Understanding the concept of finding the minimum unit of a quadratic function is essential for various applications in mathematics, economics, and science. The function c(x) = 0.6x² - 168x + 30,389 is a quadratic function that we can analyze to determine its minimum value. In this article, we will delve into the methods used to find the minimum unit of this function, the significance of the coefficients, and the practical applications of quadratic functions.
The quadratic function has a parabolic shape, and its graph can open upwards or downwards depending on the sign of the coefficient of the x² term. In our case, since 0.6 is positive, the parabola opens upwards, indicating that there is indeed a minimum point. Finding this minimum point involves using calculus or the vertex formula for quadratic equations. Let’s explore these methods in detail.
Additionally, understanding how to find the minimum of a quadratic function is not just an academic exercise; it has real-world implications in fields such as finance for cost minimization and in engineering for optimizing resources. This article aims to provide a comprehensive guide for anyone looking to master this topic.
Table of Contents
What is a Quadratic Function?
A quadratic function is a polynomial function of degree two, typically written in the standard form:
c(x) = ax² + bx + c
where:
- a, b, and c are constants,
- x is the variable.
The graph of a quadratic function is a parabola. The direction in which it opens (upward or downward) is determined by the sign of the coefficient a. If a > 0, the parabola opens upwards, and if a < 0, it opens downwards.
Understanding the Coefficients
In our function c(x) = 0.6x² - 168x + 30,389, we have:
- a = 0.6
- b = -168
- c = 30,389
The coefficient a determines the width and direction of the parabola. The coefficient b affects the position of the vertex along the x-axis, while c is the y-intercept of the function.
Finding the Minimum Value
The minimum value of a quadratic function occurs at the vertex of the parabola. The x-coordinate of the vertex can be found using the formula:
x = -b/(2a)
Substituting our values:
x = -(-168)/(2 * 0.6) = 168/(1.2) = 140
Now that we have the x-coordinate of the vertex, we can find the minimum value by substituting x back into the function:
c(140) = 0.6(140)² - 168(140) + 30,389
Calculating c(140)
Calculating the values step-by-step:
- 0.6 * 19600 = 11760
- -168 * 140 = -23520
- 11760 - 23520 + 30389 = 18629
Thus, the minimum unit of the function c(x) occurs at c(140) = 18,629.
Using the Vertex Formula
Another way to find the minimum value is using the vertex form of a quadratic function:
c(x) = a(x-h)² + k
where (h, k) is the vertex of the parabola.
To convert our function into vertex form, we can complete the square:
c(x) = 0.6(x² - 280x) + 30,389
Now, to complete the square:
c(x) = 0.6((x - 140)² - 19600) + 30,389
Expanding this gives:
c(x) = 0.6(x - 140)² - 11760 + 30,389 = 0.6(x - 140)² + 18629
The vertex is (140, 18629), confirming our previous calculations.
Using Calculus to Find the Minimum
For those familiar with calculus, we can find the minimum value by taking the derivative of the function and setting it to zero:
c'(x) = 1.2x - 168
Setting the derivative to zero:
1.2x - 168 = 0
x = 140
Then, we substitute x back into the original function to find the minimum value:
c(140) = 18629
This approach confirms our findings using algebraic methods.
Applications of Quadratic Functions
Quadratic functions are widely used in various fields. Some practical applications include:
- Cost minimization in business operations
- Projectile motion in physics
- Design optimization in engineering
- Finance for profit maximization and loss minimization
Example Problems
Here are a couple of example problems that illustrate the use of quadratic functions:
- Problem 1: A company finds that its profit can be modeled by the function p(x) = -2x² + 120x - 1000. Find the maximum profit.
- Problem 2: A ball is thrown into the air, and its height can be modeled by the function h(t) = -4.9t² + 20t + 1. At what time does the ball reach its maximum height?
Conclusion
In summary, we explored the quadratic function c(x) = 0.6x² - 168x + 30,389 and determined that its minimum value occurs at x = 140, yielding a minimum unit of 18,629. Understanding the principles of quadratic functions is crucial for applications in various fields, from economics to engineering. We encourage you to practice these methods and apply them to real-world problems.
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