Maximum Or Minimum Value Of F(x)=-2x^2+32x-3 A Step-by-Step Guide

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Finding the maximum or minimum value of a quadratic function is a fundamental concept in mathematics with widespread applications in various fields, such as physics, engineering, and economics. This article delves into a comprehensive exploration of how to determine the maximum or minimum value of the quadratic function f(x) = -2x² + 32x - 3. We will dissect the properties of quadratic functions, explore different methods for finding the vertex, and provide practical examples to solidify your understanding. By the end of this guide, you will be equipped with the knowledge and skills to confidently tackle similar problems and apply these concepts in real-world scenarios.

Understanding Quadratic Functions

At its core, a quadratic function is a polynomial function of degree two, generally expressed in the form f(x) = ax² + bx + c, where a, b, and c are constants and a ≠ 0. The graph of a quadratic function is a parabola, a U-shaped curve that opens either upwards or downwards. The direction in which the parabola opens is determined by the sign of the coefficient a. If a > 0, the parabola opens upwards, indicating a minimum value; if a < 0, the parabola opens downwards, indicating a maximum value. In our specific case, f(x) = -2x² + 32x - 3, the coefficient a is -2, which is negative. This immediately tells us that the parabola opens downwards, and thus, the function has a maximum value.

The Vertex: The Key to Maximum and Minimum Values

The vertex of a parabola is the point where the curve changes direction. It is the turning point of the parabola and represents either the minimum value (for parabolas opening upwards) or the maximum value (for parabolas opening downwards) of the quadratic function. The coordinates of the vertex are typically denoted as (h, k), where h represents the x-coordinate and k represents the y-coordinate. The y-coordinate, k, is the maximum or minimum value of the function.

Methods for Finding the Vertex

There are several methods to find the vertex of a quadratic function, each with its own advantages and suitability depending on the given information. We will explore two common methods: completing the square and using the vertex formula.

Method 1: Completing the Square

Completing the square is a technique used to rewrite a quadratic expression in the form a(x - h)² + k, where (h, k) is the vertex of the parabola. This method is particularly useful for understanding the structure of the quadratic function and directly identifying the vertex. Let's apply this method to our function, f(x) = -2x² + 32x - 3.

  1. Factor out the coefficient of the x² term: f(x) = -2(x² - 16x) - 3

    This step isolates the x² and x terms, making it easier to complete the square within the parentheses.

  2. Complete the square inside the parentheses: To complete the square, we need to add and subtract the square of half the coefficient of the x term. The coefficient of the x term is -16, so half of it is -8, and the square of -8 is 64. We add and subtract 64 inside the parentheses: f(x) = -2(x² - 16x + 64 - 64) - 3

    Adding and subtracting the same value doesn't change the overall expression, but it allows us to rewrite the quadratic expression as a perfect square trinomial.

  3. Rewrite the perfect square trinomial: The expression x² - 16x + 64 is a perfect square trinomial and can be rewritten as (x - 8)²: f(x) = -2((x - 8)² - 64) - 3

  4. Distribute and simplify: Distribute the -2 and combine the constant terms: f(x) = -2(x - 8)² + 128 - 3 f(x) = -2(x - 8)² + 125

  5. Identify the vertex: Now the function is in the vertex form f(x) = a(x - h)² + k, where h = 8 and k = 125. Therefore, the vertex of the parabola is (8, 125).

By completing the square, we have successfully rewritten the quadratic function in vertex form. This form directly reveals the vertex of the parabola, which is the point where the maximum or minimum value occurs. In this case, the vertex is (8, 125), indicating that the maximum value of the function is 125, which occurs when x = 8.

Method 2: Using the Vertex Formula

An alternative and often quicker method to find the vertex is by using the vertex formula. For a quadratic function in the form f(x) = ax² + bx + c, the x-coordinate of the vertex, h, is given by:

h = -b / 2a

Once we find the x-coordinate, we can substitute it back into the original function to find the y-coordinate, k, which represents the maximum or minimum value.

Let's apply the vertex formula to our function, f(x) = -2x² + 32x - 3.

  1. Identify the coefficients: In our function, a = -2, b = 32, and c = -3.

  2. Calculate the x-coordinate of the vertex (h): h = -b / 2a = -32 / (2 * -2) = -32 / -4 = 8

  3. Calculate the y-coordinate of the vertex (k): Substitute h = 8 back into the original function: k = f(8) = -2(8)² + 32(8) - 3 k = -2(64) + 256 - 3 k = -128 + 256 - 3 k = 125

  4. Identify the vertex: The vertex is (8, 125).

Using the vertex formula, we arrive at the same conclusion as with completing the square: the vertex of the parabola is (8, 125). This confirms that the maximum value of the function f(x) = -2x² + 32x - 3 is 125, which occurs at x = 8.

Interpreting the Results and Real-World Applications

The vertex (8, 125) provides crucial information about the quadratic function f(x) = -2x² + 32x - 3. It tells us that the maximum value of the function is 125, and this maximum value is attained when x = 8. The parabola opens downwards because the coefficient of the x² term is negative, further solidifying that the vertex represents a maximum point.

Practical Applications of Finding Maximum and Minimum Values

The concept of finding maximum and minimum values of quadratic functions has numerous real-world applications. Here are a few examples:

  • Projectile Motion: In physics, the trajectory of a projectile (like a ball thrown in the air) can be modeled by a quadratic function. The maximum height reached by the projectile corresponds to the vertex of the parabola. Understanding this allows us to calculate the optimal launch angle and initial velocity to achieve maximum range or height.

  • Optimization Problems: Many real-world problems involve optimizing a certain quantity, such as maximizing profit or minimizing cost. If the relationship between the variables can be modeled by a quadratic function, finding the vertex helps determine the optimal solution. For instance, a business might use quadratic functions to model the relationship between the price of a product and the revenue generated, then find the vertex to determine the price that maximizes revenue.

  • Engineering Design: Engineers often use quadratic functions to model the shape of arches, bridges, and other structures. Finding the vertex and other key points on the parabola helps ensure structural stability and efficiency. For example, the parabolic shape of a bridge arch distributes weight evenly, making it a strong and stable structure.

Visualizing the Parabola

To further enhance your understanding, it's beneficial to visualize the parabola represented by f(x) = -2x² + 32x - 3. You can sketch the graph by plotting a few points, including the vertex (8, 125) and the x-intercepts (where the parabola crosses the x-axis). The x-intercepts can be found by setting f(x) = 0 and solving for x. Alternatively, you can use graphing software or online tools to generate the graph. The visual representation will clearly show the downward-opening parabola with its peak at the vertex, illustrating the maximum value of the function.

Conclusion

In this comprehensive guide, we explored how to find the maximum or minimum value of the quadratic function f(x) = -2x² + 32x - 3. We learned that the vertex of the parabola represents the maximum or minimum point of the function, and we discussed two methods for finding the vertex: completing the square and using the vertex formula. Both methods led us to the same conclusion: the vertex is (8, 125), indicating that the maximum value of the function is 125, which occurs when x = 8.

Furthermore, we highlighted the practical applications of these concepts in various fields, such as physics, optimization problems, and engineering design. By mastering the techniques for finding the vertex of a quadratic function, you gain a valuable tool for solving a wide range of real-world problems. Remember, the key to success is practice. Try applying these methods to other quadratic functions and exploring different scenarios to solidify your understanding. With consistent effort, you'll become proficient in analyzing and interpreting quadratic functions and their applications.