Range Of Quadratic Function F(x) = (x-4)(x-2)

Hey guys! Let's dive into finding the range of the quadratic function f(x) = (x-4)(x-2). This is a super common type of problem in mathematics, and understanding how to solve it will definitely help you out. We'll break it down step by step, so don't worry if it seems a bit tricky at first. By the end of this article, you'll be a pro at finding the range of quadratic functions. We'll explore the concepts behind quadratic functions, how to find the vertex, and ultimately, how to determine the range. So, grab your thinking caps, and let's get started!

Understanding Quadratic Functions

First off, let's make sure we're all on the same page about what a quadratic function actually is. A quadratic function is basically a polynomial function of degree two. That means the highest power of x in the function is 2. The general form of a quadratic function is f(x) = ax² + bx + c, where a, b, and c are constants, and a is not equal to 0 (otherwise, it wouldn't be quadratic anymore!). In our case, f(x) = (x-4)(x-2), we can expand this to see it in the general form. When you multiply it out, you get f(x) = x² - 6x + 8. See? The highest power of x is 2, so it's definitely a quadratic function. Now, why is this important? Well, quadratic functions have a very specific shape when you graph them – a parabola. A parabola is a U-shaped curve, and it can either open upwards or downwards, depending on the sign of the coefficient a. If a is positive, the parabola opens upwards, and if a is negative, it opens downwards. This shape is super crucial for determining the range because the range is all the possible y-values the function can take. For a parabola that opens upwards, there's a minimum y-value (the bottom of the U), and the range will be all y-values greater than or equal to that minimum. For a parabola that opens downwards, there's a maximum y-value (the top of the U), and the range will be all y-values less than or equal to that maximum. In our function, f(x) = x² - 6x + 8, the coefficient of x² (which is a) is 1, which is positive. So, our parabola opens upwards. This means we're looking for a minimum y-value, and the range will be all real numbers greater than or equal to that minimum. Got it? Great! Let's move on to finding that minimum value.

Finding the Vertex

The vertex of a parabola is the point where the parabola changes direction. It's either the lowest point (minimum) if the parabola opens upwards or the highest point (maximum) if it opens downwards. Since we know our parabola opens upwards, the vertex will be the minimum point. Finding the vertex is key to finding the range because the y-coordinate of the vertex is the minimum y-value of the function. There are a couple of ways to find the vertex. One way is to use the formula for the x-coordinate of the vertex, which is x = -b / 2a. Remember the general form f(x) = ax² + bx + c? Well, a and b are the coefficients we need here. In our function, f(x) = x² - 6x + 8, a = 1 and b = -6. Plugging these values into the formula, we get x = -(-6) / (2 * 1) = 6 / 2 = 3. So, the x-coordinate of the vertex is 3. Now, to find the y-coordinate, we just plug this x-value back into the original function: f(3) = (3)² - 6(3) + 8 = 9 - 18 + 8 = -1. Therefore, the vertex of our parabola is (3, -1). Another way to find the vertex is by completing the square. Completing the square involves rewriting the quadratic function in vertex form, which is f(x) = a(x - h)² + k, where (h, k) is the vertex. Let's try it for our function: f(x) = x² - 6x + 8. To complete the square, we take half of the coefficient of the x term (which is -6), square it ((-3)² = 9), and add and subtract it inside the function: f(x) = x² - 6x + 9 - 9 + 8. Now, we can rewrite the first three terms as a perfect square: f(x) = (x - 3)² - 9 + 8. Simplifying, we get f(x) = (x - 3)² - 1. Ta-da! Now it's in vertex form, and we can see that the vertex is indeed (3, -1). No matter which method you use, finding the vertex is a crucial step in determining the range. We now know that the minimum y-value of our function is -1. Let's use it to find the range.

Determining the Range

Alright, we've found the vertex, which is (3, -1). We also know that our parabola opens upwards because the coefficient of x² is positive. This means the y-coordinate of the vertex, which is -1, is the minimum value of the function. So, what does that tell us about the range? Remember, the range is the set of all possible y-values the function can take. Since our parabola opens upwards and the minimum y-value is -1, the function can take any y-value greater than or equal to -1. In other words, the range of f(x) = (x-4)(x-2) is all real numbers greater than or equal to -1. We can write this in a few different ways. In set notation, we can write it as {y | y ≥ -1}, which means "the set of all y such that y is greater than or equal to -1." We can also write it in interval notation as [-1, ∞), which means the same thing. The square bracket on the -1 side means that -1 is included in the range, and the infinity symbol with a parenthesis means that the range extends infinitely upwards. So, there you have it! We've successfully found the range of the quadratic function f(x) = (x-4)(x-2). It's all real numbers greater than or equal to -1. Now, let’s recap the steps we took to solve this problem. First, we understood what a quadratic function is and how its shape (a parabola) affects its range. Then, we found the vertex of the parabola, which gave us the minimum y-value. Finally, we used the minimum y-value and the direction the parabola opens to determine the range.

Practice Problems

Now that we've walked through an example together, it's your turn to try a few on your own! Practice is key to mastering any math concept, and finding the range of quadratic functions is no exception. Here are a couple of practice problems for you guys to tackle. Try to follow the same steps we used in the example above: expand the function (if necessary), determine if the parabola opens upwards or downwards, find the vertex, and then determine the range. Problem 1: Find the range of the function g(x) = 2x² + 4x - 3. Problem 2: Find the range of the function h(x) = -(x + 1)(x - 3). Don't worry if you get stuck! Take your time, review the steps we discussed, and try to work through it. If you're still having trouble, you can always look up solutions online or ask a friend or teacher for help. The important thing is to keep practicing and keep learning. Remember, finding the range of a quadratic function involves a few key steps: understanding quadratic functions and parabolas, finding the vertex (either using the formula x = -b / 2a or completing the square), and using the vertex and the direction of the parabola to determine the range. With enough practice, you'll be able to solve these problems in your sleep! And who knows, maybe you'll even start seeing parabolas everywhere you go!

Conclusion

So, finding the range of a quadratic function might seem a little daunting at first, but as you can see, it's totally manageable if you break it down into smaller steps. We started by understanding the basic shape of a parabola and how it relates to the range of the function. Then, we learned how to find the vertex, which is the key to unlocking the range. Finally, we used the vertex and the direction the parabola opens to determine the range. The key takeaway here is that the range of a quadratic function depends on two things: the vertex and whether the parabola opens upwards or downwards. If you can remember these two things, you'll be well on your way to mastering this type of problem. And remember, practice makes perfect! The more you work with quadratic functions, the more comfortable you'll become with finding their ranges. So, keep practicing, keep learning, and keep rocking those math problems! You've got this, guys!