Finding The Minimum Value Of F(x) = X² - 6 A Step-by-Step Guide
Hey guys! Today, we're diving into a super interesting problem that involves finding the minimum value of a function and then evaluating the function at that minimum point. It might sound a bit complex at first, but trust me, we'll break it down step by step so it's easy to understand. So, grab your thinking caps, and let's get started!
Understanding the Problem
Our mission, should we choose to accept it (and we do!), is to tackle this question: If the function f(x) = x² - 6 reaches its minimum value at a, what is the value of f(a)? We've even got some options to choose from: A) -6, B) 12, C) 30, and D) 0. To crack this, we need to understand what a minimum value is and how to find it for this particular function. We'll explore the concept of finding the minimum value of a function, especially for quadratic functions like the one we're dealing with.
The function f(x) = x² - 6 is a quadratic function. Quadratic functions, my friends, are those cool U-shaped curves you might remember from algebra class. The general form of a quadratic function is f(x) = ax² + bx + c, where a, b, and c are constants. In our case, a = 1, b = 0, and c = -6. Now, here's the key: when a is positive (like it is in our function), the parabola opens upwards, meaning it has a minimum point. This minimum point is what we're after! It's like finding the bottom of the U. For those of you who love visuals, think of it as the lowest point on the graph of the function. We're trying to find where this point is and what the function's value is at that point. This involves understanding the symmetry of parabolas and how the constants in the quadratic function affect its shape and position. The vertex form of a quadratic equation, f(x) = a(x - h)² + k, is particularly useful here, as (h, k) represents the vertex of the parabola. In our case, rewriting f(x) = x² - 6 in vertex form is straightforward, allowing us to quickly identify the minimum point.
Finding the Minimum Value
So, how do we find this minimum value? Well, for quadratic functions in the form f(x) = x² + c (notice our function fits this mold!), the minimum value occurs when x = 0. Why? Because squaring any other number will give you a positive result, making the function's value larger. The smallest possible value for x² is zero, and that happens when x is zero. This is a neat little trick to remember for these types of functions. To understand this better, think about the graph of y = x². It's a parabola sitting right on the x-axis, with its lowest point (the vertex) at the origin (0, 0). When we subtract 6 from x², we're just shifting the whole parabola down 6 units. This means the lowest point also shifts down, but it still occurs at x = 0. So, the minimum x-value remains the same, but the minimum y-value changes. Remember, the y-value represents the value of the function, f(x).
In our function, f(x) = x² - 6, the minimum value occurs at x = 0. This is our a! Now, to find the minimum value of the function, we simply plug x = 0 into the function. This gives us f(0) = 0² - 6 = -6. So, the function reaches its minimum value of -6 when x is 0. Isn't that cool? We've found the first part of our puzzle. We know that a = 0, and the minimum value of the function is -6. Now, we just need to find f(a), which is the value of the function at this minimum point. This involves a simple substitution and evaluation, but it's a crucial step to complete our solution. Understanding why the minimum occurs at x = 0 is also key. It's because the square of any real number is non-negative, and we're subtracting a constant from it. Therefore, the smallest value of the expression is when the squared term is zero.
Calculating f(a)
Now that we know the minimum value occurs at a = 0, we need to find f(a), which is f(0). We already did this in the previous step, but let's reiterate to make it super clear. We substitute a (which is 0) into our function: f(a) = f(0) = 0² - 6. This simplifies to f(0) = -6. So, there you have it! f(a) = -6. We've successfully found the value of the function at its minimum point. This step is crucial because it directly answers the question posed in the problem. We're not just finding the x-value where the minimum occurs; we're finding the actual minimum value of the function itself. This involves a straightforward calculation, but it's important to understand what we're calculating and why. We're essentially finding the y-coordinate of the vertex of the parabola, which represents the minimum value of the function. This process highlights the importance of understanding function notation and how to evaluate functions at specific values.
The Answer
Looking back at our options, A) -6, B) 12, C) 30, and D) 0, the correct answer is A) -6. We found that the function f(x) = x² - 6 reaches its minimum value at x = 0, and the value of the function at this point, f(0), is -6. We've successfully navigated this problem by understanding the properties of quadratic functions, identifying the minimum point, and evaluating the function at that point. Give yourselves a pat on the back, guys! You've tackled a problem that combines algebraic concepts and problem-solving skills. This is a great example of how math problems can be solved systematically by breaking them down into smaller, manageable steps. From understanding the nature of quadratic functions to performing simple substitutions, each step played a crucial role in arriving at the correct answer. This also reinforces the importance of reviewing and understanding the question carefully to ensure that the final answer addresses the specific request.
Key Takeaways
So, what have we learned today? We've seen that for quadratic functions in the form f(x) = x² + c, the minimum value occurs at x = 0. We've also practiced evaluating functions at specific points. More importantly, we've learned how to approach a problem step-by-step, breaking it down into smaller, more manageable parts. This is a skill that's super useful not just in math, but in life in general! Remember, understanding the properties of functions is key to solving these types of problems. Knowing that the minimum of x² occurs at 0 is a fundamental concept. And being able to substitute values into a function is a skill you'll use again and again in mathematics. Keep practicing, and you'll become a pro at these types of problems in no time! Always remember to carefully read the problem, identify the key information, and break the problem down into smaller steps. With a little practice and a solid understanding of the underlying concepts, you can conquer any math challenge that comes your way! And never be afraid to ask for help or clarification when you need it. Math is a collaborative journey, and we're all in this together!
In conclusion, by understanding the nature of quadratic functions and applying basic algebraic principles, we successfully found the minimum value of f(x) = x² - 6 and evaluated the function at that point. This exercise not only provided the correct answer but also reinforced essential problem-solving strategies that can be applied to various mathematical challenges. Keep up the great work, and remember, every problem is an opportunity to learn and grow!
