二次函数公式
函数称为代数函数,可以定义为多项式函数的根。这些是包含有限项的代数表达式,其中包含常量和变量。代数函数有很多种类型,其中一些是线性函数、二次函数、三次函数等。示例:
- 2x ⇢ 线性函数
- x 2 + 2x + 3 ⇢ 二次函数
- x 3 + x 2 + 6x + 9 ⇢ 三次函数
二次函数公式
方程中变量的最高指数为 2 的函数称为二次函数。二次函数是具有一个或多个变量的多项式函数。二次方程的标准形式是f(x) = ax 2 + bx + c ,这表示给定方程中的至少一项是平方的。在上述等式中,a、b、c是常数项, x是变量。
f(x) = ax 2 + bx + c,
其中a不等于0,a、b、c为实数。
多项式的次数 = 2。
二次函数的例子:
- 4x 2 + 3x + 5
- 6x 2 + x + 7
- 7x 2 + 5x
- 9x 2
对于任何方程,都有n 个根,其中 n 是多项式的次数。二次方程的根是,
x = (-b ± √ (b² – 4ac) )/2a
二次方程根的推导
ax2 + bx + c = 0
- Start with ax2 + bx + c=0
- Divide the equation by a x2 + (b/a) + c/a = 0
- Put c/a on other side x2 + (b/a) x = -c/a
- Add (b/2a)2 to both sides x2 + (b/a)x + (b/2a)2 = -c/a + (b/2a)2
- The left hand side is now in the x2 + 2dx + d2 format, where “d” is “b/2a”
So re-write it this way:
“Complete the Square” (x+b/2a)2 = -c/a + (b/2a)2
二次方程有不同类型的根,
为了找到根的性质,我们使用了一个称为判别式的术语。该术语被称为判别式,因为它根据其符号确定二次方程的根。
根的性质有3种,
- 真实而独特的根源
对于真实的和不同的根,判别式应该是正的,即b 2 – 4ac > 0。方程的曲线在两个不同的点与 x 轴相交。
- 实等根
对于实数和等根,判别式为零,即b 2 – 4ac = 0。方程的曲线仅在一个点与 x 轴相交。
- 复根
对于复根,判别式为负,即b 2 – 4ac = 0。方程的曲线不与 x 轴相交。
示例问题
问题 1:将二次函数f(x) = (x – 9)(x + 3)写成ax 2 + bx + c的一般形式。
解决方案:
Given, the function as (x – 9)(x + 3)
= x2 + 3x – 9x – 27
= x2 – 6x – 27 is the general form.
问题 2:找出方程 4x 2 + 5x + 9 = 0 的一般形式的常数 a、b、c。
回答:
Given, the equation is 4x2 + 5x + 9 = 0, the general equation of an quadratic equation is ax2 + bx + c = 0.
Therefore from the reference of the general equation, a = 4, b = 5, c = 9.
问题 3:将二次函数f(x) = (x + 8)(x – 3) 写成 ax 2 + bx + c 的一般形式。
解决方案:
Given, the function as (x + 8)(x – 3)
= x2 – 3x + 8x – 24
= x2 + 5x – 24 is the general form
问题 4:求方程 2x 2 – 4x + 2 = 0 的根。
解决方案:
Here a = 2, b = -4, c = 2, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.
discriminant = b2 – 4ac = (-4)2 – 4(2)(2) = 16 – 16 = 0, which is equal to zero. So, it has real and equal roots.
roots = (−b ± √(b2 − 4ac)) / 2a
= (-(-4) ± √(0) ) / 2(4) )
= 1 is the root of the equation.
问题 5:求方程 4x 2 – 3x + 3 的根。
解决方案:
Here a = 4, b = -3, c = 3, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.
discriminant = b2 – 4ac = (-3)2 – 4(4)(3) = 9 – 48 = -39, which is negative. So, it has complex roots.
roots = (−b ± √(b2 − 4ac)) / 2a
= (-(-3) ± √-39 / 2(4) )
= (3 ± 39i)/8 are the roots of the quadratic equation.
问题 6:求二次方程 6x 2 – 8x + 2 = 0 的根。
解决方案:
Here a = 6, b = -8, c = 2, to find the roots of the equation, first we need to find the discriminant value which helps us to know the nature of roots.
discriminant = b2 – 4ac = (-8)(-8) – 4(6)(2) = 64 – 48 = 16, which is positive. So, it has real and distinct roots.
x = (-b ± √ (b² – 4ac) )/2a
= (-(-8) ± √((-8)(-8) – 4(6)(2))) / 2(6)
= ( 8 ± √16) / 12
= ( 8 ± 4)12
= 1/3 and 1 are the roots of the equation.