- A quadratic function \(f(x) = ax^2 + bx + c\) has a minimum at its vertex \((h, k)\) only if \(a > 0\).
- Using the vertex formula \(h = -\frac{b}{2a}\) and \(k = f(h)\), we test each option.
- Evaluating the functions:
- \(f(x) = -\frac{1}{2}x^2 + 4x - 11\) has a maximum (\(a < 0\)).
- \(f(x) = -2x^2 + 16x - 35\) has a maximum (\(a < 0\)).
- \(f(x) = \frac{1}{2}x^2 - 4x + 5\) has a minimum at \(h = 4, k = -3\).
- \(f(x) = 2x^2 - 16x + 35\) has a minimum at \(h = 4, k = 3\).
- The correct function is \(\boxed{f(x)=\frac{1}{2} x^2-4 x+5}\).
Explanation
- Understanding the Quadratic Function Properties
A quadratic function in the form \(f(x) = ax^2 + bx + c\) has a vertex at the point \((h, k)\). The x-coordinate of the vertex is given by the formula \(h = -\frac{b}{2a}\), and the y-coordinate is given by \(k = f(h)\). A quadratic function has a minimum value at its vertex if and only if the coefficient \(a\) is positive (the parabola opens upwards). If \(a\) is negative, the function has a maximum value instead. We are given four functions and need to determine which one has a minimum at \((4, -3)\).
- Evaluating the Functions
We evaluate the four given functions one by one:
- \(f(x) = -\frac{1}{2}x^2 + 4x - 11\): Here, \(a = -\frac{1}{2}\). Since \(a < 0\), the parabola opens downwards and has a maximum, not a minimum. We can eliminate this option.
- \(f(x) = -2x^2 + 16x - 35\): Here, \(a = -2\). Since \(a < 0\), this parabola also opens downwards and has a maximum. We can eliminate this option.
- \(f(x) = \frac{1}{2}x^2 - 4x + 5\): Here, \(a = \frac{1}{2} > 0\), which means it has a minimum. Let's calculate the vertex:
- \(h = -\frac{-4}{2(1/2)} = \frac{4}{1} = 4\).
- \(k = f(4) = \frac{1}{2}(4)^2 - 4(4) + 5 = \frac{1}{2}(16) - 16 + 5 = 8 - 16 + 5 = -3\).
This matches our target vertex \((4, -3)\).
- \(f(x) = 2x^2 - 16x + 35\): Here, \(a = 2 > 0\), which means it has a minimum. Let's calculate the vertex:
- \(h = -\frac{-16}{2(2)} = \frac{16}{4} = 4\).
- \(k = f(4) = 2(4)^2 - 16(4) + 35 = 2(16) - 64 + 35 = 32 - 64 + 35 = 3\).
The vertex is \((4, 3)\), which does not match our target.
- Conclusion
Based on our analysis, only the function \(f(x) = \frac{1}{2}x^2 - 4x + 5\) has a positive leading coefficient and a vertex at \((4, -3)\). Therefore, this is the correct function. The final answer is \(\boxed{f(x)=\frac{1}{2} x^2-4 x+5}\).
Examples
Quadratic functions are fundamental in modeling real-world situations involving optimization. For example, if a company wants to minimize its production costs, it can model the cost function as a quadratic equation. By finding the vertex of that function, the company can determine the exact production quantity that results in the lowest possible cost, thus maximizing efficiency and profit. Understanding how to locate the minimum point is a key skill for resource management in engineering, business, and economics.