Parabola y = - X & # 178; and straight line y = 3x + m both pass through point (2, n) (1), try to find m, n 2) If a parabola with a downward opening and a Y-axis just passes (m, n), can you determine the expression of the parabola?

Parabola y = - X & # 178; and straight line y = 3x + m both pass through point (2, n) (1), try to find m, n 2) If a parabola with a downward opening and a Y-axis just passes (m, n), can you determine the expression of the parabola?

1. Take x = 2 into y = - X & # 178; = - 2 & # 178; = - 4, so n = - 4. So (2, - 4) is also the point on the straight line y = 3x + m, take (2, - 4) into the linear equation y = 3x + m, get: - 4 = 3 × 2 + m, so: M = - 10, so: M = - 10, n = - 42. Let the expression of this parabola be y = ax & # 178; take point (m, n) as (- 10
(1) Take point (2, n) into y = - X & # to get n = - 4, then substitute y = 3x + m to get m = - 10
2) Point (- 10, - 4)
Let y = ax & # 178; + B, because the opening is downward, so a is negative, only 100A + B = - 4, other unknown questions: the answer to the second question is y = - 1 / 25X & # 178; or y = - 1 / 50x & # 178; - 2
The known set a = {x | x ^ 2-2x-24
=I did it, but I'm not sure. It's the right answer.
……。
After factorization
A＝｛x|-4
This is very easy. You just need to decompose the second formula into (x-a) * (x-3a) = 0
Suppose a > o he a
What's the difference between factorization of algebra and cross factorization,
Factorization is one of the important means of algebraic identity transformation. Factorization is based on learning four arithmetic operations of integral. It not only has direct application in division and simple operation of polynomials, but also provides necessary foundation for learning reduction and general division of fractions, solution (Group) and identity transformation of three solution functions, Therefore, it is of great significance to learn factorization well for the follow-up study of algebra knowledge, Therefore, this section focuses on the concept of factorization. The method of seeking factorization by integral multiplication is a process of reverse thinking, which is unfamiliar to junior high school students and difficult to accept. Moreover, this section does not cover the specific method of factorization, so we should understand the relationship between factorization and integral multiplication, It is difficult to find the method of factorization by using the relationship between them
Although cross multiplication is difficult to learn, once we learn it and use it to solve problems, it will bring us a lot of convenience. Here are my personal opinions on cross multiplication
1. The method of cross multiplication: multiplication on the left side of cross equals the coefficient of quadratic term, multiplication on the right side equals the constant term, cross multiplication and addition equals the coefficient of primary term
2. The use of cross phase multiplication: (1) to decompose the factor by cross phase multiplication. (2) to solve quadratic equation with one variable by cross phase multiplication
3. Advantages of cross phase multiplication: the speed of solving problems by cross phase multiplication is relatively fast, which can save time, and the amount of calculation is not large, so it is not easy to make mistakes
4. The disadvantages of cross multiplication: 1. Some problems are easy to solve by cross multiplication, but not every problem is easy to solve by cross multiplication. 2. Cross multiplication is only suitable for quadratic trinomial problems. 3. Cross multiplication is difficult to learn
5. Examples of solving cross phase multiplication problems:
1) Use cross multiplication to solve some simple and common problems
Example 1 decomposes M & sup2; + 4m-12 into factors
Analysis: in this problem, the constant term - 12 can be divided into - 1 × 12, - 2 × 6, - 3 × 4, - 4 × 3, - 6 × 2, - 12 × 1. When - 12 is divided into - 2 × 6, it is in line with this problem
Because 1-2
1 ╳ 6
So M & sup2; + 4m-12 = (m-2) (M + 6)
Example 2 decomposes 5x & sup2; + 6x-8 into factors
Analysis: in this problem, 5 can be divided into 1 × 5, - 8 can be divided into - 1 × 8, - 2 × 4, - 4 × 2, - 8 × 1
Because 1 2
5 ╳ -4
So 5x & sup2; + 6x-8 = (x + 2) (5x-4)
Example 3 solves the equation x & sup2; - 8x + 15 = 0
Analysis: if X & sup2; - 8x + 15 is regarded as a quadratic trinomial about X, then 15 can be divided into 1 × 15,3 × 5
Because 1-3
1 ╳ -5
So the original equation is deformable (x-3) (X-5) = 0
So X1 = 3, X2 = 5
Example 4. Solve the equation 6x & sup2; - 5x-25 = 0
Analysis: take 6x & sup2; - 5x-25 as a quadratic trinomial of X, then 6 can be divided into 1 × 6,2 × 3, - 25 can be divided into - 1 × 25, - 5 × 5, - 25 × 1
Because 2-5
3 ╳ 5
So the original equation can be changed to form (2x-5) (3x + 5) = 0
So X1 = 5 / 2, X2 = - 5 / 3
2) Use cross multiplication to solve some difficult problems
Example 5: Factoring 14x & sup2; - 67xy + 18y & sup2
Analysis: if 14x & sup2; - 67xy + 18y & sup2; is regarded as a quadratic trinomial about X, then 14 can be divided into 1 × 14,2 × 7,18y & sup2; and y.18y, 2y.9y, 3y.6y
Solution: because 2 - 9y
7 ╳ -2y
So 14x & sup2; - 67xy + 18y & sup2; = (2x-9y) (7x-2y)
Example 6 decomposes 10x & sup2; - 27xy-28y & sup2; - x + 25y-3 into factors
Analysis: in this problem, we should arrange this polynomial into the form of quadratic trinomial
Solution 1: 10x & sup2; - 27xy-28y & sup2; - x + 25y-3
=10x²-（27y+1）x -（28y²-25y+3） 4y -3
7y ╳ -1
=10x²-（27y+1）x -（4y-3）（7y -1）
=[2x -（7y -1）][5x +（4y -3）] 2 -（7y – 1）
5 ╳ 4y - 3
=（2x -7y +1）（5x +4y -3）
Note: in this problem, 28y & sup2; - 25y + 3 is decomposed into (4y-3) (7y-1) by cross phase multiplication, and then 10x & sup2; - (27y + 1) x - (4y-3) (7y-1) is decomposed into [2x - (7y-1)] [5x + (4y-3)]
Solution 2: 10x & sup2; - 27xy-28y & sup2; - x + 25y-3
=（2x -7y）（5x +4y）-（x -25y）- 3 2 -7y
=[（2x -7y）+1] [（5x -4y）-3] 5 ╳ 4y
=（2x -7y+1）（5x -4y -3） 2 x -7y 1
5 x - 4y ╳ -3
Note: in this problem, first, 10x & sup2; - 27xy-28y & sup2; is decomposed into (2x - 7Y) (5x + 4Y) by cross phase multiplication, and then (2x - 7Y) (5x + 4Y) - (x - 25y) - 3 is decomposed into [(2x - 7Y) + 1] [5x - 4Y) - 3] by cross phase multiplication
Example 7: the solution of X equation: X & sup2; - 3ax + 2A & sup2; – ab - B & sup2; = 0
Analysis: 2A & sup2; – ab-b & sup2; can be factorized by cross phase multiplication
x²- 3ax + 2a²–ab -b²=0
x²- 3ax +（2a²–ab - b²）=0
x²- 3ax +（2a+b）（a-b）=0 1 -b
2 ╳ +b
[x-（2a+b）][ x-（a-b）]=0 1 -（2a+b）
1 ╳ -（a-b）
So X1 = 2A + B, X2 = a-b
There are many ways of factoring
The cross method is the simplest and most commonly used one
Factorization is: square difference formula. Complete square
Cross decomposition is one of them
Can we judge the congruence of two triangles by the condition that "the diagonal correspondence of two sides and one of them is equal"? Why?
The point is the map!
May not be congruent, that is to say not necessarily congruent
As shown in the figure below,
In triangle AOB and triangle AOC:
Ao shared edge is naturally equal, ab = AC is another equal edge,
If the third condition is equal, the angles of AB and AC are equal, and the angles o are also equal,
However, triangle AOB and triangle AOC are not congruent
 
One of the following formulas that can be factorized by the square difference formula is ()
①9x²+4y²
②9x²+(-4y)²
③-9x²-4y²
④-9x²+4y²
The answer is:
（4） -9x²+4y² = (2y+3x)(2y-3x)
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Organize knowledge, help others, and please yourself
"Mathematical infinite" team welcome you
②9x²+(-4y)²=（3x+2y）（3x-2y）
④-9x²+4y²=（2y+3x）（2y-3x）
Are there two obtuse triangle congruents whose diagonals of two sides and one of them are equal?
Give reasons
It's better to have a picture
Not necessarily, if the angle between the two sides of a triangle is obtuse, and the angle between the two sides of another triangle is acute, it is not equal
1.1/9x & # 178; - 1 / 25y & # 178; 2. X4 power - 16 3. - 9x & # 178; + 24xy-16y & # 178; 4. X4 power - 18x & # 178; + 81 5.4a4 power + 12a & # 178; B & # 178; + 9b4 power
1/9x^2-1/25y^2=(1/3x+1/5y)(1/3x-1/5y)x^4-16=(x^2+4)(x^2-4)=(x^2+4)(x+2)(x-2)-9x^2+24xy-16y^2=-(9x^2-24xy+16y^2)=-(3x-4y)^2x^4-18x^2+81=(x^2-9)^2=(x+3)^2(x-3)^24a^4+12a^2b^2+9b^4=(2a^2+3b^2)^2
Mathematics talent team for you
1.1/9X²-1/25y² = （1/3X - 1/5Y ）（1/3X + 1/5Y）
2. X4 power - 16 = (X & # 178; + 4) (X & # 178; - 4) = (X & # 178; + 4) (x + 2) (X-2)
3.-9x²+24xy-16y² = -（3X - 4Y）²
4. X4 power - 18x & # 1... Expansion
Mathematics talent team for you
1.1/9X²-1/25y² = （1/3X - 1/5Y ）（1/3X + 1/5Y）
2. X4 power - 16 = (X & # 178; + 4) (X & # 178; - 4) = (X & # 178; + 4) (x + 2) (X-2)
3.-9x²+24xy-16y² = -（3X - 4Y）²
4. X4 power - 18x & # 178; + 81 = (X & # 178; - 9) &# 178; = (x + 3) &# 178; (x-3) &# 178;
5.4a4 power + 12a & # 178; B & # 178; + 9b4 power = (2a & # 178; + 3B & # 178;) &# 178; put away
Two triangles are congruent if the angles between the two sides and them are equal. Can we judge the congruence of two triangles by the condition of "the opposite angles of the two sides and one side are equal"? Reason! Why?
Whether the two triangles of "the opposite angles of two sides and one side are equal" are congruent or not can be converted into:
Can "the diagonally constant sides and one side" determine a triangle
Let's discuss: we know that in ⊿ ABC, edges AB, BC and ∠ C are constant. Is ⊿ ABC the only triangle?
1, when ∠ a ≠ 90 & # 178
It is obvious that ⊿ ABC and ⊿ DBC meet the conditions;
2,∠A=90²
Let B be the center of the circle, let AB be the radius, draw an arc, and cross AC at a. obviously, ⊿ ABC is the only one that meets the condition
According to the results of 1 and 2, it can be concluded that: 1
1. The two triangles of "the opposite angles of two sides and one side are equal" are not necessarily congruent,
2. Special case: in two right triangles, two right triangles are congruent if both sides and one of the acute angles are equal
No, because the two sides of the triangle and the diagonal of one side know that we can't determine the triangle. Even triangles are not sure how to be congruent
That's the angle. There's no such formula, only the angle (AAS).
2xy-xz
6x^2+9x
4x^2-9
x^2-4x+4
x^2-16y^2
9x^2-6x+1
2x^2y-8xy+8y
3x（a-b）-2y（b-a）
a^2（y-3）+（3-y）
x^2（x-y)-(x-y)
4xy-4x^2y-y
(3a+2)^2-(a-6）^2
x^4-8x^2+16
I've raised my reward score
x(2y-z)
3x(2x + 3)
4(x-3/2)(x+3/2)
(x - 2)^2
(x -4y)(x+4y)
(3x -1)^2
2y(x -2)^2
(a-b)(3x+2y)
(y-3)(a-1)(a+1)
(x-y)(x-1)(x+1)
-y(2x-1)^2
First, we use the square difference formula to get: (3a + 2 + a-6) [(3a + 2) - (a-6)], and sort out, 8 (A-1) (a + 4)
(x^2-4)^2=(x+2)^2*(x-2)^2
Too few points, too many questions and too simple
If we know that two triangles are congruent with the same diagonal on both sides and one side, why is this sentence wrong
Please be clear
\x0dAC=AD \x0dAB