Analyzing The Rational Function: X² + 1 Over (2x - 1)(x + 2)²

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Hey guys! Let's dive deep into a fascinating math problem: analyzing the rational function (X² + 1) / ((2X - 1)(X + 2)²). This kind of function pops up all over the place in calculus, pre-calculus, and even in real-world applications like physics and engineering. Understanding it will give you some serious math cred. So, what makes this function tick? We'll explore its key characteristics, like its domain, any potential holes in the graph, vertical and horizontal asymptotes, and how it behaves as X approaches certain values. We'll even touch on how you might go about graphing it. Ready to break it down? Let's get started, shall we?

This particular function is a ratio of two polynomials. The numerator, X² + 1, is a quadratic (degree 2), and the denominator, (2X - 1)(X + 2)², is also a polynomial, although it looks a bit more complex. Recognizing this is the first step. The denominator, when fully expanded, would be a cubic polynomial (degree 3). The shape of the graph and where it's undefined are highly influenced by what's in the numerator and denominator. We'll examine what happens when the denominator becomes zero, that is where vertical asymptotes or holes might appear. Additionally, the behavior of the function as X goes to positive or negative infinity provides valuable insights into its long-term behavior, which means the horizontal asymptotes tell a lot about the limit of the function when the X values are extremely large or small.

This function's analysis involves several key steps. First, determining the domain is essential. The domain is the set of all possible input values (X-values) for which the function is defined. Since we can't divide by zero, we need to figure out what X-values make the denominator equal to zero. Next, we investigate any potential discontinuities. This could include points where the graph has a hole (removable discontinuity) or jumps to infinity (non-removable discontinuity). We'll also find vertical asymptotes. Finally, we'll want to determine horizontal asymptotes, which describes what happens to the function as X gets very large, and look for x-intercepts (where the function crosses the x-axis). Understanding these features allows us to sketch a reasonably accurate graph without necessarily plotting hundreds of points. This approach is a fundamental part of function analysis in pre-calculus and calculus, helping us to understand the behavior of functions.

Determining the Domain

Alright, let's start by determining the domain of our rational function. The domain is all the X-values that we're allowed to plug into the function without causing any mathematical trouble. In the world of rational functions, the biggest trouble-maker is division by zero. So, our job is to figure out which X-values make the denominator equal to zero. This would mean the function is undefined at those points.

Our denominator is (2X - 1)(X + 2)². To find the values that make this zero, we set each factor equal to zero and solve for X. Firstly, we'll analyze (2X - 1) and set it equal to zero:

2X - 1 = 0

Adding 1 to both sides, we get:

2X = 1

Dividing both sides by 2, we get:

X = 1/2

So, X = 1/2 makes the first factor zero. Next, we analyze (X + 2)² and set it to zero.

(X + 2)² = 0

Taking the square root of both sides, we have:

X + 2 = 0

Subtracting 2 from both sides:

X = -2

So, X = -2 makes the second factor (and the entire denominator) zero. Therefore, the function is undefined at X = 1/2 and X = -2. That means the domain of the function is all real numbers except for 1/2 and -2. You can write this in interval notation as: (-∞, -2) ∪ (-2, 1/2) ∪ (1/2, ∞).

We can conclude that the domain of the function is all real numbers except X = 1/2 and X = -2. These are the points where the function is undefined. It's like these X-values have been blacklisted; the function just can't handle them. This is a super important step, as it shapes the entire graph. Vertical asymptotes will occur at those values, or, potentially, removable discontinuities (holes) if there is a factor that cancels out. This understanding of the domain is the bedrock of understanding the overall behavior of this function. Remember, the domain tells you the allowed values for X and gives us the initial clues about the shape of the graph.

Finding Vertical Asymptotes and Holes

Now, we're moving on to identifying the vertical asymptotes and any potential holes in the graph. Vertical asymptotes are the vertical lines that the graph approaches but never quite touches. They occur where the denominator of a rational function equals zero, and the numerator does not. Holes (removable discontinuities) occur when a factor in the denominator cancels out with a factor in the numerator.

From the previous section, we know that the denominator, (2X - 1)(X + 2)², becomes zero at X = 1/2 and X = -2. Let's first check X = 1/2. At X = 1/2, the numerator is (1/2)² + 1 = 1/4 + 1 = 5/4, which is not zero. That means we have a vertical asymptote at X = 1/2.

Now, let's check X = -2. At X = -2, the numerator is (-2)² + 1 = 4 + 1 = 5, which is also not zero. Hence, we'll have a vertical asymptote at X = -2. So, in our function, because the numerator does not become zero at either X = 1/2 or X = -2, we'll get two vertical asymptotes. There are no holes (removable discontinuities) because there are no common factors between the numerator and denominator that can be cancelled.

In summary: We have vertical asymptotes at X = 1/2 and X = -2, and no holes in the graph. This information will dramatically influence how we sketch the graph. As X approaches 1/2 or -2, the function's value will shoot off to positive or negative infinity (or possibly both, depending on which side you're approaching from). This is where the graph really starts to show its personality, and understanding the asymptotes is critical for understanding the function's long-term trends and avoiding a major headache when sketching the graph.

Determining Horizontal Asymptotes

Let's move on to figure out the horizontal asymptotes. These are the horizontal lines that the graph approaches as X goes to positive or negative infinity. They tell us about the function's end behavior – what happens to the function's output (Y-value) as the input (X-value) gets extremely large or extremely small. Calculating the horizontal asymptote will provide the final touch to the graph characteristics.

To find the horizontal asymptotes, we compare the degrees of the numerator and denominator. Our numerator is X² + 1, which is a degree 2 polynomial. The denominator, (2X - 1)(X + 2)², is a degree 3 polynomial (when fully expanded, the highest power of X is 3). When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is always Y = 0. In other words, as X approaches infinity (positive or negative), the function approaches zero.

So, for our function, the horizontal asymptote is Y = 0 (the x-axis). This means the graph will eventually get closer and closer to the x-axis as X gets very large in either direction. This does not mean that the function never crosses the x-axis (that’s for the x-intercepts that we'll look at next); it only means that eventually, it will approach the axis but never touch it as X approaches infinity.

Knowing the horizontal asymptote is extremely helpful when sketching the graph. It provides essential information about the overall behavior of the function. This knowledge helps to visualize the function's limits at infinity, essentially giving us the finishing touch when putting the graph into the final form.

Finding X-Intercepts

Next up, let's figure out the x-intercepts. These are the points where the graph crosses the x-axis, meaning the Y-value is zero. To find the x-intercepts, we set the function equal to zero and solve for X. That means, we need to set the numerator equal to zero, as a fraction can be zero only if its numerator is zero.

So, we have: (X² + 1) / ((2X - 1)(X + 2)²) = 0

For the fraction to be zero, the numerator must be zero:

X² + 1 = 0

Subtracting 1 from both sides:

X² = -1

Since there is no real number whose square is negative, there are no real solutions to this equation. That means that the numerator, X² + 1, can never equal zero for any real value of X.

Therefore, our function has no x-intercepts. This means the graph does not cross the x-axis. This is a key characteristic to remember when you are sketching the graph. It will help you to visualize how the graph stays above or below the x-axis depending on the asymptote position.

Graphing the Function

Now that we have all the key features, we can sketch a graph of our function. This function has the following key characteristics:

  • Vertical asymptotes at X = 1/2 and X = -2.
  • No holes.
  • Horizontal asymptote at Y = 0 (the x-axis).
  • No x-intercepts.

Here's how we can imagine the graph:

  1. Draw the Asymptotes: Start by drawing the vertical lines X = 1/2 and X = -2 and the horizontal line Y = 0 (the x-axis) on a coordinate plane. These lines will guide the shape of the curve.
  2. Analyze Behavior: To understand the function's behavior, analyze how the graph approaches the asymptotes. The graph will approach the vertical asymptotes from either positive or negative infinity. We'll need to determine the function's sign (positive or negative) in the intervals created by the asymptotes. We'll do this by picking test values for each interval (-∞, -2), (-2, 1/2), and (1/2, ∞) and seeing if the function is positive or negative there.
  3. Test Values: Pick test values in each interval to understand how the graph is positioned relative to the x-axis. For example:
    • For (-∞, -2): Let's use X = -3. Then f(-3) = ( (-3)² + 1 ) / ( (2(-3) - 1) ((-3) + 2)² ) = 10 / (-7) = -10/7. Therefore, the graph is negative.
    • For (-2, 1/2): Let's use X = 0. Then f(0) = (0² + 1) / ((2(0) - 1) ((0) + 2)²) = 1 / -4. Therefore, the graph is negative.
    • For (1/2, ∞): Let's use X = 1. Then f(1) = (1² + 1) / ((2(1) - 1) ((1) + 2)²) = 2/9. Therefore, the graph is positive.
  4. Sketch the Curve: Using the information about the asymptotes and the sign of the function in each interval, we can now sketch the curve. The graph will approach the vertical asymptotes, following the sign we determined in the test values. It will approach the horizontal asymptote (the x-axis) as X approaches infinity. Since there are no x-intercepts, the graph does not cross the x-axis. The graph starts below the x-axis, crosses to the other side of the asymptote, and approaches the x-axis from the top.

In general: The graph will have a general shape that reflects all the features we have analyzed. By combining the asymptotes, the intercepts, and the sign analysis, you can sketch the graph with accuracy and understanding. This process really brings the function to life, and helps you visualize its properties and behavior. Remember, the more you practice graphing functions, the easier it will get, and the more familiar you'll become with how they work.

This is an example of the process of analyzing and graphing a rational function. Practice with other functions to become a master. The math world is full of such amazing concepts, and it's all about practice and learning! Keep the questions coming, and happy calculating, math fans!