; on the right side, y decreases with the increase of X. Put it away
Y = ax ^ 2 + BX + C (a is not equal to 0) is the general formula of quadratic function of one variable
Opening direction: a > 0 opening upward, A0 vertex is the maximum a
Using quadratic function to solve quadratic absolute value equation of one variable
If X & #178; + 2a|x| + 4A & #178; - 3 = 0, and there is only one real root, what is the value of a?
Because if x is a root, then - x is also a root,
If there is only one real root, it can only be x = 0
Because there are 4A and 178; - 3 = 0
A = ± √ 3 / 2
In this case, the equation is reduced to: | x | (| x | + 2a) = 0
To make x = 0 its only root, a > = 0 is required
So we can only take a = √ 3 / 2
Solving the following practical problems with one variable and one degree equation
1. Bicycle jam around the city. The fastest person meets the slowest person 48 minutes after the start. It is known that the speed of the fastest person is three and two times that of the slowest person. A week around the city is 20 kilometers. Find the speed of two people
2. My father is going on a business trip, so he bought some dumplings for my mother and me before leaving. It is known that each bag of dumplings costs 12 yuan, a total of 50 pieces. My mother and I have to eat five times. We have to eat 25 pieces each time. How many bags should we buy? How much did we spend?
First of all, I will help you write the equation 1 yuan once
1. Set the slowest speed to x km / h
So the fastest person's speed is 3.5 × X 48 minutes converted into hours = 48 / 60 hours
When the fastest person meets the slowest person, that is, he is 20km more than the slowest person by one lap
3.5×X×（48/60）＝X×（48/60）＋20
Just solve the equation
2. I want to buy x bags altogether
5×25＝50×X
The solution is x = 2.5 bags
Because it must be full enough to eat, so when rounding x, it should carry all, that is, 3 bags
It costs 12 * 3 = 36 yuan
1. Let the slow speed be x, and the equation (3.5x-x) * 48 = 20
1. Let the slow speed be XKM / min, and the fast speed be 7 / 2xkm / min,
48 * (7 / 2x-x) = 20, x = 1 / 6, so the slow is 1 / 6km / min, and the block is 7 / 12km / min
2. Suppose you want to buy x bag, 50x is greater than or equal to 25 * 5, so x is equal to 3, money = 12 * 3 = 36
1. Suppose the slowest person's speed is x km / min, then the fastest person's speed is 7x / 2km / min
48*(7X/2)=20+48X,
So 48 * (5x / 2) = 20,
So x = 1 / 6,
So 7x / 2 = 7 / 12,
That is, the speed of the slowest person is 1 / 6 km / min, and the speed of the fastest person is 7 / 12 km / min;
2.5*25=125,125/50=2.5,
So I had to buy 3 bags, which cost 12 * 3 =... To start
1. Suppose the slowest person's speed is x km / min, then the fastest person's speed is 7x / 2km / min
48*(7X/2)=20+48X,
So 48 * (5x / 2) = 20,
So x = 1 / 6,
So 7x / 2 = 7 / 12,
That is, the speed of the slowest person is 1 / 6 km / min, and the speed of the fastest person is 7 / 12 km / min;
2.5*25=125,125/50=2.5,
So I had to buy 3 bags, which cost 12 * 3 = 36 yuan. Put it away
1. Set the slowest speed to x km / min
48 * 3 and 1 / 2 x-48x = 20
120x=20
X = 1 / 6
I want to buy x bags
50x=5*25
50x=125
x=2.5
2.5 is about 3
3*12=36
I want to buy three bags. 36 yuan.
It is known that the divisor is X3 + 3x2-1, the quotient is x and the remainder is - 1, then the divisor is X______ .
[x3+3x2-1-（-1）]÷x，=（x3+3x2）÷x，=x2+3x．
The meeting problem of mathematical equation
If the two vehicles start at the same time, they will meet eight hours later; if vehicle a starts four and one-third hours first, then vehicle B will meet six hours later to find the speed of vehicle a and vehicle B
With binary linear equation, the solution process please write in detail, this problem is actually very simple, is too large, not easy to calculate, want to see if there is any simple algorithm, thank you!
Let a's velocity be x and B's velocity be y, then 8x + 8y = 1456, (13 / 3 + 6) x + 6y = 1456. If 8 (x + y) = 1456 is substituted by 13 / 3x + 6 (x + y) = 1456, then 13 / 3x + (6 / 8) × 1456 = 1456, 13 / 3x = 1456 / 4, x = (1456 × 3) / (13 × 4) = 112 × 3 / 4 = 84, y = 1456 / 8-84 = 98
A speed x b y
8×（X+Y）=1456
(13/3+6) X+6Y=1456
X=84
Y=98
Let the speed of car a be XKM / h and that of car B be YKM / h
8（x+y）=1456
13/3x+6(x+y)=1456
Solve the equation and ask: how about the number There is no process, but at least it has to be counted
Given that P (x, y) is a point on the circle X & # 178; + Y & # 178; = 4, find the maximum and minimum of Y + 2 / x + 2 √ 3
Let (y + 2) / (x + 2 √ 3) = K
Then y = K (x + 2 √ 3) - 2
Substituting: X & # 178; + Y & # 178; = 4
We get: (1 + K & # 178;) x & # 178; + (4 √ 3K & # 178; - 4K) x + (12K & # 178; - 8 √ 3K) = 0
△=16k²（3k²-2√3k+1）-4（1+k²）（12k²-8√3k）
=-32k²+32√3k
≥0
The solution is 0 ≤ K ≤√ 3
So the minimum value of (y + 2) / (x + 2 √ 3) is 0 and the maximum value is √ 3
Here, I use the pure algebraic method, but I can also use the geometric method
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How to solve the meeting problem of one variable one time application problem
The distance between two places is 490 km. A and B start from two places and run in opposite directions. If they start at the same time, they will meet in 7 hours. If a drives for 7 hours first, B starts again. As a result, B starts and meets in 2 hours. The speed of two cars is calculated
Let the speed of car a be X
Because the combined speed of the two cars is 490 / 7 = 70 (km / h)
So the speed of car B is 70-x (km / h)
The equation is: 7x + 2 * 70 = 490
X=50
So the speed of car B is 70-50 = 20 (km / h)
Let the speed of car a and B be V A and V B respectively
7 * (V A + v b) = 490 V A + v b = 70 (1) V A = 70-v B
9V a + 2V B = 490 (2)
(1) Substituting (2)
9 (70-v b) + 2V B = 490
7V B = 630-490 = 140
Ψ v b = 20 V A = 50
If the speed of car a is respectively x km / h, then the speed of car B is (490-7x) / 7 km / h = (70-x) km / h,
If a starts for 7 hours, B starts again, and B starts for 2 hours. That is to say, a will drive for 7 hours first, and B will drive for 2 hours after departure.)
7x+2X+2(70-X)=490
7X+2X+140-2X=490
7X=490-140
X = 50... Expand
If the speed of car a is respectively x km / h, then the speed of car B is (490-7x) / 7 km / h = (70-x) km / h,
If a starts for 7 hours, B starts again, and B starts for 2 hours. That is to say, a will drive for 7 hours first, and B will drive for 2 hours after departure.)
7x+2X+2(70-X)=490
7X+2X+140-2X=490
7X=490-140
X = 50 ° Stow
F (x) = x & # 178; - 3x + 2 find the maximum value, and explain whether it is the maximum value or the minimum value
f(x)=x²-3x+2
=x²-3x+9/4-¼
=(x-3/2)²-¼
≥-¼
The minimum value is - (?) 188;
First, find the derivative f '(x) = 2x-3, Let f' (x) = 0, then x = 3 / 2, so f (x) has an extreme value when x = 3 / 2, because f '(x) is an increasing function, so there is a minimum value. Take x = 3 / 2 into f (x) to calculate the minimum value!
This problem is relatively simple, if there is no interval limit, then the default is x ∈ R (that is, to solve in the range of all real numbers). You can solve it through the formula.
First, it's a function with an opening up, so there's no maximum, only a minimum. The formula is as follows
f(x)=x²-3x+2
=x²-3x+9/4-¼
=(x-3/2)²-¼
So the minimum value is -, # 188;.
Of course, you can also expand it through
This problem is relatively simple, if there is no interval limit, then the default is x ∈ R (that is, to solve in the range of all real numbers). You can solve it through the formula.
First, it's a function with an opening up, so there's no maximum, only a minimum. The formula is as follows
f(x)=x²-3x+2
=x²-3x+9/4-¼
=(x-3/2)²-¼
So the minimum value is -, # 188;.
Of course, you can also draw an image to further confirm. Drawing an image can also solve the problem.
Hope to help you, do not understand can ask. Put it away
If there is a minimum value, the minimum value can be obtained at the vertex, and the fixed-point coordinates (- B / 2a, (4ac-b ^) / 4A) are substituted into the data to be (3 / 2, - 1 / 4), then the minimum value is when x = 3 / 2, y = - 1 / 4, and the minimum value is - 1 / 4.
In the maintenance project of Wuhan Jianghan No.1 Bridge, it is planned to complete a project jointly by two engineering teams, Party A and Party B. from the data of the two engineering teams, it can be known that if the two engineering teams cooperate, it will be completed within 24 days; if the two engineering teams cooperate for 18 days, Party A will do it alone for 10 days