Unveiling $P(x)$: A Comprehensive Guide

Hey guys! Let's dive into the fascinating world of rational functions. Today, we're going to break down the function $P(x) = \frac{3x^2 + 15x}{6x^2 + 6x - 36}$. Don't worry if it looks a bit intimidating at first; we'll go step by step, making sure everything is super clear and easy to understand. We'll explore how to simplify the function, find its domain, identify any holes, locate vertical and horizontal asymptotes, and even sketch a graph. This isn't just about crunching numbers; it's about understanding the behavior of functions and how they work. Are you ready to get started? Let's get this show on the road! This function is a classic example of a rational function, which is simply a fraction where both the numerator and the denominator are polynomials. Understanding these functions is crucial in algebra and calculus. Simplifying $P(x)$ involves factoring both the numerator and denominator and then canceling out any common factors. The domain of a function is the set of all possible input values (x-values) for which the function is defined. Holes occur when there are common factors in the numerator and denominator that cancel out. Vertical asymptotes are vertical lines that the graph of the function approaches but never touches. Horizontal asymptotes describe the behavior of the function as x approaches positive or negative infinity. Finally, sketching a graph helps visualize the function's behavior, including its asymptotes, holes, and intercepts. This comprehensive approach will give you a solid understanding of rational functions. So, let's start by simplifying the function. Ready? Let's do this!

Simplifying the Rational Function

Alright, let's simplify $P(x) = \frac{3x^2 + 15x}{6x^2 + 6x - 36}$. The first thing we want to do is factor both the numerator and the denominator. This will help us identify any common factors that we can cancel out. Factoring is like detective work, where you're trying to find the hidden factors within each polynomial. For the numerator, $3x^2 + 15x$, we can factor out a common factor of $3x$. This leaves us with $3x(x + 5)$. Easy peasy, right? Now, let's move on to the denominator, $6x^2 + 6x - 36$. Here, we can start by factoring out a common factor of 6. That gives us $6(x^2 + x - 6)$. Next, we need to factor the quadratic expression inside the parentheses, $x^2 + x - 6$. This factors into $(x + 3)(x - 2)$. So, the completely factored form of the denominator is $6(x + 3)(x - 2)$. Now, let's rewrite the entire function with the factored numerator and denominator: $P(x) = \frac{3x(x + 5)}{6(x + 3)(x - 2)}$.

We're not done yet, guys! We can simplify further by canceling out the common factor of 3 between the numerator and denominator. This leaves us with $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$. Great job, everyone! This is the simplified form of our function. Simplifying rational functions is important because it makes it easier to find the domain, identify holes, and sketch the graph. By factoring and canceling out common factors, we've reduced the complexity of the function, making it easier to analyze its behavior. Understanding the steps involved in simplification is a fundamental skill in algebra and calculus. Now that we've successfully simplified the function, we can move on to the next step: finding the domain. Remember, the domain is the set of all possible input values (x-values) for which the function is defined. Let's find the domain of our simplified function, $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$.

Determining the Domain of $P(x)$

Let's determine the domain of our simplified function, $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$. The domain of a rational function consists of all real numbers except for the values that make the denominator equal to zero. Why? Because division by zero is undefined! So, to find the domain, we need to identify the values of $x$ that make the denominator zero. In our simplified function, the denominator is $2(x + 3)(x - 2)$.

To find these values, we set the denominator equal to zero and solve for $x$: $2(x + 3)(x - 2) = 0$. First, we can divide both sides by 2, which doesn't change anything, because $0 / 2 = 0$. Now, we have $(x + 3)(x - 2) = 0$. This equation is satisfied when either $(x + 3) = 0$ or $(x - 2) = 0$. Solving for $x$ in each case, we get $x = -3$ or $x = 2$. These are the values that make the denominator zero. Therefore, the domain of $P(x)$ is all real numbers except $x = -3$ and $x = 2$. In interval notation, we can write the domain as $(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$. This means that the function is defined for all values of $x$ from negative infinity to -3, from -3 to 2, and from 2 to positive infinity. Understanding the domain is essential for sketching the graph and understanding the behavior of the function. It tells us where the function is defined and where it's not. Identifying the domain helps prevent errors and ensures the function is properly analyzed. Remember, the domain excludes values that cause the denominator to be zero. Now that we've determined the domain, let's investigate the possibility of holes in the graph of the function. Let's see if our function has any holes!

Identifying Holes in the Graph

Alright, let's find out if the graph of $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$ has any holes. Holes in the graph of a rational function occur when there are common factors in both the numerator and the denominator that cancel out. In our simplified function, the numerator is $x(x + 5)$ and the denominator is $2(x + 3)(x - 2)$. Notice anything that cancels out? Nope! There are no common factors in the numerator and denominator that cancel out. This means there are no holes in the graph of this function. If, in the initial factored form of $P(x) = \frac{3x(x + 5)}{6(x + 3)(x - 2)}$, a factor had canceled out, we'd have a hole at the x-value that made that factor equal to zero. For example, if we had a factor $(x-a)$ in both the numerator and denominator, there would be a hole at $x = a$. The y-coordinate of the hole can be found by plugging the x-value into the simplified form of the function. Since we didn't have any cancellations in our simplified form, we have no holes. This means that at every x-value, the function has a defined output, except at the values excluded from the domain. Identifying holes is an important part of graphing rational functions. Holes represent points where the function is undefined but would otherwise be continuous. Knowing where these holes are helps to give you an accurate picture of the function's behavior. We can move on now that we've determined there are no holes. Next up are the vertical asymptotes! Let's get to them!

Finding Vertical Asymptotes

Time to find the vertical asymptotes of our function, $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$. Vertical asymptotes are vertical lines that the graph of the function approaches but never actually touches. They occur at the x-values where the denominator of the simplified function is equal to zero, and the numerator is not zero at that x-value. In our simplified function, the denominator is $2(x + 3)(x - 2)$. We've already found that the denominator is zero when $x = -3$ and $x = 2$ (these are the values that we excluded from the domain). Now, we need to check if the numerator is also zero at these x-values.

The numerator is $x(x + 5)$. If we plug in $x = -3$, we get $-3(-3 + 5) = -3(2) = -6$, which is not zero. If we plug in $x = 2$, we get $2(2 + 5) = 2(7) = 14$, also not zero. Since the numerator is not zero at these x-values, we have vertical asymptotes at $x = -3$ and $x = 2$. So, the lines $x = -3$ and $x = 2$ are the vertical asymptotes of the graph of $P(x)$. This means that as $x$ approaches -3 or 2 from either side, the value of $P(x)$ will approach positive or negative infinity. This is a key feature when sketching the graph. Vertical asymptotes significantly impact the shape of the graph. They represent the boundaries where the function's value increases or decreases without bound. By correctly identifying these asymptotes, we gain valuable insight into the behavior of our rational function. Ready for more? Let's move on to the horizontal asymptotes!

Determining Horizontal Asymptotes

Okay, let's find the horizontal asymptotes of the function $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$. Horizontal asymptotes describe the behavior of the function as $x$ approaches positive or negative infinity. Finding these asymptotes helps us understand what the function does as the input values become very large or very small. To find the horizontal asymptotes, we need to compare the degrees of the numerator and the denominator of the simplified rational function. Remember, the degree of a polynomial is the highest power of the variable in the polynomial.

In our simplified function, $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$, the numerator is $x(x + 5)$, which expands to $x^2 + 5x$. The degree of the numerator is 2. The denominator is $2(x + 3)(x - 2)$, which expands to $2(x^2 + x - 6) = 2x^2 + 2x - 12$. The degree of the denominator is also 2. When the degrees of the numerator and denominator are the same, the horizontal asymptote is the line $y = \frac{a}{b}$, where $a$ is the leading coefficient of the numerator and $b$ is the leading coefficient of the denominator. In our case, the leading coefficient of the numerator is 1 (from $x^2$) and the leading coefficient of the denominator is 2 (from $2x^2$). Therefore, the horizontal asymptote is $y = \frac{1}{2}$. This means that as $x$ goes to positive or negative infinity, the graph of $P(x)$ approaches the horizontal line $y = \frac{1}{2}$. So, our horizontal asymptote is $y = \frac{1}{2}$. Knowing the horizontal asymptote is essential for understanding the end behavior of the function. It tells us the value that the function approaches as $x$ becomes extremely large or small. Let's move on to sketching the graph, and we'll bring everything together.

Sketching the Graph of $P(x)$

Let's sketch the graph of our function $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$. Now that we've done all the hard work, it's time to put it all together. To sketch the graph, we will use the information we have gathered so far. We will plot the asymptotes, the holes, and some key points like intercepts.

First, let's recap everything we've found:

• Simplified Function: $P(x) = \frac{x(x + 5)}{2(x + 3)(x - 2)}$
• Domain: $(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$
• Holes: None
• Vertical Asymptotes: $x = -3$ and $x = 2$
• Horizontal Asymptote: $y = \frac{1}{2}$

Now, let's find the x-intercepts. The x-intercepts are the points where the graph crosses the x-axis, which occurs when $P(x) = 0$. So we set the numerator equal to zero to find the x-intercepts: $x(x + 5) = 0$. This gives us $x = 0$ and $x = -5$. Thus, the x-intercepts are (0, 0) and (-5, 0). Next, let's find the y-intercept. The y-intercept is the point where the graph crosses the y-axis, which occurs when $x = 0$. Plugging in $x = 0$ into our simplified function, we get $P(0) = \frac{0(0 + 5)}{2(0 + 3)(0 - 2)} = 0$. So, the y-intercept is (0, 0). Now, we have enough information to start sketching. We draw the vertical asymptotes at $x = -3$ and $x = 2$ as dashed vertical lines. We draw the horizontal asymptote at $y = \frac{1}{2}$ as a dashed horizontal line. We plot the x-intercepts at (0, 0) and (-5, 0). Since there are no holes, we don't need to worry about any missing points. With the asymptotes and intercepts in place, we can sketch the graph. The graph will approach the asymptotes but never cross them. The graph is above the x-axis for $x < -5$, crosses the x-axis at -5, and goes below the x-axis for $-5 < x < -3$. The function will approach positive infinity as x approaches -3 from the left, cross the x-axis at 0 and goes below the x-axis for $0 < x < 2$ and approaches negative infinity as x approaches 2 from the left. After $x=2$ it crosses the x-axis at 0 and goes above the x-axis. Using this information, and potentially a few additional points, we can sketch the graph. This visual representation allows us to see the function's behavior clearly. The x-intercepts and y-intercepts help to define the shape and placement of the curve. Combining all the information obtained, we can get a thorough understanding of the function's characteristics. Voila, we have sketched the graph! It shows the function's behavior, its asymptotes, and intercepts. You've successfully analyzed and graphed $P(x)$!

Conclusion

Awesome work, everyone! Today, we successfully simplified the function $P(x) = \frac{3x^2 + 15x}{6x^2 + 6x - 36}$, determined its domain, confirmed it has no holes, found its vertical and horizontal asymptotes, and sketched its graph. We've explored all the essential features of this rational function. You should now have a strong grasp of how to analyze and understand rational functions in general. Remember, practice is key! The more you work with rational functions, the more comfortable and confident you'll become. Keep up the amazing work, and don't hesitate to revisit these steps anytime. Until next time, keep exploring the awesome world of math! Keep practicing, and you'll become a pro in no time! Remember to always check your answers and ensure that they align with the function's properties. Understanding rational functions provides a solid foundation for more complex mathematical concepts. Great job, and see you next time!