Find The Minimum Value Of $f(x)=x^2-48x+2304$

In the realm of mathematics, understanding the behavior of functions is paramount, especially when we encounter quadratic functions like $f(x)=x^2-48x+2304$. These functions, characterized by their parabolic graphs, possess a distinct turning point – either a minimum or a maximum. For the given function, $f(x)=x^2-48x+2304$, since the coefficient of the $x^2$ term is positive (which is 1), we know that the parabola opens upwards, meaning it will have a minimum value. The core question then becomes: what is this minimum value, and how do we find it? This article will delve deep into the methods for determining the minimum value of this specific quadratic function, exploring the underlying mathematical principles and providing a clear, step-by-step approach. We will uncover that finding this minimum value is not just an abstract mathematical exercise but can have practical applications in various fields, from economics to physics. The journey to understanding the minimum value of $f(x)=x^2-48x+2304$ is a journey into the heart of quadratic analysis.

Understanding Quadratic Functions and Their Minimum Values

A quadratic function is a polynomial function of degree two, typically expressed in the standard form $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants and $a \neq 0$. The graph of a quadratic function is a parabola. The direction in which the parabola opens is determined by the sign of the coefficient $a$. If $a > 0$, the parabola opens upwards, and the function has a minimum value at its vertex. If $a < 0$, the parabola opens downwards, and the function has a maximum value at its vertex. For our specific function, $f(x) = x^2 - 48x + 2304$, we have $a=1$, $b=-48$, and $c=2304$. Since $a=1$ is positive, the parabola opens upwards, confirming that we are indeed looking for a minimum value. The vertex of the parabola represents this extreme point. The coordinates of the vertex $(h, k)$ tell us that the minimum (or maximum) value of the function is $k$, and this occurs at $x=h$. Therefore, the task boils down to accurately calculating the coordinates of the vertex for $f(x)=x^2-48x+2304$.

There are several algebraic methods to find the vertex of a parabola. One common approach is to use the vertex formula. The x-coordinate of the vertex, denoted by $h$, can be found using the formula $h = -b / (2a)$. Once we have the x-coordinate, we can find the corresponding y-coordinate, $k$, by substituting this value of $h$ back into the function, i.e., $k = f(h)$. This method is straightforward and relies on the direct application of a well-established formula. Another powerful technique is completing the square. This method involves rewriting the quadratic function from its standard form into vertex form, $f(x) = a(x-h)^2 + k$. By manipulating the expression $x^2 - 48x + 2304$ into this form, the values of $h$ and $k$ become immediately apparent. The term $(x-h)^2$ is always non-negative, so when $a > 0$, the minimum value of $f(x)$ is $k$, occurring when $x=h$ (making the $(x-h)^2$ term zero). Each of these methods offers a unique perspective on how to identify the minimum value, and understanding them provides a robust toolkit for solving problems involving quadratic functions.

Method 1: Using the Vertex Formula

The vertex formula provides a direct route to finding the minimum value of a quadratic function. For a function in the form $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex is given by $h = -b / (2a)$. In our case, $f(x) = x^2 - 48x + 2304$, we have $a=1$ and $b=-48$. Plugging these values into the formula, we get:

$h = -(-48) / (2 * 1)$ $h = 48 / 2$ $h = 24$

So, the x-coordinate of the vertex is 24. This means that the minimum value of the function occurs when $x=24$. To find the actual minimum value (the y-coordinate of the vertex, $k$), we substitute $x=24$ back into the original function:

$f(24) = (24)^2 - 48(24) + 2304$

Let's calculate this step-by-step:

$(24)^2 = 576$ $48(24) = 1152$

So, the expression becomes:

$f(24) = 576 - 1152 + 2304$

Now, perform the subtraction and addition:

$f(24) = -576 + 2304$ $f(24) = 1728$

Therefore, the minimum value of the function $f(x) = x^2 - 48x + 2304$ is 1728, and it occurs at $x=24$. This method is efficient and relies on a fundamental property of quadratic functions. The vertex formula directly gives us the location of the minimum or maximum, and substituting that location back into the function yields the extreme value itself. It's a powerful shortcut that avoids the need for more complex calculus or algebraic manipulations when only the vertex is required. The symmetry of the parabola around its vertex is what makes this formula work. By finding the midpoint of the roots (if they exist) or by considering the derivative (which we'll touch upon later), we can pinpoint this critical point. In essence, the vertex formula distills these concepts into a simple and elegant calculation, making it an indispensable tool for anyone working with quadratic equations.

Method 2: Completing the Square

Completing the square is another robust method to find the minimum value of a quadratic function. This technique involves rewriting the function in its vertex form, $f(x) = a(x-h)^2 + k$. For our function, $f(x) = x^2 - 48x + 2304$, we start by focusing on the $x^2$ and $x$ terms.

We want to transform $x^2 - 48x$ into a perfect square trinomial. A perfect square trinomial has the form $(x-h)^2 = x^2 - 2hx + h^2$. By comparing $x^2 - 48x$ with $x^2 - 2hx$, we can see that $-48x$ must correspond to $-2hx$. Thus, $-48 = -2h$, which implies $h = 24$. To complete the square, we need to add $h^2$ to the expression. In this case, $h^2 = (24)^2 = 576$.

So, we can rewrite the function as follows:

$f(x) = (x^2 - 48x + 576) - 576 + 2304$

Notice that we added and subtracted 576. Adding 576 completes the square, turning $x^2 - 48x + 576$ into $(x-24)^2$. Now, we combine the constant terms:

$f(x) = (x-24)^2 + (-576 + 2304)$ $f(x) = (x-24)^2 + 1728$

This is the vertex form of the quadratic function. From this form, $f(x) = a(x-h)^2 + k$, we can directly identify that $a=1$, $h=24$, and $k=1728$. Since $a=1$ is positive, the minimum value of the function is $k$, which is 1728. This minimum occurs when $(x-24)^2 = 0$, meaning $x=24$. Completing the square is a fundamental technique in algebra that not only helps in finding the vertex but also in solving quadratic equations and understanding the structure of conic sections. It provides a deeper insight into the transformation of functions and how algebraic manipulation can reveal their intrinsic properties. The elegance of this method lies in its ability to systematically uncover the vertex form, directly revealing the axis of symmetry and the minimum or maximum value of the function. It’s a testament to the power of rearranging mathematical expressions to simplify their analysis and interpretation.

Method 3: Using Calculus (Optional)

For those familiar with calculus, finding the minimum value of a function can be approached using derivatives. The derivative of a function gives the slope of the tangent line at any point. At a minimum or maximum point (a critical point), the slope of the tangent line is zero. Therefore, we can find the minimum value by setting the first derivative of the function equal to zero and solving for $x$.

First, let's find the derivative of $f(x) = x^2 - 48x + 2304$ with respect to $x$. Using the power rule for differentiation (which states that the derivative of $x^n$ is $nx^{n-1}$), we get:

$f'(x) = d/dx (x^2 - 48x + 2304)$ $f'(x) = d/dx (x^2) - d/dx (48x) + d/dx (2304)$ $f'(x) = 2x - 48 + 0$ $f'(x) = 2x - 48$

Now, to find the critical point, we set the derivative equal to zero:

$2x - 48 = 0$ $2x = 48$ $x = 48 / 2$ $x = 24$

This tells us that the critical point occurs at $x=24$. To confirm that this is a minimum, we can use the second derivative test. The second derivative of $f(x)$ is the derivative of $f'(x)$:

$f''(x) = d/dx (2x - 48)$ $f''(x) = 2$

Since the second derivative, $f''(x) = 2$, is positive, the function is concave up at $x=24$, confirming that this point is indeed a minimum. Finally, to find the minimum value, we substitute $x=24$ back into the original function $f(x)$:

$f(24) = (24)^2 - 48(24) + 2304$ $f(24) = 576 - 1152 + 2304$ $f(24) = 1728$

The calculus method, while requiring knowledge of derivatives, offers a systematic way to find extrema for a wide range of functions, not just quadratics. The fact that all three methods yield the same result reinforces the accuracy of our calculation. The derivative represents the instantaneous rate of change of the function, and where this rate of change is zero, the function is momentarily flat, indicating a peak or a valley. For parabolas, this flat point is always the vertex. The second derivative then tells us about the curvature of the function at that point – a positive second derivative means it curves upwards like a smiley face, hence a minimum, and a negative one means it curves downwards like a frowny face, indicating a maximum.

Conclusion: The Minimum Value Revealed

Through the application of three distinct mathematical approaches – the vertex formula, completing the square, and calculus – we have consistently arrived at the same conclusion regarding the function $f(x) = x^2 - 48x + 2304$. Each method, while employing different principles, illuminated the path to identifying the function's minimum value. The vertex formula provided a direct calculation for the x-coordinate of the vertex, which we then used to find the minimum y-value. Completing the square transformed the standard form into the vertex form, directly revealing the minimum value and the x-at-which-it-occurs. Finally, calculus offered a more general approach by analyzing the function's rate of change, using the derivative to locate the critical point and the second derivative to confirm it as a minimum.

In all instances, we found that the minimum value of the function $f(x) = x^2 - 48x + 2304$ is 1728, and this minimum occurs at $x = 24$. This understanding of quadratic functions and their extrema is fundamental in various mathematical and scientific disciplines. For instance, in economics, quadratic functions can model cost or revenue, and finding the minimum cost or maximum revenue is a crucial business objective. In physics, projectile motion is often described by quadratic equations, and determining the maximum height or minimum trajectory point is essential for analysis. The ability to confidently calculate these values empowers problem-solving across diverse fields.

For further exploration into the fascinating world of quadratic equations and functions, you might find the resources at Wolfram MathWorld to be incredibly insightful. They offer a comprehensive and detailed look at the mathematical properties and applications of these fundamental concepts.