Analyzing G(x) = (x^3 - 2x^2 - 15x) / (x^2 + X - 6)

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Analyzing the Rational Function g(x) = (x³ - 2x² - 15x) / (x² + x - 6)

Hey guys! Let's dive into a fun math problem today. We're going to thoroughly analyze the rational function g(x) = (x³ - 2x² - 15x) / (x² + x - 6). Rational functions can seem intimidating, but by breaking them down step by step, we can understand their behavior and characteristics. We will cover everything from finding the domain and intercepts to identifying asymptotes and sketching the graph. So, grab your calculators and let's get started!

1. Domain of g(x)

First things first, let's talk about the domain of g(x). The domain is basically all the possible 'x' values that we can plug into our function without causing it to explode (mathematically speaking, of course!). For rational functions like this one, the main thing we need to watch out for is division by zero. A zero in the denominator will make the function undefined, so we need to find those sneaky 'x' values that make the denominator zero and exclude them from our domain. So, we need to find the values of x for which the denominator, x² + x - 6, equals zero. Factoring the denominator gives us (x + 3)(x - 2) = 0. This tells us that the denominator is zero when x = -3 or x = 2. Therefore, these values must be excluded from the domain. In other words, we can plug in any real number for x except for -3 and 2. We can express the domain in several ways. We can use set notation: {x | x ∈ ℝ, x ≠ -3, x ≠ 2}. This reads as “the set of all x such that x is a real number and x is not equal to -3 or 2”. Another way to write the domain is using interval notation: (-∞, -3) ∪ (-3, 2) ∪ (2, ∞). This notation expresses the domain as the union of three intervals, excluding the troublesome values -3 and 2. Understanding the domain is crucial because it sets the stage for the rest of our analysis. We know where the function is defined and where it’s not, which helps us interpret its behavior and sketch its graph more accurately.

2. Intercepts of g(x)

Next up, let's find the intercepts of g(x). Intercepts are the points where the graph of the function crosses the x-axis (x-intercepts) and the y-axis (y-intercept). They give us key points to plot when we're sketching the graph, and they also provide valuable information about the function's behavior. To find the x-intercepts, we need to determine where the function equals zero. A rational function is zero when its numerator is zero (and the denominator is not zero at the same time, of course!). So, we set the numerator x³ - 2x² - 15x equal to zero and solve for x. Factoring out an x from the numerator, we get x(x² - 2x - 15) = 0. Now we need to factor the quadratic x² - 2x - 15. This factors into (x - 5)(x + 3) = 0. So, the numerator becomes x(x - 5)(x + 3) = 0. This equation has three solutions: x = 0, x = 5, and x = -3. However, remember our domain! x = -3 is not in the domain of the function because it makes the denominator zero. Therefore, we discard x = -3 as an x-intercept. So, our x-intercepts are x = 0 and x = 5. These correspond to the points (0, 0) and (5, 0) on the graph. To find the y-intercept, we set x = 0 in the function and solve for g(0). Plugging in x = 0 into the original function, we get g(0) = (0³ - 2(0)² - 15(0)) / (0² + 0 - 6) = 0 / -6 = 0. So, the y-intercept is (0, 0). Notice that the x-intercept and y-intercept coincide at the origin in this case. Finding the intercepts is a crucial step in analyzing any function, as they provide us with specific points where the function crosses the axes, aiding in the visualization and sketching of the graph.

3. Asymptotes of g(x)

Now let's delve into the fascinating world of asymptotes! Asymptotes are like invisible guide rails that the graph of our function approaches but never quite touches (or, in some cases, crosses). They give us valuable information about the function's behavior as x gets very large (positive or negative) or as x approaches certain values. There are three main types of asymptotes we need to consider: vertical, horizontal, and slant (or oblique) asymptotes. Vertical asymptotes occur where the denominator of the rational function equals zero, but the numerator does not. We already know from our domain analysis that the denominator, x² + x - 6, is zero when x = -3 and x = 2. Let's check if the numerator is also zero at these points. We know the factored form of the numerator is x(x - 5)(x + 3). When x = -3, the numerator is indeed zero. This indicates a hole (or removable discontinuity) at x = -3, not a vertical asymptote. When x = 2, the numerator is 2(2 - 5)(2 + 3) = 2(-3)(5) = -30, which is not zero. Therefore, we have a vertical asymptote at x = 2. Vertical asymptotes are crucial as they indicate where the function's value approaches infinity or negative infinity. To find horizontal asymptotes, we examine the behavior of the function as x approaches positive and negative infinity. We compare the degrees of the numerator and denominator. The degree of the numerator, x³ - 2x² - 15x, is 3, and the degree of the denominator, x² + x - 6, is 2. Since the degree of the numerator is exactly one greater than the degree of the denominator, we know there will be a slant asymptote, and no horizontal asymptote. To find the slant asymptote, we perform polynomial long division of the numerator by the denominator. Dividing x³ - 2x² - 15x by x² + x - 6 gives us x - 3 with a remainder of -6x - 18. So, we can write the function as: g(x) = x - 3 + (-6x - 18) / (x² + x - 6). As x approaches infinity, the term (-6x - 18) / (x² + x - 6) approaches zero. Therefore, the slant asymptote is the line y = x - 3. Understanding asymptotes is essential for sketching the graph of the function accurately. They tell us how the function behaves at the extreme values of x and near the points where the function is undefined.

4. Holes (Removable Discontinuities)

As we briefly mentioned earlier, our function has a hole, also known as a removable discontinuity. This happens when a factor cancels out from both the numerator and the denominator. We saw that both the numerator and the denominator have a factor of (x + 3). Let's factor the function completely to see this more clearly: g(x) = x(x - 5)(x + 3) / ((x + 3)(x - 2)). The (x + 3) factor cancels out, but only for x ≠ -3. This means that the function behaves like the simplified expression g(x) = x(x - 5) / (x - 2) everywhere except at x = -3. At x = -3, there's a hole in the graph. To find the coordinates of the hole, we plug x = -3 into the simplified expression: g(-3) = -3(-3 - 5) / (-3 - 2) = -3(-8) / (-5) = -24 / 5. So, there is a hole at the point (-3, -24/5) or (-3, -4.8). When sketching the graph, we represent a hole with an open circle at that point. Recognizing and accounting for holes is crucial for accurately representing the function's behavior.

5. Simplified Form of g(x)

Now, let's take a look at the simplified form of g(x). This form helps us analyze the function's behavior more easily, especially after we've identified and dealt with any holes. After canceling the common factor (x + 3), our function simplifies to: g(x) = x(x - 5) / (x - 2), for x ≠ -3. This simplified form is equivalent to the original function everywhere except at x = -3, where the original function is undefined, and the simplified form would give us the y-coordinate of the hole. Analyzing this simplified form, we can confirm our previous findings. The x-intercepts are still at x = 0 and x = 5 (from the numerator). The vertical asymptote is still at x = 2 (from the denominator). And the slant asymptote remains y = x - 3. The simplified form allows us to work with a more manageable expression, making it easier to analyze the function's behavior between intercepts and asymptotes. For example, we can determine the sign of g(x) in different intervals by looking at the signs of the factors in the simplified form.

6. Sketching the Graph of g(x)

Alright guys, the moment we've been building up to – let's sketch the graph of g(x)! We've gathered all the essential information: the domain, intercepts, asymptotes, and holes. Now we'll put it all together to create a visual representation of the function. Here’s a step-by-step approach to sketching the graph:

  1. Plot the Intercepts: We found x-intercepts at (0, 0) and (5, 0) and a y-intercept at (0, 0). Mark these points on the coordinate plane.
  2. Draw the Asymptotes: Draw a dashed vertical line at the vertical asymptote x = 2. Also, draw a dashed line for the slant asymptote y = x - 3. These lines will guide the behavior of the graph.
  3. Mark the Hole: Indicate the hole at (-3, -4.8) with an open circle.
  4. Analyze Intervals: Divide the x-axis into intervals based on the x-intercepts and vertical asymptotes. Our key points are -3 (hole), 0, 2 (vertical asymptote), and 5. The intervals are: (-∞, -3), (-3, 0), (0, 2), (2, 5), and (5, ∞). Choose a test point in each interval and plug it into the simplified form of g(x) to determine the sign of the function in that interval. This will tell you whether the graph is above or below the x-axis in each interval.
  5. Sketch the Graph: In each interval, sketch the curve, making sure it approaches the asymptotes and passes through the intercepts correctly. Remember to approach the hole but not actually include the point. The sign analysis will guide you in drawing the curve in the correct direction. As x approaches the vertical asymptote x = 2, the function will approach either positive or negative infinity. The slant asymptote will guide the graph's behavior as x approaches positive or negative infinity.

By following these steps, we can create a fairly accurate sketch of the graph of g(x). If you want an even more precise graph, you could calculate a few more points in each interval, but the key features will be captured by this method.

Conclusion

So there you have it! We've thoroughly analyzed the rational function g(x) = (x³ - 2x² - 15x) / (x² + x - 6). We found the domain, intercepts, asymptotes, and holes, and used all that information to sketch the graph. Analyzing rational functions can seem tricky at first, but by breaking it down into smaller steps, we can understand their behavior and create accurate representations. Keep practicing, guys, and you'll become masters of rational functions in no time! Remember, math is an adventure, so keep exploring and have fun with it!