由于各种原因，您可能需要能够定义所选二次函数的最大值或最小值。如果你的原始函数写成一般形式，你可以找到最大值或最小值， f(x)=ax2+bx+c{displaystyle f(x)=ax^{2}+bx+c}
, or in standard form, f(x)=a(x−h)2+k{displaystyle f(x)=a(x-h)^{2}+k}
. Finally, you may also wish to use some basic calculus to define the maximum or minimum of any quadratic function.
Steps
Method 1 of 3: Beginning with the General Form of the Function
步骤 1. 以通用形式设置功能。
二次函数是具有 x2{displaystyle x^{2}}
term. It may or may not contain an x{displaystyle x}
term without an exponent. There will be no exponents larger than 2. The general form is f(x)=ax2+bx+c{displaystyle f(x)=ax^{2}+bx+c}
. If necessary, combine similar terms and rearrange to set the function in this general form.
- For example, suppose you start with f(x)=3x+2x−x2+3x2+4{displaystyle f(x)=3x+2x-x^{2}+3x^{2}+4}
. Combine the x2{displaystyle x^{2}}
terms and the x{displaystyle x}
terms to get the following in general form:
- f(x)=2x2+5x+4{displaystyle f(x)=2x^{2}+5x+4}
步骤 2. 确定图形的方向。
二次函数产生抛物线图。抛物线要么向上开口，要么向下开口。如果 a{displaystyle a}
, the coefficient of the x2{displaystyle x^{2}}
term, is positive, then the parabola opens upward. If a{displaystyle a}
is negative, then the parabola opens downward. Look at the following examples:
- For f(x)=2x2+4x−6{displaystyle f(x)=2x^{2}+4x-6}
, a=2{displaystyle a=2}
so the parabola opens upward.
- For f(x)=−3x2+2x+8{displaystyle f(x)=-3x^{2}+2x+8}
, a=−3{displaystyle a=-3}
so the parabola opens downward.
- For f(x)=x2+6{displaystyle f(x)=x^{2}+6}
, a=1{displaystyle a=1}
so the parabola opens upward.
- If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value.
步骤 3. 计算 -b/2a。
−b2a{displaystyle -{frac {b}{2a}}} 的值
tells you the x{displaystyle x}
value of the vertex of the parabola. When the quadratic function is written in its general form of ax2+bx+c{displaystyle ax^{2}+bx+c}
, use the coefficients of the x{displaystyle x}
and x2{displaystyle x^{2}}
terms as follows:
- For a function f(x)=x2+10x−1{displaystyle f(x)=x^{2}+10x-1}
, a=1{displaystyle a=1}
and b=10{displaystyle b=10}
. Therefore, find the x-value of the vertex as:
- x=−b2a{displaystyle x=-{frac {b}{2a}}}
- x=−10(2)(1){displaystyle x=-{frac {10}{(2)(1)}}}
- x=−102{displaystyle x=-{frac {10}{2}}}
- x=−5{displaystyle x=-5}
- x=−b2a{displaystyle x=-{frac {b}{2a}}}
- As a second example, consider the function f(x)=−3x2+6x−4{displaystyle f(x)=-3x^{2}+6x-4}
. In this example, a=−3{displaystyle a=-3}
and b=6{displaystyle b=6}
. Therefore, find the x-value of the vertex as:
- x=−b2a{displaystyle x=-{frac {b}{2a}}}
- x=−6(2)(−3){displaystyle x=-{frac {6}{(2)(-3)}}}
- x=−6−6{displaystyle x=-{frac {6}{-6}}}
- x=−(−1){displaystyle x=-(-1)}
- x=1{displaystyle x=1}
步骤 4. 找到相应的 f(x) 值。
将刚才计算的x的值插入到函数中，求出f(x)对应的值。这将是函数的最小值或最大值。
- 对于上面的第一个例子， f(x)=x2+10x−1{displaystyle f(x)=x^{2}+10x-1}
, you calculated the x-value for the vertex to be x=−5{displaystyle x=-5}
. Enter −5{displaystyle -5}
in place of x{displaystyle x}
in the function to find the maximum value:
- f(x)=x2+10x−1{displaystyle f(x)=x^{2}+10x-1}
- f(−5)=(−5)2+10(−5)−1{displaystyle f(-5)=(-5)^{2}+10(-5)-1}
- f(−5)=25−50−1{displaystyle f(-5)=25-50-1}
- f(−5)=−26{displaystyle f(-5)=-26}
- For the second example above, f(x)=−3x2+6x−4{displaystyle f(x)=-3x^{2}+6x-4}
, you found the vertex to be at x=1{displaystyle x=1}
. Insert 1{displaystyle 1}
in place of x{displaystyle x}
in the function to find the maximum value:
- f(x)=−3x2+6x−4{displaystyle f(x)=-3x^{2}+6x-4}
- f(1)=−3(1)2+6(1)−4{displaystyle f(1)=-3(1)^{2}+6(1)-4}
- f(1)=−3+6−4{displaystyle f(1)=-3+6-4}
- f(1)=−1{displaystyle f(1)=-1}
步骤 5. 报告您的结果。
回顾你被问到的问题。如果你被要求提供顶点的坐标，你需要同时报告 x{displaystyle x}
and y{displaystyle y}
(or f(x){displaystyle f(x)}
) values. If you are only asked for the maximum or minimum, you only need to report the y{displaystyle y}
(or f(x){displaystyle f(x)}
) value. Refer back to the value of the a{displaystyle a}
coefficient to be sure if you have a maximum or a minimum.
- For the first example, f(x)=x2+10x−1{displaystyle f(x)=x^{2}+10x-1}
, the value of a{displaystyle a}
is positive, so you will be reporting the minimum value. The vertex is at (−5, −26){displaystyle (-5, -26)}
, and the minimum value is −26{displaystyle -26}
- For the second example, f(x)=−3x2+6x−4{displaystyle f(x)=-3x^{2}+6x-4}
, the value of a{displaystyle a}
is negative, so you will be reporting the maximum value. The vertex is at (1, −1){displaystyle (1, -1)}
, and the maximum value is −1{displaystyle -1}
Method 2 of 3: Using the Standard or Vertex Form
步骤 1. 以标准或顶点形式编写二次函数。
一般二次函数的标准形式，也可以称为顶点形式，如下所示：
- f(x)=a(x−h)2+k{displaystyle f(x)=a(x-h)^{2}+k}
- If your function is already given to you in this form, you just need to recognize the variables a{displaystyle a}
, h{displaystyle h}
and k{displaystyle k}
. If your function begins in the general form f(x)=ax2+bx+c{displaystyle f(x)=ax^{2}+bx+c}
, you will need to complete the square to rewrite it in vertex form.
- To review how to complete the square, see Complete the Square.
步骤 2. 确定图形的方向。
就像用一般形式写成的二次函数一样，你可以通过查看系数 a{displaystyle a} 来判断抛物线的方向
. If a{displaystyle a}
in this standard form is positive, then the parabola opens upward. If a{displaystyle a}
is negative, then the parabola opens downward. Look at the following examples:
- For f(x)=2(x+1)2−4{displaystyle f(x)=2(x+1)^{2}-4}
, a=2{displaystyle a=2}
, which is positive, so the parabola opens upward.
- For f(x)=−3(x−2)2+2{displaystyle f(x)=-3(x-2)^{2}+2}
, a=−3{displaystyle a=-3}
, which is negative, so the parabola opens downward.
- If the parabola opens upward, you will be finding its minimum value. If the parabola opens downward, you will find its maximum value.
步骤 3. 确定最小值或最大值。
当函数以标准形式编写时，找到最小值或最大值就像说明变量 k{displaystyle k} 的值一样简单
. For the two example functions given above, these values are:
- For f(x)=2(x+1)2−4{displaystyle f(x)=2(x+1)^{2}-4}
, k=−4{displaystyle k=-4}
. This is the minimum value of the function because this parabola opens upward.
- For f(x)=−3(x−2)2+2{displaystyle f(x)=-3(x-2)^{2}+2}
, k=2{displaystyle k=2}
. This is the maximum value of the function, because this parabola opens downward.
步骤 4. 找到顶点。
如果询问最小值或最大值的坐标，点将是 (h, k){displaystyle (h, k)}
. Note, however, that in the standard form of the equation, the term inside the parentheses is (x−h){displaystyle (x-h)}
, so you need the opposite sign of the number that follows the x{displaystyle x}
- For f(x)=2(x+1)2−4{displaystyle f(x)=2(x+1)^{2}-4}
, the term inside the parentheses is (x+1), which can be rewritten as (x-(-1)). Thus, h=−1{displaystyle h=-1}
. Therefore, the coordinates of the vertex for this function are (−1, −4){displaystyle (-1, -4)}
- For f(x)=−3(x−2)2+2{displaystyle f(x)=-3(x-2)^{2}+2}
, the term inside the parentheses is (x-2). Therefore, h=2{displaystyle h=2}
. The coordinates of the vertex are (2, 2).
Method 3 of 3: Using Calculus to Derive the Minimum or Maximum
步骤 1. 从一般表格开始。
将二次函数写成一般形式， f(x)=ax2+bx+c{displaystyle f(x)=ax^{2}+bx+c}
. If necessary, you may need to combine like terms and rearrange to get the proper form.
- Begin with the sample function f(x)=2x2−4x+1{displaystyle f(x)=2x^{2}-4x+1}
步骤 2. 使用幂律求一阶导数。
使用基本的一年级微积分，您可以找到一般二次函数的一阶导数为 f′(x)=2ax+b{displaystyle f^{prime }(x)=2ax+b}
- For the sample function f(x)=2x2−4x+1{displaystyle f(x)=2x^{2}-4x+1}
, find the derivative as:
- f′(x)=4x−4{displaystyle f^{prime }(x)=4x-4}
步骤 3. 将导数设置为零。
回想一下，函数的导数告诉您函数在所选点处的斜率。当斜率为零时，函数的最小值或最大值出现。因此，要找到最小值或最大值出现的位置，请将导数设置为零。继续上面的示例问题：
- f′(x)=4x−4{displaystyle f^{prime }(x)=4x-4}
- 0=4x−4{displaystyle 0=4x-4}
步骤 4. 求解 x。
当导数为零时，使用代数的基本规则重新排列函数并求解 x 的值。此解决方案将告诉您函数顶点的 x 坐标，即最大值或最小值出现的位置。
- 0=4x−4{displaystyle 0=4x-4}
- 4=4x{displaystyle 4=4x}
- 1=x{displaystyle 1=x}
步骤 5. 将 x 的求解值插入到原始函数中。
函数的最小值或最大值将是 f(x){displaystyle f(x)}
at the selected x{displaystyle x}
position. Insert your value of x{displaystyle x}
into the original function and solve to find the minimum or maximum.
- For the function f(x)=2x2−4x+1{displaystyle f(x)=2x^{2}-4x+1}
at x=1{displaystyle x=1}
- f(1)=2(1)2−4(1)+1{displaystyle f(1)=2(1)^{2}-4(1)+1}
- f(1)=2−4+1{displaystyle f(1)=2-4+1}
- f(1)=−1{displaystyle f(1)=-1}
步骤 6. 报告您的解决方案。
该解决方案为您提供了最大或最小点的顶点。对于这个示例函数， f(x)=2x2−4x+1{displaystyle f(x)=2x^{2}-4x+1}
, the vertex occurs at (1, −1){displaystyle (1, -1)}
. The coefficient a{displaystyle a}
is positive, so the function opens upward. Therefore, the minimum value of the function is the y-coordinate of the vertex, which is −1{displaystyle -1}
