-
Notifications
You must be signed in to change notification settings - Fork 1
/
Copy path45b2c5d0-c08c-4262-8d8a-1bf24da7b3ae.html
288 lines (285 loc) · 17.9 KB
/
45b2c5d0-c08c-4262-8d8a-1bf24da7b3ae.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
<h4>Activity (20 minutes)</h4>
<p>This activity introduces students to finding a specific quadratic equation that has certain zeros and passes through
a particular point. In previous lessons, students found a quadratic function that has certain zeros in its simplest
standard form. In this activity, students will find a more specific quadratic function using the given point.</p>
<p>This is accomplished by using the intercept form, \(f(x) = a(x − p)(x − q)\), of a quadratic equation.
The roots of the function and the given point are substituted into the function to find the value of \(a\). This value
of the multiplier, \(a\), is then distributed through the equation to find the specific quadratic equation that meets
our given criteria.</p>
<h4>Launch</h4>
<p>In this activity, questions 1–3 provide a step-through introduction to finding the specific equation of a
quadratic equation that has certain zeros and passes through a given point. Give students time to complete the first
three questions and then pause for a class discussion.</p>
<p>While discussing the first three questions, emphasize that the function \(m(x) = x^2− 3x − 40\) is a
function that has the zeros –5 and 8. As we found in question 3, \(m(x) = 2x^2− 6x − 80\) is another
function that has the zeros –5 and 8. Only the function \(m(x) = 2x^2− 6x − 80\) passes through the
point \((5, -60)\) and is the specific function we want. Show the graphs of these two functions using <a href="https://k12.openstax.org/contents/raise/resources/0107b2bfda94c3432e1996a46a5cbfad3035b0cf" target="_blank">Desmos</a> to help visualize the difference between these two quadratic equations. Emphasize that
these quadratic equations are related since they both have the same zeros and axis of symmetry (\(x = 1.5)\). Only one
of them, however, passes through the point \((5, -60)\).</p>
<p>Allow students to complete questions 4–6 in pairs. </p>
<h4>Student Activity</h4>
<p>When finding an exact function that has particular zeros and passes through a certain point, we use the intercept
form of a quadratic function:</p>
<p>\(f(x) = a(x − p)(x − q)\)</p>
<p>The variables \(p\) and \(q\) represent the zeros or roots of the function. The variable \(a\) accounts for any
constant that might have been factored out when solving.</p>
<p>Remember, if needed in the following problems, use the "^" symbol to enter exponents.</p>
<ol class="os-raise-noindent">
<li>Find a function, \(m(x)\), in intercept form, that has the zeros –5 and 8.</li>
</ol>
<p><strong>Answer:</strong> The intercept form is \(m(x) = a(x − (-5))(x − 8)\). </p>
<p class="os-raise-noindent">\(= a(x + 5)(x − 8)\) </p>
<p>Let's use the intercept form of the function above to find the exact function that passes through \((5, -60)\).</p>
<ol class="os-raise-noindent" start="2">
<li>First, find the value of \(a\). To do this, substitute the values of the point into the intercept form of the
function.</li>
</ol>
<p><strong>Answer:</strong><br>
\(m(x) = a(x + 5)(x − 8)\) <br>
\(-60 = a((5) + 5)((5) − 8)\) <br>
\(-60 = a(10)(-3)\) <br>
\(-60 = -30a\)<br>
\(2 = a\)</p>
<ol class="os-raise-noindent" start="3">
<li>Substitute the value of \(a\) into the intercept form of the function and multiply to find the standard form.</li>
</ol>
<p><strong>Answer:</strong><br>
\(m(x) = a(x + 5)(x − 8)\) <br>
\(m(x) = 2(x + 5)(x − 8)\)<br>
\(m(x) = 2(x^2 + 5x − 8x − 40)\)<br>
\(m(x) = 2(x^2 − 3x − 40)\)<br>
\(m(x) = 2x^2 − 6x − 80\)</p>
<p>This is the function with zeros –5 and 8 that passes through \((5, -60)\).</p>
<ol class="os-raise-noindent" start="4">
<li>Write the quadratic function, \(n(x)\), in standard form, that has the zeros 9 and –2 and passes through the
point \((0, -36)\).</li>
</ol>
<p><strong>Answer:</strong> \(n(x) = 2x^2 − 14x − 36\)</p>
<ol class="os-raise-noindent" start="5">
<li>Write the quadratic function, \(p(x)\), in standard form, that has the zeros –6 and –4 and passes
through the point \((-5, 3)\).</li>
</ol>
<p><strong>Answer:</strong> \(p(x) = −3x^2 − 30x − 72\)</p>
<ol class="os-raise-noindent" start="6">
<li>Write the quadratic function, \(c(x)\), in standard form, that has the zeros 1 and 7 and passes through the point
\((9, 16)\).</li>
</ol>
<p><strong>Answer:</strong> \(c(x) = x^2 − 8x + 7\)</p>
<h4>Video: Finding a Quadratic Function from Its Zeros and a Point</h4>
<p>Watch the following video to learn more about finding a quadratic function from its zeros and a point. </p>
<div class="os-raise-d-flex-nowrap os-raise-justify-content-center">
<div class="os-raise-video-container"><video controls="true" crossorigin="anonymous">
<source src="https://k12.openstax.org/contents/raise/resources/edd0a3e9d4afdfbd30566216317a5425c3ae450b">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/9ec7085ed4a5348ba5bda6db2a8c2f66caeb427e" srclang="en_us">
https://k12.openstax.org/contents/raise/resources/edd0a3e9d4afdfbd30566216317a5425c3ae450b
</video></div>
</div>
<br>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for English Language Learners</p>
<p class="os-raise-extrasupport-name">MLR 8 Discussion Supports: Representing, Conversing</p>
</div>
<div class="os-raise-extrasupport-body">
<p>Ask students to closely examine and then compare the two functions, \(m(x) = x^2 - 3x - 40\) and \(m(x) = 2x^2 - 6x - 80\). Encourage students to look at the algebraic, tabular, and graphic representations of both functions. Display sentence frames “One thing that is the same is…” and “One thing that is different is…” Give students time to make sure that everyone in the class can explain or justify that each function has the same zeros. Next, ask students to explain why these quadratic equations are related. Provide the sentence frame, “Calling the equations related reminds me of … “ to support students as they explain their thinking. Some students may benefit from the opportunity to rehearse what they will say with their partner before they share with the whole class.<br>
</p>
<p class="os-raise-text-italicize">Design Principle(s): Support sense-making</p>
</div>
</div>
<br>
<div class="os-raise-extrasupport">
<div class="os-raise-extrasupport-header">
<p class="os-raise-extrasupport-title">Support for Students with Disabilities</p>
<p class="os-raise-extrasupport-name">Representation: Internalize Comprehension</p>
</div>
<div class="os-raise-extrasupport-body">
<p> To support working memory, provide students with sticky notes or encourage them to use a T-chart to compare and contrast the equations. Help students understand that related equations have the same zeros and axis of symmetry, but the tables of values, vertices, etc. will differ. Then, have them include how these graphical differences impact the values in each form of the quadratic equations.<br>
</p>
<p class="os-raise-text-italicize">Supports accessibility for: Memory, Organization</p>
</div>
</div>
<br>
<br>
<h4>Activity Synthesis</h4>
<p>To summarize key ideas from the activity, discuss the following questions:</p>
<ul class="os-raise-noindent">
<li> “In this activity, what does it mean for two quadratic equations to be related?” (Two quadratic
functions are related if they have the same zeros and the same axis of symmetry.) </li>
<li> “How can you use the intercept form of a quadratic equation such as \(f(x) = a(x − 5)(x − 4)\)
and a point that it passes through to find the value of \(a\)?” (Substitute the point into the function and
solve for the value of \(a\). Substitute the value of \(a\) back into the equation and multiply to find the standard
form of the quadratic equation.) </li>
</ul>
<h3> 8.11.3: Self Check </h3>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their
understanding of the concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Write the quadratic function, \(z(x)\), in standard form, that has the zeros –1 and 8 and passes through the
point \((0, 16)\).</p>
<table class="os-raise-textheavytable">
<thead>
<tr>
<th scope="col">
Answers
</th>
<th scope="col">
Feedback
</th>
</tr>
</thead>
<tbody>
<tr>
<td>
<p>\(z(x) = −2x^2 + 14x + 16\)</p>
</td>
<td>
<p>That's correct! Check yourself: Substitute the points \((-1, 0)\), \((8, 0)\), and \((0, 16)\) into the
function and check that each is true. Since each one is true, the answer is correct.</p>
</td>
</tr>
<tr>
<td>
<p>\(z(x) = 3x^2 − 21x − 24\)</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This is an incorrect value \(a\), in which \(a = 3\). The correct
value of \(a\) is –2. The answer is \(z(x) = −2x^2 + 14x +16\).</p>
</td>
</tr>
<tr>
<td>
<p>\(z(x) = 2x^2 − 14x − 16\)</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This is an incorrect value \(a\), in which \(a = 2\). The correct
value of \(a\) is –2. The answer is \(z(x) = −2x^2 + 14x +16\).</p>
</td>
</tr>
<tr>
<td>
<p>\(z(x) = x^2 + 7x + 8\)</p>
</td>
<td>
<p>Incorrect. Let's try again a different way: This is the standard form of the function without finding the
value of \(a\). Use the intercept form and the point \((0, 16)\) to find the value of \(a\). The answer is
\(z(x) = −2x^2 + 14x +16\).</p>
</td>
</tr>
</tbody>
</table>
<br>
<h3>8.11.3: Additional Resources</h3>
<p class="os-raise-text-bold"><em>The following content is available to students who would like more support based on
their experience with the self check. Students will not automatically have access to this content, so you may wish
to share it with those who could benefit from it. </em></p>
<h4>Finding a Quadratic Function from Its Zeros and a Point</h4>
<p>Think about how you would solve the quadratic equation \(x^2 − 3x + 2 = 0\).</p>
<ul class="os-raise-noindent">
<li> The factored form is \((x − 1)(x − 2) = 0\). </li>
</ul>
<ul class="os-raise-noindent">
<li> The solutions are \(x = 1\) and \(x = 2\). </li>
</ul>
<p>Now think about how you would solve \(2x^2 − 6x + 4 = 0\).</p>
<ul class="os-raise-noindent">
<li> Factor out the GCF, 2. </li>
</ul>
<p>\(2(x^2 − 3x + 2) = 0\)</p>
<p>It looks similar to the previous quadratic equation.</p>
<ul class="os-raise-noindent">
<li> The factored form is \(2(x − 1)(x − 2) = 0\). </li>
</ul>
<ul class="os-raise-noindent">
<li> Once again, the solutions are \(x = 1\) and \(x = 2\). </li>
</ul>
<p>Since they have the same solutions, these two quadratic equations are related. They have the same zeros and the same
axis of symmetry. </p>
<p>When we are reversing this process, going from the zeros of a quadratic function to finding its standard form, it is
possible there is a multiplier, such as 2 in the example above, that we do not know about.</p>
<p>To account for this, we use the variable \(a\) as the multiplier until we find out exactly what it is.</p>
<p>We can incorporate \(a\) into the function using the intercept form, \(f(x) = a(x − p)(x − q)\). The
variables \(p\) and \(q\) represent the zeros or roots of the function.</p>
<p>Let's look at an example to help us understand.</p>
<p><strong>Example 1</strong></p>
<p>Write the quadratic function, \(f(x)\), that has the zeros –5 and 4 and passes through the point \((-4, -16)\).
</p>
<p>We know the zeros are –5 and 4, so the factored form of the quadratic function is:</p>
<p class="os-raise-noindent">\(f(x) = (x + 5)(x − 4)\).</p>
<p>The intercept form is \(f(x) = a(x + 5)(x − 4)\).</p>
<p>We are given one more piece of information that allows us to find the value of \(a\). We know that the function
passes through the point \((-4, -16)\).</p>
<p>We can substitute –4 for \(x\) and –16 for the value of the function and solve for \(a\).</p>
<p class="os-raise-noindent"> \(f(x) = a(x + 5)(x − 4)\)</p>
<p class="os-raise-noindent">\(-16 = a((-4) + 5)((-4) − 4)\)</p>
<p class="os-raise-noindent">\(-16 = a(1)(-8)\)</p>
<p class="os-raise-noindent">\(-16 = -8a\)</p>
<p class="os-raise-noindent">\(2 = a\)</p>
<p>Since we know \(a = 2\), we can substitute 2 for \(a\) into the intercept form of the quadratic function.</p>
<p class="os-raise-noindent">\(f(x) = a(x + 5)(x − 4)\)</p>
<p class="os-raise-noindent">\(f(x) = 2(x + 5)(x − 4)\)</p>
<p class="os-raise-noindent">\(f(x) = 2(x^2 + 5x − 4x − 20)\)</p>
<p class="os-raise-noindent">\(f(x) = 2(x^2 + x − 20)\)</p>
<p class="os-raise-noindent">\(f(x) = 2x^2 + 2x − 40\)</p>
<p>The specific function that has zeros –5 and 4 and passes through the point \((-4, -16)\) is:</p>
<p class="os-raise-noindent">\(f(x) = 2x^2 + 2x − 40\).</p>
<p>Let's graph the functions to check our work.</p>
<p><img alt="TWO GRAPHED PARABOLAS, ONE RED AND ONE GREEN, BOTH OPENING UP AND BOTH HAVING \(x\)-intercepts OF NEGATIVE 5 AND 4. THE GREEN PARABOLA HAS A \(y\)-intercepts OF NEGATIVE 20. THE RED PARABOLA HAS A \(y\)-intercepts OF NEGATIVE 40 AND PASSES THROUGH THE POINT (NEGATIVE 4, NEGATIVE 16)." height="374" src="https://k12.openstax.org/contents/raise/resources/9db19e4cf408223871873e1c128afb7dc9fe58a9" width="374"></p>
<p>Notice that \(f(x) = x^2 + x − 20\) and \(f(x) = 2x^2 + 2x − 40\) have the same zeros. These quadratic
functions are related.</p>
<p>Only \(f(x) = 2x^2 + 2x − 40\) passes through \((-4, -16)\) and is the specific function we were looking for at
the start.</p>
<div class="os-raise-graybox">
<p>It is very important to note that this is why, in the previous lesson, we were looking for “a function that
has zeros at –5 and 4” and not “the function that has zeros at –5 and 4.” </p>
<p>This difference in wording is very important since there are infinitely many functions that are related by their
solutions.</p>
<p>We can only determine an exact specific function if we have the additional information of a point through which the
function passes.</p>
</div>
<br>
<p>Let's try one more example. </p>
<p class="os-raise-text-bold">Example 2</p>
<p>Write the quadratic function, \(g(x)\), in standard form, that has the zeros –7 and –3 and passes through
the point \((-2, -5)\).</p>
<p>Since the zeros are –7 and –3, the factored form of the function is \(g(x) = (x + 7)(x + 3)\).</p>
<p>The intercept form of the function is \(g(x) = a(x + 7)(x + 3)\).</p>
<p>We substitute values using the point \((-2, -5)\) to find the value of \(a\).</p>
<p class="os-raise-noindent">\(g(x) = a(x + 7)(x + 3)\)</p>
<p class="os-raise-noindent">\(-5 = a((-2) + 7)((-2) + 3)\)</p>
<p class="os-raise-noindent">\(-5 = a(5)(1)\)</p>
<p class="os-raise-noindent">\(-5 = 5a\)</p>
<p class="os-raise-noindent">\(-1 = a\)</p>
<p>Substitute the value of \(a\) into the intercept form of the function.</p>
<p class="os-raise-noindent">\(g(x) = a(x + 7)(x + 3)\)</p>
<p class="os-raise-noindent">\(g(x) = -1(x + 7)(x + 3)\)</p>
<p class="os-raise-noindent">\(g(x) = -1(x^2 + 7x + 3x + 21)\)</p>
<p class="os-raise-noindent">\(g(x) = -1(x^2 + 10x + 21)\)</p>
<p class="os-raise-noindent">\(g(x) = -x^2 − 10x − 21\)</p>
<p>The specific function that has zeros –7 and –3 and passes through the point \((-2, -5)\) is:</p>
<p class="os-raise-noindent">\(g(x) = -x^2 − 10x − 21\).</p>
<p>Let's graph the functions to check our work.</p>
<p><img alt="GRAPH OF TWO PARABOLAS, ONE RED AND ONE GREEN, BOTH WITH \(x\)-intercepts OF NEGATIVE 7 AND NEGATIVE 3. THE GREEN PARABOLA OPENS UP. THE RED PARABOLA OPENS DOWN AND PASSES THROUGH THE POINT (NEGATIVE 2, NEGATIVE 5)." height="377" src="https://k12.openstax.org/contents/raise/resources/8b08c2407b483e6a5be7ecd1891958e36ef32031" width="377"></p>
<p>Again, notice that \(g(x) = x^2 + 10x + 21\) and \(g(x) = −x^2 − 10x − 21\) have the same zeros.
They are related quadratic functions.</p>
<p>Even though the leading coefficient is negative and the function has been reflected over the \(x\)-axis, only \(g(x)
= −x^2 − 10x − 21\) passes through \((-2, -5)\).</p>
<h4>Try It: Finding a Quadratic Function from Its Zeros and a Point</h4>
<p>Write the quadratic function, \(h(x)\), that has the zeros 2 and 6 and passes through the point \((7, 15)\).</p>
<p>Write down your answer, then select the<strong> solution</strong> button to compare your work.</p>
<h5>Solution</h5>
<p>Here is how to find a function when given its zeros and a point:</p>
<p>The zeros are 2 and 6.</p>
<p>The intercept form is \(h(x) = a(x − 2)(x − 6)\).</p>
<p>Use the point \((7, 15)\) to find \(a\).</p>
<p class="os-raise-noindent">\(h(x) = a(x − 2)(x − 6)\)</p>
<p class="os-raise-noindent">\(15 = a((7) − 2)((7) − 6)\)</p>
<p class="os-raise-noindent">\(15 = a(5)(1)\)</p>
<p class="os-raise-noindent">\(3 = a\)</p>
<p>Substitute 3 for the value of \(a\) in the intercept form.</p>
<p class="os-raise-noindent">\(h(x) = a(x − 2)(x − 6)\)</p>
<p class="os-raise-noindent">\(h(x) = 3(x^2 − 2x − 6x + 12)\)</p>
<p class="os-raise-noindent">\(h(x) = 3(x^2 − 8x + 12)\)</p>
<p class="os-raise-noindent">\(h(x) = 3x^2 − 24x + 36\)</p>
<p>The specific quadratic with zeros of 2 and 6 that passes through the point \((7, 15)\) is \(h(x) = 3x^2 − 24x +
36\).</p>