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

Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
ChapterP: Prerequisites
SectionP.7: A Library Of Parent Functions
Problem 47E
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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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