Textbook Question
Identify each equation without completing the square.
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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 51
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 a hyperbola centered at the origin. This is because it matches the standard form of a hyperbola with a horizontal transverse axis: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) where \(a^2 = 9\) and \(b^2 = 16\).
To graph the hyperbola, first identify the vertices on the x-axis at \((\pm a, 0)\), which are \((\pm 3, 0)\) in this case. These points are where the hyperbola intersects the x-axis.
Next, find the asymptotes of the hyperbola, which are lines that the graph approaches but never touches. For this hyperbola, the asymptotes are given by the equations \(y = \pm \frac{b}{a} x\), or \(y = \pm \frac{4}{3} x\).
To determine the domain, consider the values of \(x\) for which the expression under the square root (when solving for \(y\)) is non-negative. Since \(\frac{x^2}{9} - 1 \geq 0\), solve \(\frac{x^2}{9} \geq 1\) which leads to \(|x| \geq 3\). Thus, the domain is \((-\infty, -3] \cup [3, \infty)\).
To find the range, solve the equation for \(y\): \(y^2 = 16 \left( \frac{x^2}{9} - 1 \right)\). Since \(y^2 \geq 0\), the range depends on \(x\). For each \(x\) in the domain, \(y\) can take values \(\pm \sqrt{16 \left( \frac{x^2}{9} - 1 \right)}\). Therefore, the range is all real numbers except values between \(-0\) and \(0\) when \(|x| < 3\), so the range is \((-\infty, \infty)\) but with \(y=0\) only at \(x=\pm 3\).
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hyperbola Equation and Its Graph
The given equation represents a hyperbola, a type of conic section characterized by two separate branches. It is defined by the difference of squared terms divided by constants equal to 1. Understanding the standard form helps in sketching the graph, identifying vertices, asymptotes, and the orientation of the hyperbola.
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Graph Hyperbolas at the Origin
Domain of a Relation
The domain of a relation is the set of all possible x-values for which the relation is defined. For the hyperbola, the domain is restricted by the values of x that keep the expression under the square root or denominator valid, ensuring the equation holds true and the graph exists.
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Range of a Relation
The range is the set of all possible y-values that the relation can take. For the hyperbola, the range depends on the values of y that satisfy the equation for given x-values, often found by analyzing the graph or solving the equation for y in terms of x.
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Related Practice
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
Graph each ellipse and give the location of its foci. 9(x − 1)²+4(y+3)² = 36
Textbook Question
Identify each equation without completing the square. 4x2 - 9y2 - 8x - 36y - 68 = 0
Textbook Question
In Exercises 43–50, convert each equation to standard form by completing the square on x and y. Then graph the hyperbola. Locate the foci and find the equations of the asymptotes.