Vertex Coordinates: F(x) = 5x^2 - 6x + 3

Hey guys! Today, we're diving into the fascinating world of quadratic functions and tackling a common question: How do we find the vertex coordinates? Specifically, we'll be working with the function f(x) = 5x² - 6x + 3. Don't worry, it's not as scary as it looks! We'll break it down step-by-step, so you'll be a vertex-finding pro in no time. Understanding the vertex is crucial because it represents the minimum or maximum point of the parabola, which is the graph of a quadratic function. Let's get started!

Understanding Quadratic Functions and the Vertex

Before we jump into the calculations, let's make sure we're all on the same page about quadratic functions. A quadratic function is a polynomial function of degree two, generally written in the standard form: f(x) = ax² + bx + c, where a, b, and c are constants and a is not equal to zero. The graph of a quadratic function is a parabola, a U-shaped curve. The vertex of the parabola is the point where the parabola changes direction. If a is positive, the parabola opens upwards, and the vertex represents the minimum point of the function. Conversely, if a is negative, the parabola opens downwards, and the vertex represents the maximum point.

In our example, f(x) = 5x² - 6x + 3, we have a = 5, b = -6, and c = 3. Since a is positive, we know the parabola opens upwards, and we're looking for the minimum point. There are a couple of ways to find the vertex coordinates, and we'll explore the most common and efficient method: using the vertex formula. This formula directly gives us the x-coordinate of the vertex, which we can then use to find the y-coordinate. Grasping this concept is essential as the vertex dictates the extreme values of the quadratic function. Think of it as the turning point, the spot where the function shifts from decreasing to increasing or vice versa. This characteristic makes the vertex a key feature in various applications, from physics problems involving projectile motion to optimization problems in economics and engineering. So, paying close attention to the vertex will unlock a deeper understanding of quadratic functions and their practical uses. Remember, the ability to quickly and accurately identify the vertex is a valuable skill, making the methods we're about to discuss incredibly useful in a wide range of scenarios.

Method 1: Using the Vertex Formula

The vertex formula is your best friend when it comes to finding the vertex coordinates quickly and easily. The formula states that the x-coordinate of the vertex, often denoted as h, is given by: h = -b / 2a. Once we have h, we can find the y-coordinate of the vertex, often denoted as k, by plugging h back into the original function: k = f(h). Let's apply this to our function, f(x) = 5x² - 6x + 3.

1. Identify a, b, and c: As we mentioned earlier, a = 5, b = -6, and c = 3.
2. Calculate the x-coordinate (h): Using the formula h = -b / 2a, we get h = -(-6) / (2 * 5) = 6 / 10 = 3/5. So, the x-coordinate of the vertex is 3/5.
3. Calculate the y-coordinate (k): Now, we need to find k = f(h) = f(3/5). Let's plug 3/5 into our function: f(3/5) = 5(3/5)² - 6(3/5) + 3 = 5(9/25) - 18/5 + 3 = 9/5 - 18/5 + 15/5 = 6/5. Therefore, the y-coordinate of the vertex is 6/5.

So, the vertex coordinates are (3/5, 6/5). This means the minimum point of the parabola represented by f(x) = 5x² - 6x + 3 is located at (3/5, 6/5). Remember, the vertex formula is your go-to tool for efficiently finding the vertex. Mastering its application ensures you can quickly determine the extreme point of any quadratic function. This skill is particularly valuable in real-world applications, such as optimizing business processes or analyzing physical trajectories. For instance, imagine you're trying to maximize the profit of a product. By modeling the profit as a quadratic function, the vertex formula can pinpoint the production level that yields the highest profit. Similarly, in physics, the trajectory of a projectile, like a ball thrown in the air, follows a parabolic path. Knowing the vertex allows you to calculate the maximum height the ball reaches. The beauty of the vertex formula lies in its simplicity and directness. It avoids the need for more complex methods like completing the square or calculus-based approaches, making it accessible and practical for a wide range of problems. By understanding and applying the vertex formula, you're not just solving mathematical equations; you're gaining a powerful tool for analyzing and optimizing scenarios across diverse fields.

Method 2: Completing the Square (Alternative Method)

While the vertex formula is the quickest way, it's also helpful to understand another method: completing the square. This method transforms the quadratic function into vertex form, which directly reveals the vertex coordinates. The vertex form of a quadratic function is f(x) = a(x - h)² + k, where (h, k) are the vertex coordinates. Let's see how this works with f(x) = 5x² - 6x + 3.

1. Factor out 'a' from the x² and x terms: In our case, a = 5, so we factor it out: f(x) = 5(x² - (6/5)x) + 3.
2. Complete the square inside the parentheses: To complete the square, we take half of the coefficient of the x term, square it, and add it inside the parentheses. The coefficient of the x term is -6/5, so half of it is -3/5, and squaring it gives us 9/25. Now we add and subtract 9/25 inside the parentheses: f(x) = 5(x² - (6/5)x + 9/25 - 9/25) + 3.
3. Rewrite the perfect square trinomial: The first three terms inside the parentheses now form a perfect square trinomial: f(x) = 5((x - 3/5)²) - 5(9/25) + 3.
4. Simplify: Distribute the 5 and simplify: f(x) = 5(x - 3/5)² - 9/5 + 3 = 5(x - 3/5)² + 6/5.

Now, our function is in vertex form: f(x) = 5(x - 3/5)² + 6/5. Comparing this to the general vertex form f(x) = a(x - h)² + k, we can see that h = 3/5 and k = 6/5. So, the vertex coordinates are (3/5, 6/5), which matches our result from the vertex formula. Completing the square provides a deeper understanding of how quadratic functions are structured. While it might seem more involved than the vertex formula, this method is invaluable in various mathematical contexts, such as solving quadratic equations and graphing parabolas. Imagine, for instance, you're faced with a quadratic equation that doesn't readily factor. Completing the square can transform it into a form where you can easily isolate the variable and find the solutions. Furthermore, understanding this technique enhances your ability to manipulate and simplify algebraic expressions, a crucial skill in more advanced mathematical studies. The key takeaway is that completing the square isn't just about finding the vertex; it's about mastering a fundamental algebraic tool that opens doors to solving a wider range of mathematical problems. By investing time in understanding this method, you're building a strong foundation for tackling more complex challenges in the future.

Conclusion

So, there you have it! We've successfully found the vertex coordinates of the function f(x) = 5x² - 6x + 3 using both the vertex formula and the method of completing the square. The vertex is located at (3/5, 6/5). Remember, the vertex represents the minimum point of this parabola since the coefficient of the x² term is positive. Whether you prefer the quickness of the vertex formula or the deeper understanding gained from completing the square, you now have two powerful tools in your arsenal for analyzing quadratic functions. Both methods are valuable, and choosing the right one often depends on the specific context of the problem. The vertex formula provides a direct route to the solution, making it ideal for situations where you need the answer quickly and efficiently. On the other hand, completing the square offers a more thorough understanding of the function's structure, which can be beneficial when you need to manipulate the equation or solve related problems. In essence, mastering both methods gives you the flexibility to tackle a wider range of quadratic function problems with confidence. Think of it as having two different tools in your toolbox – each suited for different tasks but both essential for achieving the desired outcome. So, practice both methods, explore their nuances, and become a true master of quadratic functions! Keep practicing, and you'll become a pro at finding vertices in no time! Now go out there and conquer those parabolas!