Precalculus: Mathematics for Calculus (Standalone Book) 7 Edition
1
Chapter 1: Chapter 1
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2
Chapter 2: Chapter 2
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3
Chapter 3: Chapter 3
764 questions
4
Chapter 4: Chapter 4
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5
Chapter 5: Chapter 5
520 questions
6
Chapter 6: Chapter 6
479 questions
7
Chapter 7: Chapter 7
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8
Chapter 8: Chapter 8
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9
Chapter 9: Chapter 9
370 questions
10
Chapter 10: Chapter 10
707 questions
11
Chapter 11: Chapter 11
463 questions
12
Chapter 12: Chapter 12
511 questions
13
Chapter 13: Chapter 13
260 questions
Question
Conic Sections - Parabolas - Chapter 11.1, Problem 9E
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Text solution
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Now, when rearranging the equation, it will turn out to be
The reason why this is, is because we want to isolate our . So add , divide by , and you will have your equation.
Now because this is an equation, the parabola will be open towards the right, horizontally. The only option is III or VI. But #5 is III, therefore #9 is VI. To prove this, find a point. One point is (1.125, 3). Graph III, does not have this point, Graph VI does.
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| Question Text | Conic Sections - Parabolas - Chapter 11.1, Problem 9E |
| Answer Type | Text solution:1 |
