Unveiling The Secrets: Graphing Rational Functions

Hey math enthusiasts! Let's dive into the exciting world of graphing rational functions. We'll be dissecting the function f(x) = (-3(x-4)(x+4)(2x-6)^2) / ((2x+6)(x-4)(2x-5)(x+4)) and uncovering all its hidden features. This is where things get interesting, so buckle up, guys!

Understanding the Function's Foundation

Alright, before we jump into the nitty-gritty, let's take a quick look at what we're dealing with. The given function, f(x) = (-3(x-4)(x+4)(2x-6)^2) / ((2x+6)(x-4)(2x-5)(x+4)), is a rational function. That means it's a fraction where both the numerator and denominator are polynomials. These types of functions can be a bit tricky, but they're also super interesting because their graphs often have unique characteristics like asymptotes and holes. Understanding these features is key to accurately sketching the graph. The first step we need to take is to simplify the equation as much as possible. This makes it easier to see the important features, like the zeros (where the function crosses the x-axis), the vertical asymptotes (where the function approaches infinity), and any holes in the graph. Let's get started!

To begin simplifying, we want to factor out the common terms in both the numerator and the denominator. Notice that we have (x-4) and (x+4) in both the numerator and the denominator. Also, we can factor a 2 from the (2x-6) term in the numerator, which turns into 2(x-3). Since this is squared, it becomes 4(x-3)^2. Let's rewrite the function incorporating these simplifications: f(x) = (-3(x-4)(x+4) * 4(x-3)^2) / (2(x+3)(x-4)(2x-5)(x+4)). Now, we can start canceling out the terms that appear in both the numerator and the denominator. We can cancel (x-4) and (x+4). We can also cancel a 2 from the 4 in the numerator and the 2 in the denominator. This gives us a simplified function that is much easier to work with: f(x) = (-6(x-3)^2) / ((x+3)(2x-5)). We've successfully simplified the function. From here, we can move to identify all the important features of the graph.

So, when we're dealing with rational functions, the main things we're looking for are the zeros of the function, any vertical asymptotes, any horizontal or oblique asymptotes, and any holes. Zeros are where the function crosses the x-axis. Vertical asymptotes are vertical lines that the graph approaches but never touches. Horizontal or oblique asymptotes describe the behavior of the function as x goes to positive or negative infinity. And holes are points where the function is undefined but the graph would otherwise be continuous. Let's get started!

Finding the Zeros

Alright, let's find the zeros of the function. Zeros are the x-values where the function equals zero – in other words, where the graph crosses the x-axis. To find these, we set the numerator of the simplified function equal to zero and solve for x. Remember, from our simplified function, f(x) = (-6(x-3)^2) / ((x+3)(2x-5)), the numerator is -6(x-3)^2. So, we solve: -6(x-3)^2 = 0. Dividing both sides by -6, we get (x-3)^2 = 0. Taking the square root of both sides, we find x - 3 = 0. Finally, solving for x, we get x = 3. This means there is a zero at x = 3. This tells us that the graph of our function will cross the x-axis at the point (3, 0). Great job, guys!

When we get to complex rational functions, the key is to always simplify first. Then you can see the zeros easily. Also, keep in mind that a zero exists when the numerator of the simplified function equals zero. It's a simple concept but critical for graphing. Now that we have the zero, let's move on to finding vertical asymptotes, which are another super important feature.

Identifying Vertical Asymptotes

Vertical asymptotes are the vertical lines that the graph of the function approaches but never touches. These occur where the denominator of the simplified function equals zero, but the numerator does not. Why? Because a division by zero is undefined, and in the context of a graph, that undefined behavior is represented by a vertical asymptote. Back to our simplified function f(x) = (-6(x-3)^2) / ((x+3)(2x-5)), the denominator is (x+3)(2x-5). To find the vertical asymptotes, we set the denominator equal to zero and solve for x. We do this by setting each factor in the denominator equal to zero and solving. So, first, x + 3 = 0, which gives us x = -3. Secondly, 2x - 5 = 0, which simplifies to 2x = 5, then x = 5/2 or x = 2.5. Thus, our vertical asymptotes are at x = -3 and x = 2.5. These are vertical lines where the graph of our function will get infinitely close but never cross. These vertical asymptotes are crucial for sketching the graph, as they define the function's behavior around these x-values. Neat, right?

One crucial thing to remember is that the vertical asymptotes come from the denominator of the simplified function. The zeros, on the other hand, come from the numerator. And always make sure you are using the simplified version of the function to avoid any confusion with holes in the graph. Now we have a good idea of how the graph will behave, including how it moves relative to the x-axis (zeros) and where it will shoot up or down (vertical asymptotes). Next up, let's tackle those horizontal asymptotes.

Determining Horizontal Asymptotes

Horizontal asymptotes describe the behavior of the function as x approaches positive or negative infinity. In other words, what value does the function get closer and closer to as we move far to the left or right on the x-axis? To find the horizontal asymptote, we compare the degrees of the numerator and denominator of our simplified function. Our simplified function is f(x) = (-6(x-3)^2) / ((x+3)(2x-5)). First, let's expand and simplify the numerator and denominator a little bit to make it easier to see the degrees. The numerator expands to -6(x^2 - 6x + 9), which simplifies to -6x^2 + 36x - 54. The denominator expands to 2x^2 + x - 15. Now, the simplified function is f(x) = (-6x^2 + 36x - 54) / (2x^2 + x - 15).

Now, let's compare degrees. The degree of the numerator is 2 (because of the x^2 term), and the degree of the denominator is also 2 (again, the x^2 term). When the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is -6, and the leading coefficient of the denominator is 2. So, the horizontal asymptote is y = -6/2, which simplifies to y = -3. That means that as x goes to positive or negative infinity, the graph of our function approaches the line y = -3. This helps us get a clear picture of the overall shape of the graph. Keep in mind, there is only one horizontal asymptote, so the graph won't approach different values on the left versus the right.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, there might be an oblique (or slant) asymptote, which you can find by dividing the numerator by the denominator using polynomial long division.

Checking for Holes

Holes in the graph are points where the function is undefined but would otherwise be continuous. These occur when a factor cancels out from both the numerator and the denominator before simplification. In the original, unsimplified function, f(x) = (-3(x-4)(x+4)(2x-6)^2) / ((2x+6)(x-4)(2x-5)(x+4)), we had (x-4) and (x+4) in both the numerator and the denominator. We canceled these out during simplification. However, these factors create holes in the original function.

To find the x-coordinate of the holes, we set the cancelled factors equal to zero and solve for x. For the (x-4) factor, we solve x - 4 = 0, which gives us x = 4. For the (x+4) factor, we solve x + 4 = 0, which gives us x = -4. This means that there are holes at x = 4 and x = -4. To find the y-coordinate of the holes, we plug these x-values into the simplified function. Remember, the simplified function is f(x) = (-6(x-3)^2) / ((x+3)(2x-5)). For x = 4: f(4) = (-6(4-3)^2) / ((4+3)(2(4)-5)) = -6 / (7 * 3) = -6 / 21 = -2/7. So, there's a hole at the point (4, -2/7). For x = -4: f(-4) = (-6(-4-3)^2) / ((-4+3)(2(-4)-5)) = (-6*49) / (-1 * -13) = -294 / 13. So, there's a hole at the point (-4, -294/13). Holes are essentially removable discontinuities, which means that the graph would be continuous at these points if the factors weren't canceled out.

In summary, we have found the following:

• Zero: x = 3
• Vertical Asymptotes: x = -3 and x = 2.5
• Horizontal Asymptote: y = -3
• Holes: (4, -2/7) and (-4, -294/13)

With all of this information, we can create a detailed sketch of the function's graph. You can plot the zeros, asymptotes, and holes, and then sketch the general shape of the curve, keeping in mind how the function behaves near the asymptotes and where the holes are located. Great job, guys! You've now successfully determined all the key features of the graph of the given rational function. Keep practicing, and you'll become a pro at graphing these kinds of functions!