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For each of the following, determine whether the equation defines y as a function of x. = kl 16 = 2 + x 2 2 16 + y = x y = 8 |x| − 1 - 3 5x = y Function Function Function Function Not a function Not a function Not a function Not a function

For each of the following, determine whether the equation defines y as a function of x. = kl 16 = 2 + x 2 2 16 + y = x y = 8 |x| − 1 - 3 5x = y Function Function Function Function Not a function Not a function Not a function Not a function

Algebra and Trigonometry (6th Edition)
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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### Understanding Functions: Identifying y as a Function of x **Instruction:** For each of the following, determine whether the equation defines \( y \) as a function of \( x \). **Equations and Determinations:** 1. \( 16 = \left| y \right| + x^2 \) - **Function** ⭕ - Not a function ⭕ 2. \( 16 + y^2 = x^2 \) - Function ⭕ - **Not a function** ⭕ 3. \( y = 8 \left| x - 1 \right| \) - Function ⭕ - **Not a function** ⭕ 4. \( 5x = y^3 \) - **Function** ⭕ - Not a function ⭕ **Explanation:** - For \( 16 = \left| y \right| + x^2 \), solving for \( y \) gives two possible values for most \( x \)-values, but one of the values is always negative. This does not violate the definition of \( y \) being a function of \( x \) as each \( x \) gives a single \( y \). Therefore, it is a function. - For \( 16 + y^2 = x^2 \), solving for \( y \) gives two possible values (positive and negative square roots) for each \( x \) (except \( x = 4 \) or \( x = -4 \)). Hence, it is not a function. - For \( y = 8 \left| x - 1 \right| \), each \( x \) value produces exactly one \( y \) value (because absolute values ensure non-negative outputs). Hence, it is a function. - For \( 5x = y^3 \), solving for \( y \) gives exactly one value for each \( x \) (different cube roots of the same value are not possible). Hence, it is a function.
Transcribed Image Text:### Understanding Functions: Identifying y as a Function of x **Instruction:** For each of the following, determine whether the equation defines \( y \) as a function of \( x \). **Equations and Determinations:** 1. \( 16 = \left| y \right| + x^2 \) - **Function** ⭕ - Not a function ⭕ 2. \( 16 + y^2 = x^2 \) - Function ⭕ - **Not a function** ⭕ 3. \( y = 8 \left| x - 1 \right| \) - Function ⭕ - **Not a function** ⭕ 4. \( 5x = y^3 \) - **Function** ⭕ - Not a function ⭕ **Explanation:** - For \( 16 = \left| y \right| + x^2 \), solving for \( y \) gives two possible values for most \( x \)-values, but one of the values is always negative. This does not violate the definition of \( y \) being a function of \( x \) as each \( x \) gives a single \( y \). Therefore, it is a function. - For \( 16 + y^2 = x^2 \), solving for \( y \) gives two possible values (positive and negative square roots) for each \( x \) (except \( x = 4 \) or \( x = -4 \)). Hence, it is not a function. - For \( y = 8 \left| x - 1 \right| \), each \( x \) value produces exactly one \( y \) value (because absolute values ensure non-negative outputs). Hence, it is a function. - For \( 5x = y^3 \), solving for \( y \) gives exactly one value for each \( x \) (different cube roots of the same value are not possible). Hence, it is a function.
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