Section: 解决二次函数的方法 | 2.1 解决赤道函数的方法

Section outline

Can you measure height with a stopwatch? How could that be possible?
::你能用手表测量身高吗?

This lesson is all about quadratic functions, like the one used in Physics and described later in the lesson, which allows a good approximation of the height of a tall object to be calculated with time!
::这堂课都是关于二次函数的, 就像物理课中使用的, 在课中稍后描述的, 这样就可以用时间来计算高物体高度的近似值!

Methods for Solving Quadratic Functions
::解决二次函数的方法

A quadratic function can be described as:
::二次函数可描述为:

Formally: A function $\begin{align*}f\end{align*}$ defined by $\begin{align*}f(x)=ax^{2}+bx+c\end{align*}$ , where $\begin{align*}a, b,\end{align*}$ and $\begin{align*}c\end{align*}$ are real numbers and $\begin{align*}a \ne 0\end{align*}$ .
::形式: f( x) = ax2+bx+c 定义的函数 f, 其中 a、 b 和 c 是真实数字和 a+++0 。

Informally: The defining characteristic of a quadratic function is that it is a polynomial whose highest exponent is 2.
::非正式方式:二次函数的决定性特征是它是一个多元体,其最高出处是2。

There are several ways to write quadratic functions:
::有几种方法可以写出二次函数:

standard form , the form of the quadratic function above: $\begin{align*}f(x)=ax^{2}+bx+c\end{align*}$
::标准窗体,上面的二次函数的形式:f(x)=ax2+bx+c

vertex form , commonly used for quick sketching: $\begin{align*}f(x)=a(x-h)^{2}+k\end{align*}$
::用于快速绘图的顶部窗体: f(x)=a(x-h)2+k

factored form , excellent for finding x - intercepts : $\begin{align*}f(x)=a(x-r_{1})(x-r_{2})\end{align*}$
::用于查找 x 界面的极优 : f(x) = a(x-r1)(x-r2)

You can convert between forms of quadratic functions using algebra and you will see that there are uses for each of these forms when working with quadratic functions.
::您可以使用代数转换四方函数的形态, 在使用四方函数时, 可以看到每种形态都有用途。

When graphing, the $\begin{align*}y-\end{align*}$ intercept of a quadratic function in standard form is $\begin{align*}(0, c)\end{align*}$ and it is found by substituting for $\begin{align*}x\end{align*}$ in $\begin{align*}f(x)=ax^{2}+bx+c\end{align*}$ .
::当图形化时,标准格式的二次函数的 y- interview 是 (0,c) , 通过 f(x) = ax2+bx+c 中的 x 替换而找到。

Examples
::实例

Example 1
::例1

Earlier, you were asked if you could measure height with a stopwatch.
::之前有人问你能不能用手表测量身高

What formula could be used to calculate the height of an object based on the time required for an object to fall from the top?
::可使用什么公式根据物体从顶部掉落所需的时间计算对象的高度?

One formula is $\begin{align*}h = \frac{1}{2}gt^{2}\end{align*}$

This is not completely accurate, as it ignores the effects of air resistance of an object, but offers a very close approximate for most common objects.
::这并非完全准确,因为它忽略了物体的空气抵抗效应,但为大多数常见物体提供了非常接近的近似值。

The value g is gravity. In Metric units this is $\begin{align*}9.81 m/s/s\end{align*}$
::g 值为重力。在计量单位中,该值为9.81m/秒/秒。

The value t is the time taken for the object to fall in seconds
::值 t 是天体以秒数坠落所需的时间

The result h is the distance or the height in meters.
::结果h是距离或高度(以米计)。

Example 2
::例2

Solve $\begin{align*}x^{2}-5x+6=0\end{align*}$ using the " Factor to Solve" method.
::使用“ 解决因素” 方法解决 x2 - 5x+6=0 。

The factor method is based on writing the quadratic equation in factored form . That is, as a product of two linear expressions. So our equation maybe solved by the following way:
::系数法的基础是以系数形式写出二次方程。也就是说, 作为两个线性表达式的产物。因此, 我们的方程也许可以通过以下方式解析 :

$\begin{aligned} x^{2} - 5 x + 6 & = 0 \\ (x - 3) (x - 2) & = 0 \end{aligned}$
$\begin{align*}x^{2}-5x+6 & = 0\\ (x-3)(x-2) & = 0\end{align*}$
::x2-5x+6=0(x-3)(x-2)=0

Recall, that
::回顾,

$\begin{aligned} a \cdot b = 0 if and only if a = 0 and b = 0 \end{aligned}$
$\begin{align*}a \cdot b=0 \ \text{if and only if} \ a=0 \ \text{and} \ b=0\end{align*}$
::ab=0,如果且仅在a=0和b=0的情况下

This tells us that
::这告诉我们:

$\begin{aligned} x - 3 = 0 \end{aligned}$
$\begin{align*}x-3=0\end{align*}$
::x-3=0

or
::或

$\begin{aligned} x - 2 = 0 \end{aligned}$
$\begin{align*}x-2=0\end{align*}$
::x-2=0

which give the roots (or zeros )
::给根( 或零)

$\begin{aligned} x = 3 \end{aligned}$
$\begin{align*}x=3\end{align*}$
::x=3x=3

and
::和

$\begin{aligned} x = 2 \end{aligned}$
$\begin{align*}x=2\end{align*}$
::x=2x=2

In other words, the solution set is {2, 3}.
::换句话说,设定的解决方案是{2,3}。

Example 3
::例3

Solve $\begin{align*}x^{2}-5x+6=0\end{align*}$ by the "Completing the Square" method.
::使用“ 完成广场” 方法解决 x2 - 5x+6=0 。

To solve the above equation by , first move the “ $\begin{align*}c\end{align*}$ ” term to the other side of the equation,
::为了用 . . . . . . . . . . . . . . . . . . . . .

$\begin{aligned} x^{2} - 5 x = - 6 \end{aligned}$
$\begin{align*}x^{2}-5x=-6\end{align*}$
::x2 - 5x% 6

Next, make the left-hand side a “perfect square” by adding the appropriate number. To do so, take one-half of the coefficient of $\begin{align*}x\end{align*}$ (the $\begin{align*}b\end{align*}$ coefficient) and square it and then add the result to both sides of the equation:
::下一步,通过添加合适的数字,使左手侧成为“完美的正方形 ” 。要这样做, 取x( b 系数) 系数的一半, 然后平方, 然后将结果添加到方程的两侧 :

$\begin{aligned} b & = - 5 \\ \frac{b}{2} & = \frac{- 5}{2} \\ {(\frac{b}{2})}^{2} & = \frac{25}{4} \end{aligned}$
$\begin{align*}b & = -5 \\ \frac{b}{2} & = \frac{-5}{2}\\ \left ( \frac{b}{2} \right ) ^2 & = \frac{25}{4}\end{align*}$
::b*5b252(b2)2=254

Adding to both sides of the equation,
::加上等式的两面,

$\begin{aligned} x^{2} - 5 x + \frac{25}{4} & = - 6 + \frac{25}{4} \\ x^{2} - 5 x + \frac{25}{4} & = \frac{1}{4} \\ {(x - \frac{5}{2})}^{2} & = \frac{1}{4} \end{aligned}$
$\begin{align*}x^{2}-5x+\frac{25}{4} & = -6 + \frac{25}{4}\\ x^2 -5x + \frac{25}{4} & = \frac{1}{4}\\ \left ( x-\frac{5}{2} \right )^2 & = \frac{1}{4}\end{align*}$
::x2-5x+2546+254x2-5x+254=14(x-52)2=14

this last equation can be easily solved by taking the square root of both sides,
::最后这个方程式可以很容易地通过双方的平方根来解决,

$\begin{aligned} (x - \frac{5}{2}) = \sqrt{\frac{1}{4}} = \pm \frac{1}{2} \end{aligned}$
$\begin{align*}\left(x-\frac{5}{2}\right)=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}\end{align*}$
:x-52)=1412

Hence
::因此

$\begin{aligned} x = \pm \frac{1}{2} + \frac{5}{2} \end{aligned}$
$\begin{align*}x=\pm\frac{1}{2}+\frac{5}{2}\end{align*}$
::x12+52

and the solutions are
::解决办法是:

$\begin{aligned} x = 3 \end{aligned}$
$\begin{align*}x=3\end{align*}$
::x=3x=3

and
::和

$\begin{aligned} x = 2 \end{aligned}$
$\begin{align*}x=2\end{align*}$
::x=2x=2

which are identical to answers of the factor method.
::与系数方法的答案相同。

Example 4
::例4

Solve $\begin{align*}x^{2}-5x+6=0\end{align*}$ using the Quadratic Formula .
::使用二次曲线公式解决 x2 - 5x+6=0 。

The quadratic formula works for finding the x -intercepts in all quadratic equations, therefore it is highly encouraged that you memorize the formula. The other methods can be much faster though, so it is well worth understanding each of the different methods.
::二次方程用于在所有二次方程中寻找 x 界面, 因此非常鼓励您记住公式。其他方法可以更快得多, 因此非常值得理解每种不同方法。

Recall, if $\begin{align*}ax^{2}+bx+c=0\end{align*}$ where $\begin{align*}a, b,\end{align*}$ and $\begin{align*}c\end{align*}$ are real numbers and $\begin{align*}a\ne0\end{align*}$ , then the roots of the equation can be determined by the quadratic formula.
::回想一下,如果a、b和c为实际数字和a+0的ax2+bx+c=0,则方程的根由四方形公式决定。

$\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} . \end{aligned}$
$\begin{align*}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.\end{align*}$
::xbb2 -4ac2a.

Here the coefficients are $\begin{align*}a=1, b=-5,\end{align*}$ and $\begin{align*}c=6\end{align*}$ .
::这里的系数是a=1,b=5和c=6。

Substituting into the quadratic formula, we get:
::替代二次方程式,我们得到:

$\begin{aligned} x & = \frac{- (- 5) \pm \sqrt{(- 5)^{2} - 4 (1) (6)}}{2 (1)} \\ = \frac{5 \pm \sqrt{25 - 24}}{2} \\ = \frac{5 \pm 1}{2} \\ = 3 or 2 \end{aligned}$
$\begin{align*}x & = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}\\ & = \frac{5 \pm \sqrt{25-24}}{2}\\ & = \frac{5 \pm 1}{2}\\ & = 3 \ \text{or} \ 2\end{align*}$
::x( - 5) ( - 5)2- 4(1)(6)(2)(1)=525-242=512=3或2

which is, again, identical to our two solutions above.
::这与我们上述两种解决办法完全相同。

Example 5
::例5

Solve the following equation by factoring: $\begin{align*}4x^{2}+7x=2\end{align*}$ .
::通过乘数解决下列方程式: 4x2+7x=2。

$\begin{align*}4x^{2} + 7x - 2 = 0\end{align*}$
::4x2+7x-2=0

$\begin{align*}0 = (4x - 1)(x + 2) \therefore x = \frac{1}{4}, -2\end{align*}$
::0=(4x-1)(x+2)x=14,-2)

Example 6
::例6

Solve the following equation by completing the square: $\begin{align*}\frac{2}{3}x^{2}-x+\frac{1}{3}=0\end{align*}$ .
::通过完成正方形( 23x2- x+13=0) 来解决以下方程式。

$\begin{align*}x^{2}-\frac{3}{2}x+\frac{1}{2}=0\end{align*}$
::x2 - 32x+12=0

$\begin{align*}x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{16}-\frac{1}{2}\end{align*}$
::x2-32x+916=916-12

$\begin{align*}(x-\frac{3}{4})(x-\frac{3}{4}) = \frac{1}{16}\end{align*}$
:x-34(x-34)=116)

$\begin{align*}x=\frac{1}{2}, 1\end{align*}$
::x=12,1 个

Example 7
::例7

Solve the following equation by using the quadratic formula: $\begin{align*}(z+6)^{2}+2z=0\end{align*}$ .
::使用四方形公式解决以下方程式z+6)2+2z=0。

$\begin{align*}z^{2}+14z+36=0 \therefore x = \frac{-14\pm\sqrt{14^{2}-4\cdot1\cdot36}}{2\cdot1}\end{align*}$
::z2+14z+36=0x14142-413621

$\begin{align*}x = -7\pm\sqrt{13}\end{align*}$
::x713

Review
::回顾

List three ways to write a quadratic function.
::列出写二次函数的三种方式。

Solve with the quadratic formula:
::用二次方程式解决:

$\begin{align*}m^2 - 5m - 14 = 0\end{align*}$
::m2-5-5m-14=0

$\begin{align*}b^2 - 4b + 4 = 0\end{align*}$
::b2-4b+4=0

$\begin{align*}4b^2 + 8b + 7 = 0\end{align*}$
::4b2+8b+7=0

$\begin{align*}2m^2 - 7m - 13 = -10\end{align*}$
::2米-2-7米-1310

$\begin{align*}5r^2 = 80\end{align*}$
::5r2=80

$\begin{align*}k^2 - 31 - 2k = -6 - 3k^2 - 2k\end{align*}$
::k2 - 31 - 2kk 6 - 3k2 - 2k

If you have an equation with a power of 4, explain how you could solve it using the quadratic formula.
::如果你的方程有4的功率, 请解释您如何使用二次方程来解答它。

Solve by completing the square:
::完成方块的解决方式 :

$\begin{align*}2x^2 - 12x + 26 = 10\end{align*}$
::2x2 - 12x+26=10

$\begin{align*}x^2 - 12x + 29 = -3\end{align*}$
::x2 - 12x+29+3

$\begin{align*}7x^2 - 14x -64 = -8\end{align*}$
::7x2-14x-648

Solve by the most efficient method:
::以最有效的方法解决:

$\begin{align*}x^4 + 13x^2 + 36 = 0\end{align*}$
::x4+13x2+36=0

$\begin{align*}x^4 + 16x^2 -225 = 0\end{align*}$
::x4+16x2-225=0

$\begin{align*}\frac{1}{4}x^2 - \frac{1}{3}x + 1 = 0\end{align*}$
::14x2-13x+1=0

$\begin{align*}\frac{2}{7}c^2 - \frac{1}{2}c - \frac{3}{14} = 0\end{align*}$
::27c2-12c-314=0

In the quadratic formula $\begin{align*} b^2 - 4ac\end{align*}$ is called the discriminant. The values of the discriminant tell us the nature of the solutions or roots of a quadratic equation, $\begin{align*}ax^2 + bx + c = 0\end{align*}$
::在四方公式 b2 - 4ac 中, 称为 discriminant 。 discriminant 的数值告诉我们四方方公式的溶液或根的性质 , x2+bx+c=0 。

What value(s) of the discriminant result in two unique real solutions?
::争吵有什么价值导致两种独特的实际解决办法?

What value(s) of the discriminant result in one unique real solutions?
::争吵有什么价值导致一种独特的实际解决办法?

What value(s) of the discriminant result in two unique imaginary solutions?
::争议产生两种独特的假想解决办法,其价值何在?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

Methods for Solving Quadratic Functions ::解决二次函数的方法

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Review ::回顾

Review (Answers) ::回顾(答复)

Methods for Solving Quadratic Functions
::解决二次函数的方法

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Example 6
::例6

Example 7
::例7

Review
::回顾

Review (Answers)
::回顾(答复)