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Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 7 - Conic SectionsProblem 57
Chapter 8, Problem 57

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? y2 + 6y - x + 5 = 0

Verified step by step guidance
1
Rewrite the given equation \(y^2 + 6y - x + 5 = 0\) to express \(x\) in terms of \(y\). Start by isolating \(x\): \(x = y^2 + 6y + 5\).
Complete the square for the \(y\)-terms to rewrite the equation in vertex form. Take \(y^2 + 6y\) and complete the square: \(y^2 + 6y = (y + 3)^2 - 9\).
Substitute the completed square back into the expression for \(x\): \(x = (y + 3)^2 - 9 + 5 = (y + 3)^2 - 4\).
Identify the vertex of the parabola from the equation \(x = (y + 3)^2 - 4\). The vertex is at \((-4, -3)\), and since \(x\) is expressed as a square of \((y + 3)\), the parabola opens horizontally (to the right).
Determine the domain and range: Since the parabola opens horizontally, the domain is all \(x\) values greater than or equal to the vertex's \(x\)-coordinate, and the range is all real \(y\) values. Finally, check if for each \(x\) there is only one \(y\) value to decide if the relation is a function.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Rearranging and Identifying the Parabola's Form

To analyze the given relation, rewrite the equation to isolate x or y. Completing the square for the y-terms helps express the equation in vertex form, revealing the parabola's vertex and orientation. This step is crucial for understanding the graph's shape and properties.
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Horizontal Parabolas

Domain and Range of Parabolas

The domain is the set of all possible x-values, and the range is the set of all possible y-values of the relation. For parabolas, the vertex and the direction it opens (left, right, up, or down) determine these intervals. Understanding how the parabola extends helps identify domain and range accurately.
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Domain & Range of Transformed Functions

Determining if the Relation is a Function

A relation is a function if each input (x-value) corresponds to exactly one output (y-value). For parabolas that open sideways (like this one), some x-values may have two y-values, violating the function definition. Using the vertical line test or analyzing the equation helps decide if the relation is a function.
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Relations and Functions
Related Practice
Textbook Question

In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function?

x=−4(y−1)2+3x = - 4(y - 1)^2 + 3


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Textbook Question

Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 25x²+4y² – 150x + 32y + 189 = 0

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Textbook Question

Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 4x² + y²+ 16x - 6y - 39 = 0

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Textbook Question

Convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci. 36x2 +9y2 - 216x = 0

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Textbook Question

In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function?

y=−x2+4x−3y=-x^2+4x-3


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Textbook Question

Identify each equation without completing the square.

y2+8x+6y+25=0y^2+8x+6y+25=0

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