Finding The Minimum Value Of Quadratic Function F(x) = 3x² - √8x + 32

In the realm of mathematics, pinpointing the minimum value of a function stands as a fundamental challenge with far-reaching applications across diverse fields such as physics, engineering, and economics. Among the myriad of functions encountered, quadratic functions hold a special place due to their inherent properties and the elegant methods available for determining their extreme values. This comprehensive guide delves into the intricacies of finding the minimum value of a quadratic function, employing a blend of analytical techniques and practical insights.

At its core, a quadratic function is a polynomial function of degree two, typically expressed in the form f(x) = ax² + bx + c, where 'a', 'b', and 'c' represent constant coefficients, and 'a' is non-zero. The graph of a quadratic function invariably takes the shape of a parabola, a symmetrical U-shaped curve. The parabola's orientation, whether it opens upwards or downwards, hinges on the sign of the coefficient 'a'. When 'a' is positive, the parabola opens upwards, signifying the existence of a minimum value at its vertex. Conversely, when 'a' is negative, the parabola opens downwards, indicating a maximum value at the vertex. Our primary focus here lies on quadratic functions with a positive leading coefficient ('a' > 0), as these functions possess a minimum value that we aim to uncover.

Several robust methods exist for determining the minimum value of a quadratic function. Let's explore the most prevalent techniques:

1. Completing the Square

Completing the square stands as a powerful algebraic technique that transforms a quadratic expression into a perfect square trinomial, plus a constant term. This transformation unveils the vertex form of the quadratic function, directly revealing the minimum value and the corresponding x-value at which it occurs. The vertex form is expressed as f(x) = a(x - h)² + k, where (h, k) represents the coordinates of the vertex. The minimum value, in this case, is 'k', occurring at x = h.

The process of completing the square involves a series of algebraic manipulations. First, factor out the coefficient 'a' from the x² and x terms. Next, take half of the coefficient of the x term, square it, and add and subtract it inside the parentheses. This maneuver allows you to rewrite the expression within the parentheses as a perfect square trinomial. Finally, simplify the expression to arrive at the vertex form.

2. Using the Vertex Formula

The vertex formula provides a direct and efficient means of calculating the coordinates of the vertex, and consequently, the minimum value of the quadratic function. The vertex formula states that for a quadratic function in the standard form f(x) = ax² + bx + c, the x-coordinate of the vertex (h) is given by h = -b / 2a. To obtain the y-coordinate of the vertex (k), simply substitute the value of h back into the original function, i.e., k = f(h). The minimum value of the function is then 'k', occurring at x = h.

The vertex formula stems from the process of completing the square and offers a shortcut for finding the vertex without explicitly performing the algebraic manipulations involved in completing the square.

3. Calculus Approach: Finding Critical Points

Calculus provides a powerful arsenal of tools for optimization problems, including finding the minimum value of a function. The cornerstone of this approach lies in the concept of critical points. Critical points are the points where the derivative of the function equals zero or is undefined. These points represent potential locations of local minima or maxima.

To find the minimum value of a quadratic function using calculus, first, compute the derivative of the function. For a quadratic function f(x) = ax² + bx + c, the derivative is f'(x) = 2ax + b. Next, set the derivative equal to zero and solve for x. The solution(s) represent the x-coordinates of the critical points. To determine whether a critical point corresponds to a minimum or maximum, you can employ the second derivative test. Compute the second derivative of the function. If the second derivative is positive at a critical point, then the point corresponds to a local minimum. If the second derivative is negative, then the point corresponds to a local maximum. For quadratic functions with a positive leading coefficient, the single critical point will always correspond to the minimum value.

Let's illustrate these methods with a concrete example. Consider the quadratic function f(x) = 3x² - √8x + 32. Our mission is to find the minimum value of this function.

1. Completing the Square:

Factor out the coefficient 3 from the x² and x terms: f(x) = 3(x² - (√8/3)x) + 32

Take half of the coefficient of the x term (-√8/3), square it ((√8/6)² = 8/36 = 2/9), and add and subtract it inside the parentheses: f(x) = 3(x² - (√8/3)x + 2/9 - 2/9) + 32

Rewrite the expression within the parentheses as a perfect square trinomial: f(x) = 3((x - √2/3)²) - 3(2/9) + 32

Simplify: f(x) = 3(x - √2/3)² - 2/3 + 32 f(x) = 3(x - √2/3)² + 94/3

From the vertex form, we see that the vertex is (√2/3, 94/3), and the minimum value is 94/3.

2. Using the Vertex Formula:

The x-coordinate of the vertex is h = -b / 2a = -(-√8) / (2 * 3) = √8 / 6 = √2 / 3.

The y-coordinate of the vertex is k = f(h) = f(√2/3) = 3(√2/3)² - √8(√2/3) + 32 = 3(2/9) - 4/3 + 32 = 2/3 - 4/3 + 32 = -2/3 + 32 = 94/3.

The minimum value is 94/3.

3. Calculus Approach:

Compute the derivative: f'(x) = 6x - √8

Set the derivative equal to zero and solve for x: 6x - √8 = 0 => x = √8 / 6 = √2 / 3

Compute the second derivative: f''(x) = 6, which is positive, indicating a minimum.

Substitute x = √2/3 back into the original function to find the minimum value: f(√2/3) = 3(√2/3)² - √8(√2/3) + 32 = 94/3

As evidenced by these three distinct methods, the minimum value of the function f(x) = 3x² - √8x + 32 is consistently found to be 94/3. This example underscores the versatility and reliability of these techniques in determining the minimum value of a quadratic function.

The ability to find the minimum value of a function extends beyond theoretical mathematics, finding practical applications in various real-world scenarios. Here are a few notable examples:

1. Optimization Problems: In engineering and physics, optimization problems frequently involve minimizing a certain quantity, such as energy consumption or material cost. Quadratic functions often arise in these contexts, and finding their minimum values becomes crucial for achieving optimal designs and solutions.

2. Trajectory Analysis: The trajectory of a projectile, such as a ball thrown into the air, can be modeled using a quadratic function. The minimum value of this function represents the lowest point the projectile reaches, which is essential for analyzing its motion.

3. Business and Economics: In business and economics, quadratic functions can model cost functions, profit functions, and revenue functions. Finding the minimum cost or the maximum profit often involves determining the minimum or maximum value of a quadratic function.

Finding the minimum value of a quadratic function constitutes a fundamental skill in mathematics with broad applications across diverse disciplines. Whether employing the elegance of completing the square, the directness of the vertex formula, or the rigor of calculus, the techniques discussed in this guide empower you to confidently tackle such problems. By mastering these methods, you unlock a powerful tool for optimization and problem-solving in a wide range of real-world contexts. The journey through quadratic functions unveils not only the beauty of mathematical techniques but also their practical significance in shaping our understanding of the world around us. From engineering marvels to economic models, the ability to pinpoint minimum values plays a pivotal role in achieving efficiency, optimizing performance, and making informed decisions.