- A function \(y=f(x)\) must assign exactly one output \(y\) for each input \(x\).
- Equation \(y^2 = 4x^2 - 100\), \(x = -y^2 + 60y - 400\), and \(x^2 = -6y^2 + 200\) result in multiple \(y\)-values for a single \(x\), thus failing the definition of a function.
- Equation \(y = -x^2 + 50x - 300\) explicitly defines a unique \(y\) for each \(x\), satisfying the function criteria.
- The correct equation is \(\boxed{y = -x^2 + 50x - 300}\).
Explanation
- Defining a Function of \(x\)
A mathematical function \(y = f(x)\) is a rule that assigns to each input value \(x\) exactly one output value \(y\). When we say that the weekly profit \(y\) is a function of the number of items sold \(x\), we mean that for any given number of items \(x\), there must be a unique, well-defined profit \(y\). Graphically, this is represented by the vertical line test: any vertical line in the \(xy\)-plane must intersect the graph of the function at most once. Conversely, equations that produce two or more values of \(y\) for a single value of \(x\) do not represent a function.
- Evaluating the Given Equations
Let's test each option to see if it allows us to define \(y\) as a function of \(x\):
- Option 1: \(y^2 = 4x^2 - 100\)
Solving for \(y\), we get \(y = \pm\sqrt{4x^2 - 100}\). Because of the \(\pm\) sign, for any valid \(x\), there are generally two possible values for \(y\). This fails the condition of a function.
- Option 2: \(y = -x^2 + 50x - 300\)
This equation is already explicitly solved for \(y\) in terms of \(x\). For any input \(x\), there is exactly one value of \(y\) produced by the expression \(-x^2 + 50x - 300\). This matches the definition of a function.
- Option 3: \(x = -y^2 + 60y - 400\)
This equation expresses \(x\) as a function of \(y\). If we tried to solve for \(y\), we would need to rearrange it into a quadratic form \(y^2 - 60y + (x + 400) = 0\) and use the quadratic formula: \(y = \frac{60 \pm \sqrt{3600 - 4(x + 400)}}{2}\). The \(\pm\) sign indicates two possible values for \(y\) for a single \(x\), so this does not represent \(y\) as a function of \(x\).
- Option 4: \(x^2 = -6y^2 + 200\)
Rearranging for \(y\), we get \(6y^2 = 200 - x^2\), so \(y^2 = \frac{200 - x^2}{6}\). Taking the square root, we get \(y = \pm\sqrt{\frac{200 - x^2}{6}}\). Again, this gives two possible values of \(y\) for each \(x\), failing the function condition.
- Conclusion
Based on the analysis, only the equation \(y = -x^2 + 50x - 300\) correctly defines \(y\) as a single-valued function of \(x\). Therefore, this is the equation that can represent the company's weekly profit. The final answer is \(\boxed{y = -x^2 + 50x - 300}\).
Examples
In business, profit models are crucial. For example, a store owner might model their daily profit based on how many units of a product are sold. If the price per unit is constant or if production costs vary with volume, the profit equation might be a linear or quadratic function. Understanding whether an equation represents a function allows the owner to accurately predict profit for any sales quantity, ensuring they can plan their budget and inventory effectively.