急求一篇二次函数的英文文献

数与方程是初中数学中两个最基本的概念，它们的形式虽然不同，但本质上是相互连接的，有密切关系。如：一元二次方程与二次函数。
我们知道形如ax2+bx+c=0的方程是一元二次方程，而形式为y= ax2+bx+c（a、b、c为常数，a≠0）是二次函数。它们在形式上几乎相同，差别只是一元二次方程的表达式等于0，而二次函数的表达式等于y。这种形式上的类似使得它们之间的关系格外密切，很多题型都是以此来命题。为什么会这样？主要是因为当二次函数中的变量y取0时，二次函数就变成一元二次方程。由此可见，方程中的很多知识点可以运用在函数中。下面，我们就它们间的具体运用详细的了解一下。
一、 配方法解方程与二次函数的应用关系
在解方程的四种方法就有一种用配方法来解方程的。而在二次函数中，我们经常要将一般形式 转化成 的样式，这个转化过程实际上就是对其进行配方，与方程配方相同。
例1：用配方法解方程
解：
（1）
（2）
（3）
（4）
……
例2：指出函数 的顶点坐标。
解：
（5）
（6）
（7）
（8）
∴顶点为（－2，－17）
方程中的（1）、（2）、（3）、（4）四个步骤与函数中的（5）、（6）、（7）、（8）四个步骤的方法是完全一样的。可见，方程与函数密切相关。
我们通过课本的学习可知；二次函数y= ax2+bx+c（a≠0）的图象与x轴有交点时，交点横坐标的值就是方程ax2+bx+c=0（a≠0）的根。
二、 一元二次方程根的判别式与二次函数的结合应用
在二次函数中，当函数与x轴分别有两个交点、一个交点和无交点时，该函数所对应的一元二次方程根的判别式分别是：△>0、△＝0和△<0。而在一元二次方程中有以下结论：当△>0时，方程有两个不相等的实数根；当△＝0时，方程有两个相等的实数根；当△<0时，方程无实数根。
例3：判断二次函数y= x2－4x+3与x轴的交点个数
分析：因为二次函数与x轴的交点个数可由对应方程根的判别式△来确定。若△>0，则有两个交点；若△＝0，则有一个交点；若△<0，则无交点。该题中△＝4>0，所以有两个交点。
例4：试说明函数y= x2－4x+5，无论x取何值，y>0。
分析：第一种方法：用配方法将其化成y= （x－2）2 +1的形式来说明。（但如果系数取值不好，该方法就比较麻烦）
第二种方法：用△来说明，因为△＝－4<0，所以函数与x轴无交点，又因为该函数的二次项系数a＝1>0，所以图象开口向上。于是，图象在x轴上方，因此无论x取何值，y>0。
例5：求证：不论m取什么实数，方程x2－（m2+m）x+m－2＝0必有两个不相等的实数根。
分析：这道题如果用常规做法，就是证明一元二次方程的△>0的问题。然而本题的判别式△是一个关于m的一元四次多项式，符号不易判断，这就给证明带来了麻烦，若用函数思想分析题意，设f（x）＝x2－（m2+m）x+m－2，由于它的开口向上，所以只要找到一个实数x0，使得f（x0）<0，就说明这个二次函数的图象与x轴有两个交点，问题就得到了解决。
注意观察，容易发现当x＝1时，f（1）＝1－（m2+m）+m－2＝－m2－1<0，故这个图象必与x轴有两个交点。
这就说明要证明的结论是成立的。
证明 略。
三、 一元二次方程中根与系数的关系在函数中的应用
例6：二次函数图象过点（－1，0）、（3，0），且与y轴交于（0，3），求函数解析式。
分析：此类题型的常规解法是待定系数法。然而在这里可以用根与系数的关系来解，因为（－1，0）、（3，0）实际在x轴上，所以－1和3是函数所对应方程的两个根。
解：设函数形式为
∵函数过点（0，3）
∴ c＝3
∴
又∵函数过点（－1，0）、（3，0）
即函数与x轴交点的横坐标是－1和3
∴

解得 a＝－1，b＝2
∴函数形式为y= －x2＋29x+3
很明显，此方法要比待定系数法简单。
一元二次方程与二次函数之间的密切关系还有很多巧妙的用处。在这里，我们只探讨这么多，更多的地方需要在实践中去慢慢体会。

Number of junior high school mathematics and equations are the two basic concepts, they form are different, but essentially interconnected, closely related. Such as: a quadratic equation and quadratic function.
We know that the form ax2 + bx + c = 0 the equation is a quadratic equation, but the form y = ax2 + bx + c (a, b, c are constants, a ≠ 0) is a quadratic function. They form almost the same, the difference is only a quadratic equation of the expression is equal to 0, and quadratic function of the expression is equal to y. This form makes a similar relationship between them is particularly close to a lot of Questions in this Proposition are. Why? Mainly because when a quadratic function of the variable y of 0, the quadratic function becomes quadratic equation. Thus, the equation can be used in many knowledge points in the function. Here, we use them inter-specific detailed to find out.
1, with method to solve quadratic equations and the application of relations
In the equations for the four methods have a kind used with methods to solve the equation. In the second function, we often want to convert the general form of the style, this conversion process is actually carried out their formula, the same formula with the equation.
Example 1: Solving Equations with the Method
Solution:
(1)
(2)
(3)
(4)
... ...
Example 2: pointed out that the function of the vertex coordinates.
Solution:
(5)
(6)
(7)
(8)
∴ vertices (-2, -17)
Equation (1), (2), (3), (4) four steps with the function of (5), (6), (7), (8) four steps of the method is exactly the same. Can be seen, the equation is closely related with the function.
We know that learning through textbooks; quadratic function y = ax2 + bx + c (a ≠ 0) of the image with the x-axis there is an intersection, the intersection of the horizontal value is the equation ax2 + bx + c = 0 (a ≠ 0 ) roots.
Second, a quadratic equation quadratic root of the discriminant function with the combined application
In the second function, the function and the x-axis when the two intersection points, respectively, an intersection and no intersection point, the function of a quadratic equation corresponding to the root of the discriminant are: △> 0, △ = 0 and △ < 0. In the quadratic equation has the following conclusion: When △> 0, the equation has two unequal real roots; When △ = 0, the equation has two equal real roots; when △ <0, the equation has no real roots.
Example 3: Determine the quadratic function y = x2-4x +3 x-axis with the intersection number
Analysis: as a quadratic function and the x-axis corresponds to the intersection number can be the root of the discriminant equation △ to determine. If △> 0, then there are two intersection; if △ = 0, then there is a point of intersection; if △ <0, no intersection. The title of △ = 4> 0, so there are two intersection.
Example 4: Explain the function y = x2-4x +5, no matter what value x, y> 0.
Analysis: The first method: The method with its into y = (x-2) 2 +1 in the form of instructions. (However, if the coefficient is not good, the method was more trouble)
The second method: use △ to illustrate, as △ =- 4 <0, so the function and the x-axis non-intersection, and because the function of the quadratic coefficient a = 1> 0, so opening up image. Thus, images in the x-axis side, so no matter what value x, y> 0.
Example 5: confirmation: whether to take any real number m, the equation x2-(m2 + m) x + m-2 = 0 must have two unequal real roots.
Analysis: This question if the conventional approach is to prove that quadratic equation of △> 0 problem. However, the question of discriminant △ is a one dollar four times on the m, polynomial, symbolic difficult to determine, which brought trouble to prove, if thought analyze the question with a function, let f (x) = x2-(m2 + m) x + m-2, because it's opening up, so long as to find a real number x0, so f (x0) <0, the indicator of the quadratic function of the image with the x-axis there are two intersection, the problem has been solution.
Observe, easy to find when x = 1,, f (1) = 1 - (m2 + m) + m-2 =- m2-1 <0, so the image will have two intersection points with the x-axis.
This shows that to prove the conclusion is established.
Proved slightly.
Third, a quadratic equation in the relationship between roots and coefficients of the function
Example 6: quadratic function over image point (-1,0), (3,0), and with the y-axis intersect at (0,3), find the analytic function.
Analysis: Questions in the conventional method is of such undetermined coefficient method. However, here the relationship between roots and coefficients used to solve, because the (-1,0), (3,0) actually in the x-axis, so -1 and 3 is the function corresponding to the equation of the two roots.
Solution: set function in the form of
∵ function through point (0,3)
∴ c = 3
∴
Also ∵ function through point (-1,0), (3,0)
That function with the x-axis intersection of the horizontal axis is -1 and 3
∴

Solutions have a =- 1, b = 2
∴ function of the form y =-x2 +29 x +3
Obviously, this method is simpler than the undetermined coefficient method.
Quadratic equation and quadratic function of the close relationship between the many ingenious use. Here, we only investigate so many more places need to understand in practice slowly.

Number of junior high school mathematics and the equations are two basic concepts, they form are different, but essentially interconnected, closely related. Such as: a quadratic equation and quadratic function.
We know that the form ax2 + bx + c = 0 the equation is a quadratic equation, but the form y = ax2 + bx + c (a, b, c are constants, a ≠ 0) is a quadratic function. They form almost the same, the difference is only a quadratic equation of the expression is equal to 0, and quadratic function of the expression is equal to y. This form makes a similar relationship between them is particularly close to a lot of Questions in this Proposition are. Why? Mainly because when a quadratic function of the variable y of 0, the quadratic function becomes quadratic equation. Thus, the equation can be used in many knowledge points in the function. Here, we use them inter-specific detailed to find out.
1, with method to solve quadratic equations and the application of relations
In the equations for the four methods have a kind used with methods to solve the equation. In the second function, we often want to convert the general form of the style, this conversion process is actually carried out their formula, the same formula with the equation.
Example 1: Solving Equations with the Method
Solution:
(1)
(2)
(3)
(4)
... ...
Example 2: pointed out that the function of the vertex coordinates.
Solution:
(5)
(6)
(7)
(8)
∴ vertices (-2, -17)
Equation (1), (2), (3), (4) four steps with the function of (5), (6), (7), (8) four steps of the method is exactly the same. Can be seen, the equation is closely related with the function.
We know that learning through textbooks; quadratic function y = ax2 + bx + c (a ≠ 0) of the image with the x-axis there is an intersection, the intersection of the horizontal value is the equation ax2 + bx + c = 0 (a ≠ 0 ) roots.
Second, a quadratic equation quadratic root of the discriminant function with the combined application
In the second function, the function and the x-axis when the two intersection points, respectively, an intersection and no intersection point, the function of a quadratic equation corresponding to the root of the discriminant are: △> 0, △ = 0 and △ < 0. In the quadratic equation has the following conclusion: When △> 0, the equation has two unequal real roots; When △ = 0, the equation has two equal real roots; when △ <0, the equation has no real roots.
Example 3: Determine the quadratic function y = x2-4x +3 x-axis with the intersection number
Analysis: as a quadratic function and the x-axis corresponds to the intersection number can be the root of the discriminant equation △ to determine. If △> 0, then there are two intersection; if △ = 0, then there is a point of intersection; if △ <0, no intersection. The title of △ = 4> 0, so there are two intersection.
Example 4: Explain the function y = x2-4x +5, no matter what value x, y> 0.
Analysis: The first method: The method with its into y = (x-2) 2 +1 in the form of instructions. (However, if the coefficient is not good, the method was more trouble)
The second method: use △ to explain, because △ =- 4 <0, so the function and the x-axis non-intersection, and because the function of the quadratic coefficient a = 1> 0, so opening up image. Thus, images in the x-axis side, so no matter what value x, y> 0.
Example 5: confirmation: whether to take any real number m, the equation x2-(m2 + m) x + m-2 = 0 there must be two unequal real roots.
Analysis: This question if the conventional approach is to prove that quadratic equation of △> 0 problem. However, the question of discriminant △ is a one dollar four times on the m, polynomial, symbolic difficult to determine, which brought trouble to prove, if thought analyze the question with a function, let f (x) = x2-(m2 + m) x + m-2, because it's opening up, so long as to find a real number x0, so f (x0) <0, to illustrate the quadratic function of the image with the x-axis there are two intersection, the problem has been solution.
Observe, easy to find when x = 1,, f (1) = 1 - (m2 + m) + m-2 =- m2-1 <0, so the image will have two intersection points with the x-axis.
This shows that to prove the conclusion is established.
Proved slightly.
Third, a quadratic equation in the relationship between roots and coefficients of the function
Example 6: quadratic function over image point (-1,0), (3,0), and with the y-axis intersect at (0,3), find the analytic function.
Analysis: Questions in the conventional method is of such undetermined coefficient method. However, here the relationship between roots and coefficients used to solve, because the (-1,0), (3,0) actually in the x-axis, so -1 and 3 is the function corresponding to the equation of the two roots.
Solution: set function in the form of
∵ function through point (0,3)
∴ c = 3
∴
It ∵ function pass point (-1,0), (3,0)
That function with the x-axis intersection of the horizontal axis is -1 and 3
∴

Solutions have a =- 1, b = 2
∴ function of the form y =-x2 +29 x +3
Obviously, this method is simpler than the undetermined coefficient method.
Quadratic equation and quadratic function of the close relationship between the many ingenious use. Here, we only investigate so many more places need to slowly realize in practice.