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
（1）
（2）
（3）
（4）
……
例2：指出函数 的顶点坐标.
（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,则有两个交点；若△＝0,则有一个交点；若△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△＝－40,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, the equation has two unequal real roots; When △ = 0, the equation has two equal real roots; when △ 0, then there are two intersection; if △ = 0, then there is a point of intersection; if △ 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 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 equation has two unequal real roots; When △ = 0, the equation has two equal real roots; when △ 0, then there are two intersection; if △ = 0, then there is a point of intersection; if △ 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 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)