Course: 10.3 使用图解来解析赤道等量

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使用图形来解析二次方等量

Use Graphs to Solve Quadratic Equations
::使用图形来解析二次方等量

Solving a quadratic equation means finding the $\begin{aligned} x - \end{aligned}$ values that will make the quadratic function equal zero; in other words, it means finding the points where the graph of the function crosses the $\begin{aligned} x - \end{aligned}$ axis. The solutions to a quadratic equation are also called the roots or zeros of the function, and in this section we’ll learn how to find them by graphing the function.
::解决二次方程式意味着找到能使二次函数为零的 x- 值;换句话说,这意味着找到函数图形交叉 x- 轴的点。二次方程式的解决方案也被称为函数的根或零,在本节中,我们将通过绘制函数图来学习如何找到它们。

Identify the Number of Solutions of a Quadratic Equation
::确定二次赤道解决方案的数量

Three different situations can occur when graphing a quadratic function:
::在绘制二次函数图时,可出现三种不同的情况:

Case 1: The parabola crosses the $\begin{aligned} x - \end{aligned}$ axis at two points. An example of this is $\begin{aligned} y = x^{2} + x - 6 \end{aligned}$ :
::案例1: 抛物线在两个点上横跨 x - 轴。举例来说, y=x2+x-6:

Looking at the graph, we see that the parabola crosses the $\begin{aligned} x - \end{aligned}$ axis at $\begin{aligned} x = - 3 \end{aligned}$ and $\begin{aligned} x = 2 \end{aligned}$ .
::查看图时,我们看到抛物线横穿 x++++3 和 x=2 的 x - 轴。

We can also find the solutions to the equation $\begin{aligned} x^{2} + x - 6 = 0 \end{aligned}$ by setting $\begin{aligned} y = 0 \end{aligned}$ . We solve the equation by factoring:
::我们也可以通过设置 y=0 找到公式 x2+x-6=0 的解决方案。我们通过乘法解析公式 :

$\begin{aligned} (x + 3) (x - 2) = 0 \end{aligned}$ , so $\begin{aligned} x = - 3 \end{aligned}$ or $\begin{aligned} x = 2 \end{aligned}$ .
:x+3)(x-2)=0, 所以 x3 或 x=2。

When the graph of a quadratic function crosses the $\begin{aligned} x - \end{aligned}$ axis at two points, we get two distinct solutions to the quadratic equation.
::当二次函数的图形在两个点穿过 x - 轴时, 我们得到二次方程的两个不同的解决方案。

Case 2: The parabola touches the $\begin{aligned} x - \end{aligned}$ axis at one point. An example of this is $\begin{aligned} y = x^{2} - 2 x + 1 \end{aligned}$ :
::案例2: 抛物线在一个点上触碰 x - 轴。这方面的一个例子是 y=x2-2x+1 :

We can see that the graph touches the $\begin{aligned} x - \end{aligned}$ axis at $\begin{aligned} x = 1 \end{aligned}$ .
::我们可以看到,图表在 x=1 时触摸了 x - 轴。

We can also solve this equation by factoring. If we set $\begin{aligned} y = 0 \end{aligned}$ and factor , we obtain $\begin{aligned} (x - 1)^{2} = 0 \end{aligned}$ , so $\begin{aligned} x = 1 \end{aligned}$ . Since the quadratic function is a perfect square , we get only one solution for the equation—it’s just the same solution repeated twice over.
::我们也可以通过乘数来解答这个方程式。如果我们设定y=0和因数, 我们就会获得 (x-1)2=0, 所以 x=1 。由于四方函数是一个完美的正方形, 我们只能找到一个方程式的解决方案 — — 同样的解决方案重复了两次。

When the graph of a quadratic function touches the $\begin{aligned} x - \end{aligned}$ axis at one point, the quadratic equation has one solution and the solution is called a double root .
::当二次函数的图形在一个点接触到 x - 轴时, 二次方程有一个解决方案, 解决方案被称为双根。

Case 3: The parabola does not cross or touch the $\begin{aligned} x - \end{aligned}$ axis. An example of this is $\begin{aligned} y = x^{2} + 4 \end{aligned}$ :
::案例3: 抛物线不交叉或触碰 x- 轴。例如, y=x2+4 :

If we set $\begin{aligned} y = 0 \end{aligned}$ we get $\begin{aligned} x^{2} + 4 = 0 \end{aligned}$ . This quadratic polynomial does not factor.
::如果我们设定了 Y= 0, 我们就会得到 x2+4=0。这个四边多圆性不因子。

When the graph of a quadratic function does not cross or touch the $\begin{aligned} x - \end{aligned}$ axis, the quadratic equation has no real solutions.
::当二次函数的图形没有交叉或触碰 x- 轴时, 二次方程式没有真正的解决方案。

Solve Quadratic Equations by Graphing
::通过图形绘制解析二次等量

So far we’ve found the solutions to quadratic equations using factoring. However, in real life very few functions factor easily. As you just saw, graphing a function gives a lot of information about the solutions. We can find exact or approximate solutions to a quadratic equation by graphing the function associated with it.
::到目前为止,我们已经找到了使用保理因素的二次方程式的解决方案。然而,在现实生活中,很容易找到的函数因素很少。正如你刚才所看到的,绘制函数图能提供大量解决方案信息。我们可以通过绘制与其相关的函数图来找到二次方程式的精确或近似解决方案。

Solving by Graphing
::通过图形图解

Find the solutions to the following quadratic equations by graphing.
::通过图形化找到以下二次方程的解决方案。

a) $\begin{aligned} - x^{2} + 3 = y \end{aligned}$
::a)-x2+3=y

b) $\begin{aligned} - x^{2} + x - 3 = y \end{aligned}$
::b)-x2+x-3=y

c) $\begin{aligned} y = - x^{2} + 4 x - 4 \end{aligned}$
::c)yx2+4x-4

Since we can’t factor any of these equations, we won’t be able to graph them using intercept form (if we could, we wouldn’t need to use the graphs to find the intercepts!) We’ll just have to make a table of arbitrary values to graph each one.
::由于我们无法计算任何这些方程式, 我们便无法使用截取表(如果我们能, 我们不需要用图表来找到截取数! )我们只需绘制一个任意数值表来绘制每个数值的图。

a) $\begin{aligned} - x^{2} + 3 = y \end{aligned}$
::a)-x2+3=y

$\begin{aligned} x \end{aligned}$	$\begin{aligned} y = - x^{2} + 3 \end{aligned}$
$\begin{aligned} - 3 \end{aligned}$	$\begin{aligned} y = - (- 3)^{2} + 3 = - 6 \end{aligned}$
–2	$\begin{aligned} y = - (- 2)^{2} + 3 = - 1 \end{aligned}$
–1	$\begin{aligned} y = - (- 1)^{2} + 3 = 2 \end{aligned}$
0	$\begin{aligned} y = - (0)^{2} + 3 = 3 \end{aligned}$
1	$\begin{aligned} y = - (1)^{2} + 3 = 2 \end{aligned}$
2	$\begin{aligned} y = - (2)^{2} + 3 = - 1 \end{aligned}$
3	$\begin{aligned} y = - (3)^{2} + 3 = - 6 \end{aligned}$

We plot the points and get the following graph:
::我们绘制点数并获得下图:

From the graph we can read that the $\begin{aligned} x - \end{aligned}$ intercepts are approximately $\begin{aligned} x = 1.7 \end{aligned}$ and $\begin{aligned} x = - 1.7 \end{aligned}$ . These are the solutions to the equation.
::从图中我们可以看到, X - 界面大约是 x= 1. 7 和x\\\\\\\\\\\ 1. 7。这是方程的解决方案。

b) $\begin{aligned} - x^{2} + x - 3 = y \end{aligned}$
::b)-x2+x-3=y

$\begin{aligned} x \end{aligned}$	$\begin{aligned} y = - x^{2} + x - 3 \end{aligned}$
$\begin{aligned} - 3 \end{aligned}$	$\begin{aligned} y = - (- 3)^{2} + (- 3) - 3 = - 15 \end{aligned}$
–2	$\begin{aligned} y = - (- 2)^{2} + (- 2) - 3 = - 9 \end{aligned}$
–1	$\begin{aligned} y = - (- 1)^{2} + (- 1) - 3 = - 5 \end{aligned}$
0	$\begin{aligned} y = - (0)^{2} + (0) - 3 = - 3 \end{aligned}$
1	$\begin{aligned} y = - (1)^{2} + (1) - 3 = - 3 \end{aligned}$
2	$\begin{aligned} y = - (2)^{2} + (2) - 3 = - 5 \end{aligned}$
3	$\begin{aligned} y = - (3)^{2} + (3) - 3 = - 9 \end{aligned}$

We plot the points and get the following graph:
::我们绘制点数并获得下图:

The graph curves up toward the $\begin{aligned} x - \end{aligned}$ axis and then back down without ever reaching it. This means that the graph never intercepts the $\begin{aligned} x - \end{aligned}$ axis, and so the corresponding equation has no real solutions.
::图形曲线向上向 x - 轴, 然后向下向下, 却从未达到过。这意味着图形从未拦截过 x - 轴, 因此相应的方程式没有真正的解决方案。

c) $\begin{aligned} y = - x^{2} + 4 x - 4 \end{aligned}$
::c)yx2+4x-4

$\begin{aligned} x \end{aligned}$	$\begin{aligned} y = - x^{2} + 4 x - 4 \end{aligned}$
$\begin{aligned} - 3 \end{aligned}$	$\begin{aligned} y = - (- 3)^{2} + 4 (- 3) - 4 = - 25 \end{aligned}$
–2	$\begin{aligned} y = - (- 2)^{2} + 4 (- 2) - 4 = - 16 \end{aligned}$
–1	$\begin{aligned} y = - (- 1)^{2} + 4 (- 1) - 4 = - 9 \end{aligned}$
0	$\begin{aligned} y = - (0)^{2} + 4 (0) - 4 = - 4 \end{aligned}$
1	$\begin{aligned} y = - (1)^{2} + 4 (1) - 4 = - 1 \end{aligned}$
2	$\begin{aligned} y = - (2)^{2} + 4 (2) - 4 = 0 \end{aligned}$
3	$\begin{aligned} y = - (3)^{2} + 4 (3) - 4 = - 1 \end{aligned}$
4	$\begin{aligned} y = - (4)^{2} + 4 (4) - 4 = - 4 \end{aligned}$
5	$\begin{aligned} y = - (5)^{2} + 4 (5) - 4 = - 9 \end{aligned}$

Here is the graph of this function:
::以下是此函数的图形 :

The graph just touches the $\begin{aligned} x - \end{aligned}$ axis at $\begin{aligned} x = 2 \end{aligned}$ , so the function has a double root there. $\begin{aligned} x = 2 \end{aligned}$ is the only solution to the equation.
::图形只需在 x=2 处触摸 x - 轴, 函数就具有双根。 x=2 是方程的唯一解决方案。

Analyze Quadratic Functions Using a Graphing Calculator
::使用图形计算计算器分析二次函数

A graphing calculator is very useful for graphing quadratic functions. Once the function is graphed, we can use the calculator to find important information such as the roots or the vertex of the function.
::图形化计算器对于图形化二次函数非常有用。函数一旦被图形化,我们可以使用计算器查找重要信息,如函数的根或顶点。

Using a Graphing Calculator
::使用图形计算计算器

Use a graphing calculator to analyze the graph of $\begin{aligned} y = x^{2} - 20 x + 35 \end{aligned}$ .
::使用图形计算器分析 y=x2-20x+35 的图形。

1. Graph the function.
::1. 绘制函数图。

Press the [Y=] button and enter “ $\begin{aligned} x^{2} - 20 x + 35 \end{aligned}$ ” next to $\begin{aligned} [Y_{1} =] \end{aligned}$ . Press the [GRAPH] button. This is the plot you should see:
::按 [Y=] 按钮,然后在 [Y1=] 旁边输入“x2-20x+35”。按 [GRAPH] 按钮。这是您应该看到的图案:

If this is not what you see, press the [WINDOW] button to change the window size. For the graph shown here, the $\begin{aligned} x - \end{aligned}$ values should range from -10 to 30 and the $\begin{aligned} y - \end{aligned}$ values from -80 to 50.
::如果这不是您看到的, 请按 [WINDOW] 按钮更改窗口大小。对于此处显示的图形, x - 值应该介于 - 10 至 30 之间, y - 值介于 - 80 至 50 之间。

2. Find the roots.
::2. 找出根源。

There are at least three ways to find the roots:
::至少有三种方法可以找到根源:

Use [TRACE] to scroll over the $\begin{aligned} x - \end{aligned}$ intercepts. The approximate value of the roots will be shown on the screen. You can improve your estimate by zooming in.
::使用 [TRACE] 滚动到 x- 界面上。根的大致值将在屏幕上显示。您可以通过缩放来改进估计值。

OR
::或

Use [TABLE] and scroll through the values until you find values of $\begin{aligned} y \end{aligned}$ equal to zero. You can change the accuracy of the solution by setting the step size with the [TBLSET] function.
::使用 [表并滚动到值为 y 等于零。您可以用 [TBLSET] 函数设置步数大小, 从而改变解决方案的准确性。

OR
::或

Use [2nd] [TRACE] (i.e. ‘calc’ button) and use option ‘zero’.
::使用[第二 [TRACE](即`计算 ' 按钮)并使用选项`零 ' 。

Move the cursor to the left of one of the roots and press [ENTER] .
::将光标移动到根点之一的左侧并按 [ENTER] 键。

Move the cursor to the right of the same root and press [ENTER] .
::将光标移动到同一根的右侧并按 [ENTER] 键。

Move the cursor close to the root and press [ENTER] .
::将光标移到根端并按住 [ENTER] 。

The screen will show the value of the root. Repeat the procedure for the other root.
::屏幕将显示根值。重复另一个根的程序。

Whichever technique you use, you should get about $\begin{aligned} x = 1.9 \end{aligned}$ and $\begin{aligned} x = 18 \end{aligned}$ for the two roots.
::无论你使用哪种技术,你应该得到两个根的 x=1.9和x=18。

3. Find the vertex.
::3. 找到顶部。

There are three ways to find the vertex:
::找到顶点有三种方法:

Use [TRACE] to scroll over the highest or lowest point on the graph. The approximate value of the roots will be shown on the screen.
::使用 [TRACE] 滚动到图形上的最高或最低点。根的大致值将在屏幕上显示。

OR
::或

Use [TABLE] and scroll through the values until you find values the lowest or highest value of $\begin{aligned} y \end{aligned}$ . You can change the accuracy of the solution by setting the step size with the [TBLSET] function.
::使用 [表并滚动浏览值,直到找到 y 的最低值或最高值。您可以通过使用 [TBLSET] 函数设定步数大小来改变解决方案的准确性。

OR
::或

Use [2nd] [TRACE] and use the option ‘maximum’ if the vertex is a maximum or ‘minimum’ if the vertex is a minimum.
::使用[第二 [TRACE],如果顶点为最大值,则使用`最大值 ' 选项;如果顶点为最低值,则使用`最小值 ' 选项。

Move the cursor to the left of the vertex and press [ENTER] .
::将光标移动到顶部左侧并按 [ENTER] 键。

Move the cursor to the right of the vertex and press [ENTER] .
::将光标移动到顶部右侧并按 [ENTER] 键。

Move the cursor close to the vertex and press [ENTER] .
::将光标移到顶部并按 [ENTER] 。

The screen will show the $\begin{aligned} x - \end{aligned}$ and $\begin{aligned} y - \end{aligned}$ values of the vertex.
::屏幕将显示顶点的 x - 和 y - 值。

Whichever method you use, you should find that the vertex is at (10, -65) .
::无论你使用哪种方法,你都应该发现顶点在10,65。

Solve Real-World Problems by Graphing Quadratic Functions
::通过绘制二次曲线函数图解解决现实世界问题

Here’s a real-world problem we can solve using the graphing methods we’ve learned.
::这是一个现实世界的问题, 我们可以使用我们所学的图表绘制方法来解决。

Real-World Application: Arching
::真实世界应用程序: 存档

Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the ball with respect to time is $\begin{aligned} y = - 4.9 t^{2} + 48 t \end{aligned}$ , where $\begin{aligned} y \end{aligned}$ is the height of the arrow in meters and $\begin{aligned} t \end{aligned}$ is the time in seconds since Andrew shot the arrow. Find how long it takes the arrow to come back to the ground.
::Andrew 是一个快速射箭手。他发射箭头, 选择抛物线路径。球高度相对于时间的方程式是 y4. 9t2+48t, y是箭头在米上的高度, t是 Andrew 射箭后几秒内的时间。找到箭头返回地面需要多久。

Let’s graph the equation by making a table of values.
::让我们通过绘制一个数值表来绘制方程式。

$\begin{aligned} t \end{aligned}$	$\begin{aligned} y = - 4.9 t^{2} + 48 t \end{aligned}$
0	$\begin{aligned} y = - 4.9 (0)^{2} + 48 (0) = 0 \end{aligned}$
1	$\begin{aligned} y = - 4.9 (1)^{2} + 48 (1) = 43.1 \end{aligned}$
2	$\begin{aligned} y = - 4.9 (2)^{2} + 48 (2) = 76.4 \end{aligned}$
3	$\begin{aligned} y = - 4.9 (3)^{2} + 48 (3) = 99.9 \end{aligned}$
4	$\begin{aligned} y = - 4.9 (4)^{2} + 48 (4) = 113.6 \end{aligned}$
5	$\begin{aligned} y = - 4.9 (5)^{2} + 48 (5) = 117.6 \end{aligned}$
6	$\begin{aligned} y = - 4.9 (6)^{2} + 48 (6) = 111.6 \end{aligned}$
7	$\begin{aligned} y = - 4.9 (7)^{2} + 48 (7) = 95.9 \end{aligned}$
8	$\begin{aligned} y = - 4.9 (8)^{2} + 48 (8) = 70.4 \end{aligned}$
9	$\begin{aligned} y = - 4.9 (9)^{2} + 48 (9) = 35.1 \end{aligned}$
10	$\begin{aligned} y = - 4.9 (10)^{2} + 48 (10) = - 10 \end{aligned}$

Here’s the graph of the function:
::以下是函数图 :

The roots of the function are approximately $\begin{aligned} x = 0 \end{aligned}$ sec and $\begin{aligned} x = 9.8 \end{aligned}$ sec. The first root tells us that the height of the arrow was 0 meters when Andrew first shot it. The second root says that it takes approximately 9.8 seconds for the arrow to return to the ground.
::函数的根大约为 x=0 秒和 x=9. 8 秒。第一个根告诉我们, 安德鲁第一次射箭时箭头的高度是 0 米。第二个根表示, 箭头返回地面大约需要9. 8 秒。

Example
::示例示例示例示例

Example 1
::例1

Find the solutions to $\begin{aligned} 2 x^{2} + 5 x - 7 = 0 \end{aligned}$ by graphing.
::通过图形化查找 2x2+5x-7=0 的解决方案。

Since we can’t factor this equation, we won’t be able to graph it using intercept form (if we could, we wouldn’t need to use the graphs to find the intercepts!) We’ll just have to make a table of arbitrary values to graph the equation.
::既然我们无法计算这个等式, 我们就不能用截取表(如果我们可以, 我们不需要用图表找到截取数! )我们只需要用一个任意数值表来绘制等式图。

$\begin{aligned} x \end{aligned}$	$\begin{aligned} y = 2 x^{2} + 5 x - 7 \end{aligned}$
$\begin{aligned} - 5 \end{aligned}$	$\begin{aligned} y = 2 (- 5)^{2} + 5 (- 5) - 7 = 18 \end{aligned}$
–4	$\begin{aligned} y = 2 (- 4)^{2} + 5 (- 4) - 7 = 5 \end{aligned}$
–3	$\begin{aligned} y = 2 (- 3)^{2} + 5 (- 3) - 7 = - 4 \end{aligned}$
–2	$\begin{aligned} y = 2 (- 2)^{2} + 5 (- 2) - 7 = - 9 \end{aligned}$
–1	$\begin{aligned} y = 2 (- 1)^{2} + 5 (- 1) - 7 = - 10 \end{aligned}$
0	$\begin{aligned} y = 2 (0)^{2} + 5 (0) - 7 = - 7 \end{aligned}$
1	$\begin{aligned} y = 2 (1)^{2} + 5 (1) - 7 = 0 \end{aligned}$
2	$\begin{aligned} y = 2 (2)^{2} + 5 (2) - 7 = 11 \end{aligned}$
3	$\begin{aligned} y = 2 (3)^{2} + 5 (3) - 7 = 26 \end{aligned}$

We plot the points and get the following graph:
::我们绘制点数并获得下图:

From the graph we can read that the $\begin{aligned} x - \end{aligned}$ intercepts are $\begin{aligned} x = 1 \end{aligned}$ and $\begin{aligned} x = - 3.5 \end{aligned}$ . These are the solutions to the equation.
::从图表中我们可以看到 x - 界面是 x=1 和x\\\\\ 3.5。这些是方程的解决方案。

Review
::回顾

For 1-6, find the solutions of the following equations by graphing.
::对于1-6,通过图形化找到以下方程式的解决方案。

$\begin{aligned} x^{2} + 3 x + 6 = 0 \end{aligned}$
::x2+3x+6=0

$\begin{aligned} - 2 x^{2} + x + 4 = 0 \end{aligned}$
::− 2x2+x+4=0

$\begin{aligned} x^{2} - 9 = 0 \end{aligned}$
::x2- 9=0

$\begin{aligned} x^{2} + 6 x + 9 = 0 \end{aligned}$
::x2+6x+9=0

$\begin{aligned} 10 x - 3 x^{2} = 0 \end{aligned}$
::10 - 3x2=0

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

For 7-12, find the roots of the following quadratic functions by graphing.
::对于 7-12, 通过图形化找到以下二次函数的根。

$\begin{aligned} y = - 3 x^{2} + 4 x - 1 \end{aligned}$
::y3x2+4x-1

$\begin{aligned} y = 9 - 4 x^{2} \end{aligned}$
::y=9 - 4x2

$\begin{aligned} y = x^{2} + 7 x + 2 \end{aligned}$
::y=x2+7x+2

$\begin{aligned} y = - x^{2} - 10 x - 25 \end{aligned}$
::yx2 - 10x - 25

$\begin{aligned} y = 2 x^{2} - 3 x \end{aligned}$
::y= 2x2 - 3x

$\begin{aligned} y = x^{2} - 2 x + 5 \end{aligned}$
::y=x2 - 2x+5 y=x2 - 2x+5

For 13-18, use your graphing calculator to find the roots and the vertex of each polynomial.
::对于 13-18, 使用您的图形计算器来查找每个多面形的根和顶点。

$\begin{aligned} y = x^{2} + 12 x + 5 \end{aligned}$
::y=x2+12x+5

$\begin{aligned} y = x^{2} + 3 x + 6 \end{aligned}$
::y=x2+3x+6 y=x2+3x+6

$\begin{aligned} y = - x^{2} - 3 x + 9 \end{aligned}$
::yx2 - 3x+9

$\begin{aligned} y = - x^{2} + 4 x - 12 \end{aligned}$
::yx2+4x- 12

$\begin{aligned} y = 2 x^{2} - 4 x + 8 \end{aligned}$
::y=2x2 - 4x+8

$\begin{aligned} y = - 5 x^{2} - 3 x + 2 \end{aligned}$
::y5x2 - 3x+2

Graph the equations $\begin{aligned} y = 2 x^{2} - 4 x + 8 \end{aligned}$ and $\begin{aligned} y = x^{2} - 2 x + 4 \end{aligned}$ on the same screen. Find their roots and vertices.
1. What is the same about the graphs? What is different?
  ::图表有什么不同?有什么不同?
2. How are the two equations related to each other? (Hint: factor them.)
  ::两个方程式之间有何关联? (提示: 乘以。 )
3. What might be another equation with the same roots? Graph it and see.
  ::同样的根又是什么方程?
::在同一屏幕上图形 y= 2x2- 4x+8 和 y= x2-2x+4 。查找它们的根和顶。图形是相同的吗? 是什么不同? 两个方程式是如何相互联系的? (提示: 系数。) 另一个根的方程式是什么? 图表和查看。

Graph the equations $\begin{aligned} y = x^{2} - 2 x + 2 \end{aligned}$ and $\begin{aligned} y = x^{2} - 2 x + 4 \end{aligned}$ on the same screen. Find their roots and vertices.
1. What is the same about the graphs? What is different?
  ::图表有什么不同?有什么不同?
2. How are the two equations related to each other?
  ::这两个方程式如何相互联系?
::在同一屏幕上绘制 y=x2-2x+2 和 y=x2-2x+4 的方程式。查找它们的根和顶。图形有什么不同? 不同吗 ? 这两个方程式是如何互相关联的 ?

Phillip throws a ball and it takes a parabolic path. The equation of the height of the ball with respect to time is $\begin{aligned} y = - 16 t^{2} + 60 t \end{aligned}$ , where $\begin{aligned} y \end{aligned}$ is the height in feet and $\begin{aligned} t \end{aligned}$ is the time in seconds. Find how long it takes the ball to come back to the ground.
::Phillip 抛出一个球, 它需要一个抛射路径。球的高度相对于时间的方程式是 y16t2+60t, y是脚的高度, t是秒中的时间。找出球返回地面需要多长时间。

Use your graphing calculator to solve Ex. C. You should get the same answers as we did graphing by hand, but a lot quicker!
::使用您的图形计算计算器解析 Ex. C. 。您应该得到与我们手动绘图一样的答案, 但速度要快得多 !

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

General

使用图形来解析二次方等量

Use Graphs to Solve Quadratic Equations ::使用图形来解析二次方等量

Identify the Number of Solutions of a Quadratic Equation ::确定二次赤道解决方案的数量

Solve Quadratic Equations by Graphing ::通过图形绘制解析二次等量

Solving by Graphing ::通过图形图解

Using a Graphing Calculator ::使用图形计算计算器

Real-World Application: Arching ::真实世界应用程序: 存档

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Use Graphs to Solve Quadratic Equations
::使用图形来解析二次方等量

Identify the Number of Solutions of a Quadratic Equation
::确定二次赤道解决方案的数量

Solve Quadratic Equations by Graphing
::通过图形绘制解析二次等量

Solving by Graphing
::通过图形图解

Using a Graphing Calculator
::使用图形计算计算器

Real-World Application: Arching
::真实世界应用程序: 存档

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)