Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Problem 33: Find the Directrix**
Given the equation of a parabola:
\[ x = \frac{1}{4}(y - 9)^2 - 8 \]
Determine the directrix from the following options:
- \( \circ \; y = -9 \)
- \( \circ \; y = -7 \)
- \( \circ \; x = -7 \)
- \( \circ \; x = -9 \)
**Explanation:**
The equation provided is in vertex form for a parabola that opens horizontally. The general form for such a parabola is:
\[ x = a(y - k)^2 + h \]
Where \((h, k)\) is the vertex of the parabola. In this case, the vertex is \((-8, 9)\).
This specific parabola has \(a = \frac{1}{4}\), which informs us about the direction and width of the parabola's opening. For a horizontally opening parabola, the directrix is a vertical line given by the equation:
\[ x = h - \frac{1}{4a} \]
Substituting the values:
\[ x = -8 - \frac{1}{4 \times \frac{1}{4}} \]
\[ x = -8 - 1 \]
\[ x = -9 \]
Therefore, the directrix is \(x = -9\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabe9e786-be4a-454e-b2cf-16d08b96c793%2F527f8bed-67ef-432e-af80-b6b0c5d8bb94%2Fq1lupn9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 33: Find the Directrix**
Given the equation of a parabola:
\[ x = \frac{1}{4}(y - 9)^2 - 8 \]
Determine the directrix from the following options:
- \( \circ \; y = -9 \)
- \( \circ \; y = -7 \)
- \( \circ \; x = -7 \)
- \( \circ \; x = -9 \)
**Explanation:**
The equation provided is in vertex form for a parabola that opens horizontally. The general form for such a parabola is:
\[ x = a(y - k)^2 + h \]
Where \((h, k)\) is the vertex of the parabola. In this case, the vertex is \((-8, 9)\).
This specific parabola has \(a = \frac{1}{4}\), which informs us about the direction and width of the parabola's opening. For a horizontally opening parabola, the directrix is a vertical line given by the equation:
\[ x = h - \frac{1}{4a} \]
Substituting the values:
\[ x = -8 - \frac{1}{4 \times \frac{1}{4}} \]
\[ x = -8 - 1 \]
\[ x = -9 \]
Therefore, the directrix is \(x = -9\).
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