Finding The Vertex Of F(x) = (x-8)(x-2) A Step-by-Step Guide
Hey guys! Let's dive into a super common and important topic in algebra: finding the vertex of a quadratic function. Specifically, we're going to tackle the function f(x) = (x-8)(x-2). Trust me, once you get the hang of this, you'll be able to handle any quadratic that comes your way. Understanding the vertex is crucial because it gives us the maximum or minimum point of the parabola, which can be super useful in real-world applications like figuring out the peak of a projectile's trajectory or minimizing costs in a business scenario.
Understanding Quadratic Functions and Their Vertex
Before we jump into the specifics of our function, let's quickly recap what quadratic functions are all about. Quadratic functions are those that can be written in the general form f(x) = ax² + bx + c, where 'a', 'b', and 'c' are constants, and 'a' isn't zero. The graph of a quadratic function is a parabola, which is a U-shaped curve. Now, the vertex is the point where the parabola changes direction. If the parabola opens upwards (meaning 'a' is positive), the vertex is the lowest point, or the minimum. If it opens downwards (meaning 'a' is negative), the vertex is the highest point, or the maximum. Think of it like the bottom or the top of the U. Identifying the vertex is super helpful because it tells us a lot about the function's behavior. For instance, if you're dealing with a profit function, the vertex might represent the point of maximum profit. Or, if you're modeling the height of a ball thrown in the air, the vertex would represent the highest point the ball reaches. There are a few ways to find the vertex, and we'll explore a couple of them in this article to give you a solid understanding. So, stick with me, and let's get started!
Method 1: Expanding and Using the Vertex Formula
Okay, so let's get our hands dirty with the first method for finding the vertex. This involves expanding the given function and then using a nifty formula. Our function is f(x) = (x-8)(x-2). The first step is to expand this into the standard quadratic form, which is f(x) = ax² + bx + c. To do this, we'll use the good old FOIL method (First, Outer, Inner, Last). So, we multiply the terms in the parentheses like this:
- First: x * x = x²
- Outer: x * -2 = -2x
- Inner: -8 * x = -8x
- Last: -8 * -2 = 16
Now, we add all these terms together: x² - 2x - 8x + 16. Combining the like terms, we get f(x) = x² - 10x + 16. Awesome! We've got our quadratic function in the standard form. Now, here comes the magic formula. The x-coordinate of the vertex, which we'll call h, can be found using the formula h = -b / 2a. Remember, 'a' and 'b' are the coefficients from our standard form equation. In our case, a = 1 and b = -10. So, plugging these values into the formula, we get h = -(-10) / (2 * 1) = 10 / 2 = 5. So, the x-coordinate of our vertex is 5. But we're not done yet! We need to find the y-coordinate, which we'll call k. To do this, we simply plug our x-coordinate (h = 5) back into our function: f(5) = (5)² - 10(5) + 16 = 25 - 50 + 16 = -9. So, the y-coordinate of our vertex is -9. Putting it all together, the vertex of the quadratic function f(x) = (x-8)(x-2) is (5, -9). This method is super reliable and works for any quadratic function, so it's a great one to have in your toolkit. Let's move on to another method that can be even quicker in some cases.
Method 2: Using the Intercept Form
Alright, let's explore another way to find the vertex, and this one's particularly slick when our quadratic function is given in intercept form. Remember our function, f(x) = (x-8)(x-2)? This is actually in intercept form, which looks like f(x) = a(x - r₁)(x - r₂), where r₁ and r₂ are the x-intercepts of the parabola. In our case, we can easily see that r₁ = 8 and r₂ = 2. The x-intercepts are the points where the parabola crosses the x-axis, meaning f(x) = 0. They are also sometimes called the roots or zeros of the function. Now, here's the cool part: the vertex of a parabola is always smack-dab in the middle of the two x-intercepts. This is because parabolas are symmetrical. So, to find the x-coordinate of the vertex, we can simply take the average of the x-intercepts. That means we add the intercepts and divide by 2: (8 + 2) / 2 = 10 / 2 = 5. Ta-da! We've found the x-coordinate of the vertex, which is 5. Notice that this is the same x-coordinate we found using the first method. Consistency is key, guys! Now, just like before, we need to find the y-coordinate. We do this by plugging our x-coordinate (5) back into the original function: f(5) = (5 - 8)(5 - 2) = (-3)(3) = -9. And there you have it! The y-coordinate of the vertex is -9. So, using the intercept form method, we've once again found that the vertex of the quadratic function f(x) = (x-8)(x-2) is (5, -9). This method is often faster than expanding and using the vertex formula, especially when the function is already given in intercept form. Plus, it gives you a nice visual understanding of where the vertex sits in relation to the x-intercepts. Knowing both methods gives you flexibility and helps you choose the most efficient approach for any given problem.
Comparing the Two Methods
Okay, now that we've walked through two different methods for finding the vertex of a quadratic function, let's take a step back and compare them. This will help you decide which method to use in different situations. The first method, expanding and using the vertex formula (h = -b / 2a), is a solid, reliable approach that works for any quadratic function, no matter how it's presented. It's like the trusty old Swiss Army knife of vertex-finding methods. However, it does involve a bit more algebra, especially the expansion step, which can sometimes be prone to errors if you're not careful. On the other hand, the second method, using the intercept form, is often quicker and more elegant when the function is already given in the form f(x) = a(x - r₁)(x - r₂). It leverages the symmetry of the parabola and the relationship between the vertex and the x-intercepts. This method is like the sleek sports car of vertex-finding methods – fast and efficient when the conditions are right. However, it's not always applicable. If the function isn't given in intercept form, or if it's not easily factorable, this method might not be the best choice. So, which method should you use? Well, it really depends on the problem at hand. If the function is already in intercept form, or if you can easily factor it, the intercept form method is often the way to go. But if the function is in standard form (f(x) = ax² + bx + c) or if you're just not sure how to factor it, the vertex formula method is a safe bet. The best approach is to be comfortable with both methods so you can choose the one that's most efficient for each problem. Think of it like having two different tools in your toolbox – you want to know when to use the hammer and when to use the screwdriver.
Real-World Applications of the Vertex
So, we've learned how to find the vertex of a quadratic function, which is awesome! But you might be wondering, "Why is this so important? What can we actually do with this knowledge?" Well, guys, the vertex has a ton of real-world applications! Think about anything that follows a parabolic path or has a maximum or minimum value – that's where the vertex comes into play. One classic example is projectile motion. When you throw a ball, shoot an arrow, or launch a rocket, the path it follows is (ideally) a parabola. The vertex of that parabola represents the highest point the object reaches. So, if you know the quadratic function that describes the projectile's trajectory, you can use the vertex to figure out the maximum height and how far the object travels. This is super useful in sports, engineering, and even military applications. Another area where the vertex is crucial is in optimization problems. These are problems where you want to find the maximum or minimum value of something, like profit, cost, or area. For example, a business might use a quadratic function to model the relationship between the price of a product and the profit they make. The vertex of that function would represent the price that maximizes profit. Similarly, an architect might use a quadratic function to design a structure that minimizes the amount of material needed, while still meeting certain requirements. The vertex helps them find the optimal design. Quadratic functions and their vertices also pop up in physics, economics, and even computer graphics. They're incredibly versatile tools for modeling and solving a wide range of problems. So, by mastering the art of finding the vertex, you're not just learning a math skill – you're gaining a powerful tool for understanding and solving real-world problems. Keep that in mind as you continue your mathematical journey!
Conclusion
Alright, guys, we've reached the end of our journey to find the vertex of the quadratic function f(x) = (x-8)(x-2). We've covered a lot of ground, from understanding the basics of quadratic functions and their vertices to exploring two different methods for finding the vertex and even diving into some real-world applications. Remember, the vertex is a crucial point on a parabola, representing either the maximum or minimum value of the function. We learned two main methods for finding it: expanding and using the vertex formula (h = -b / 2a) and using the intercept form. Each method has its strengths and weaknesses, so it's important to be comfortable with both. By understanding these methods, you'll be well-equipped to tackle any quadratic function that comes your way. And remember, the vertex isn't just a mathematical concept – it's a powerful tool for solving real-world problems in fields like physics, engineering, economics, and more. So, keep practicing, keep exploring, and keep applying your knowledge. You've got this! If you found this guide helpful, share it with your friends and classmates, and keep an eye out for more math explorations in the future. Happy vertex-finding!
