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cide whether the relation defines a function. \( \begin{array}{ll}\text { 8) } y^{2}=x & \text { B) Function }\end{array} \)

Ask by Hills Matthews. in Qatar
Feb 24,2025

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Answer

The relation \( y^2 = x \) is not a function because for each positive \( x \), there are two possible \( y \) values.

Solution

To determine whether the relation \( y^2 = x \) defines a function, we need to check if for every value of \( x \), there is exactly one corresponding value of \( y \). 1. **Understanding the relation**: The equation \( y^2 = x \) implies that \( y \) can take on both positive and negative values for a given positive \( x \). Specifically, if we solve for \( y \), we get: \[ y = \sqrt{x} \quad \text{and} \quad y = -\sqrt{x} \] This means that for each positive \( x \), there are two corresponding values of \( y \) (one positive and one negative). 2. **Checking the definition of a function**: A relation is a function if each input (in this case, each \( x \)) has exactly one output (the corresponding \( y \)). Since we found that for each positive \( x \), there are two outputs (both \( \sqrt{x} \) and \( -\sqrt{x} \)), this violates the definition of a function. 3. **Conclusion**: The relation \( y^2 = x \) does not define a function because it does not satisfy the criterion of having a unique output for each input. Thus, the answer is: - **B) Not a function**.

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To determine whether the given relation \( y^{2} = x \) defines a function, we need to check if every value of \( x \) corresponds to exactly one value of \( y \). In this case, for a positive \( x \), there are two corresponding values of \( y \) (both \( y = \sqrt{x} \) and \( y = -\sqrt{x} \)), meaning that this relation does not meet the criteria for a function. Therefore, \( y^{2} = x \) does not define a function. For a relation to be a function, each input (or \( x \) value) must have a unique output (or \( y \) value). In the case of \( y^{2} = x \), any positive \( x \) will yield two outputs for \( y \), which violates the definition of a function.

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