- Recall that \(y\) is a function of \(x\) if each \(x\) yields exactly one \(y\).
- Analyze each equation: \(x=5\) (not a function), \(x=y^2+9\) (two \(y\) values), \(x^2=y\) (one \(y\) value), \(x^2=y^2+16\) (two \(y\) values).
- Confirm that \(x^2=y\) maps every \(x\) to a unique \(y\).
- Final answer: \(\boxed{x^2 = y}\).
Explanation
- Analyzing the definition of a function of x
To determine which equation represents a function of \(x\), we recall that a relationship \(y = f(x)\) defines \(y\) as a function of \(x\) if and only if for every input \(x\), there corresponds exactly one output \(y\). Algebraically, if we solve the equation for \(y\), we should obtain a unique expression for \(y\) in terms of \(x\). Graphically, this corresponds to the vertical line test: any vertical line \(x = c\) must intersect the graph at most once.
- Evaluating the given equations
Let's examine each given equation individually:
- \(x = 5\): This equation describes a vertical line where \(x\) is fixed at 5, and \(y\) can be any real value. Since a single value of \(x\) maps to infinitely many values of \(y\), it does not define \(y\) as a function of \(x\).
- \(x = y^2 + 9\): Solving for \(y\), we get \(y^2 = x - 9\), which implies \(y = \pm\sqrt{x - 9}\). For any \(x > 9\), this gives two distinct values for \(y\), so it is not a function of \(x\).
- \(x^2 = y\): Solving for \(y\), we get \(y = x^2\). For any given real value \(x\), this expression yields exactly one unique value for \(y\). Therefore, this equation defines \(y\) as a function of \(x\).
- \(x^2 = y^2 + 16\): Solving for \(y\), we get \(y^2 = x^2 - 16\), which implies \(y = \pm\sqrt{x^2 - 16}\). Similar to the second case, for values of \(x\) such that \(x^2 > 16\), we obtain two possible values for \(y\), so it is not a function of \(x\).
- Conclusion
Based on the analysis, the only equation that assigns a single, unique \(y\) value for every valid \(x\) input is \(x^2 = y\). Thus, the correct equation is \(x^2 = y\). The final answer is \(\boxed{x^2 = y}\).
Examples
In engineering and physics, functions are crucial for modeling systems. For example, if you are modeling the distance \(d\) an object falls over time \(t\), you use the formula \(d = \frac{1}{2}gt^2\). Here, the distance is a function of time, meaning for any specific moment in time, the object has exactly one position. Knowing how to identify if a model is a function is essential for ensuring your calculations provide predictable and unique results.