Textbook Question
In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range.
Ch. 7 - Conic Sections
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
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Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 8, Problem 53
In Exercises 51–56, graph each relation. Use the relation's graph to determine its domain and range.
Verified step by step guidance1
Recognize that the given equation \(\frac{x^2}{9} + \frac{y^2}{16} = 1\) represents an ellipse centered at the origin, where the denominators 9 and 16 correspond to the squares of the ellipse's semi-axes lengths.
Identify the lengths of the semi-major and semi-minor axes: \(a = 3\) (since \(9 = 3^2\)) and \(b = 4\) (since \(16 = 4^2\)). This means the ellipse stretches 3 units along the x-axis and 4 units along the y-axis.
To graph the ellipse, plot the points \((\pm 3, 0)\) and \((0, \pm 4)\) on the coordinate plane, which are the vertices of the ellipse along the x- and y-axes respectively.
Use the ellipse equation to express \(y\) in terms of \(x\) to understand the shape: rearrange to \(\frac{y^2}{16} = 1 - \frac{x^2}{9}\), then multiply both sides by 16 to get \(y^2 = 16 \left(1 - \frac{x^2}{9}\right)\), and finally take the square root to find \(y = \pm 4 \sqrt{1 - \frac{x^2}{9}}\).
Determine the domain and range from the ellipse: since \(x^2/9 \leq 1\), the domain is \(-3 \leq x \leq 3\); similarly, since \(y^2/16 \leq 1\), the range is \(-4 \leq y \leq 4\).
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equation of an Ellipse
An ellipse is a set of points where the sum of distances to two fixed points (foci) is constant. Its standard form equation is (x²/a²) + (y²/b²) = 1, where a and b are the lengths of the semi-major and semi-minor axes. This equation describes the shape and size of the ellipse on the coordinate plane.
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5:30
Foci and Vertices of an Ellipse
Domain and Range of a Relation
The domain of a relation is the set of all possible x-values, while the range is the set of all possible y-values. For graphs like ellipses, these correspond to the horizontal and vertical extents of the curve. Determining domain and range involves identifying the minimum and maximum values x and y can take.
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4:22Domain & Range of Transformed Functions
Graphing Conic Sections
Graphing conic sections involves plotting points that satisfy their equations and understanding their geometric properties. For ellipses, the graph is symmetric about both axes, with intercepts at ±a on the x-axis and ±b on the y-axis. Recognizing these features helps in sketching the graph accurately.
Recommended video:
3:08Geometries from Conic Sections
Related Practice
Textbook Question
Identify each equation without completing the square.
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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. 9x2 +25y² - 36x + 50y – 164 = 0
Textbook Question
Identify each equation without completing the square.
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. 9x² + 16y² – 18x + 64y – 71 = 0
Textbook Question
Identify each equation without completing the square. 4x2 - 9y2 - 8x - 36y - 68 = 0