Finding Vertical Asymptotes For F(x)=(x-4)(x+9)/(x^2-1)

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#mainkeyword Understanding Vertical Asymptotes

In the realm of mathematical analysis, vertical asymptotes play a crucial role in defining the behavior of functions, especially rational functions. A vertical asymptote is a vertical line that a function approaches but never quite touches. These asymptotes occur at x-values where the function becomes undefined, typically due to division by zero. Identifying vertical asymptotes is essential for sketching the graph of a function and understanding its limits. In this comprehensive guide, we will explore the process of finding vertical asymptotes for a given rational function, specifically focusing on the function f(x)=(x-4)(x+9)/(x^2-1). We will dissect the function, identify potential asymptotes, and confirm their existence through rigorous analysis. By the end of this discussion, you will have a firm grasp on how to locate and interpret vertical asymptotes, a skill vital for advanced mathematical studies and applications. This exploration will not only enhance your understanding of rational functions but also provide a practical approach to problem-solving in calculus and beyond.

Deconstructing the Function f(x)=(x-4)(x+9)/(x^2-1)

Before we delve into the specifics of finding vertical asymptotes, let's first deconstruct the given function: f(x) = (x-4)(x+9) / (x^2-1). This function is a rational function, which is essentially a ratio of two polynomials. The numerator is a quadratic expression obtained by multiplying (x-4) and (x+9), while the denominator is another quadratic expression, x^2-1. The key to finding vertical asymptotes lies in understanding the behavior of the function where the denominator approaches zero.

To proceed, it is beneficial to factor both the numerator and the denominator. The numerator, (x-4)(x+9), is already in factored form, making it easy to identify the roots, which are x=4 and x=-9. These roots are crucial for determining the x-intercepts of the function, but they do not directly influence the vertical asymptotes. The denominator, x^2-1, can be factored using the difference of squares formula, which states that a2-b2=(a-b)(a+b). Applying this formula, we get x^2-1=(x-1)(x+1). The factored form of the denominator is paramount for finding the vertical asymptotes, as it reveals the values of x that make the denominator zero. These values are the potential locations of the vertical asymptotes. This initial deconstruction of the function sets the stage for a more in-depth analysis of its behavior and the identification of its vertical asymptotes.

Identifying Potential Vertical Asymptotes

To pinpoint the potential vertical asymptotes of the function f(x) = (x-4)(x+9) / (x^2-1), we must focus on the denominator. As previously determined, the denominator can be factored into (x-1)(x+1). Vertical asymptotes occur where the denominator equals zero, as this leads to the function being undefined. Thus, we need to solve the equation (x-1)(x+1) = 0. This equation is satisfied when either (x-1) = 0 or (x+1) = 0. Solving these two simple equations gives us x = 1 and x = -1. These values are the potential locations of the vertical asymptotes.

However, it's crucial to note that not every value that makes the denominator zero necessarily results in a vertical asymptote. If a factor in the denominator also appears in the numerator, there might be a hole (a removable discontinuity) instead of a vertical asymptote. In our case, the factors in the numerator are (x-4) and (x+9), neither of which are the same as the factors (x-1) and (x+1) in the denominator. This means that there are no common factors between the numerator and the denominator that can be canceled out. Therefore, both x = 1 and x = -1 are strong candidates for vertical asymptotes. To confirm this, we need to examine the behavior of the function as x approaches these values from both the left and the right. This detailed analysis will solidify our understanding of the function's asymptotic behavior and provide a definitive answer regarding the vertical asymptotes.

Confirming Vertical Asymptotes Through Limit Analysis

After identifying potential vertical asymptotes at x = 1 and x = -1, it is imperative to confirm their existence through limit analysis. A vertical asymptote exists at x = a if the limit of the function as x approaches a from either the left or the right (or both) is infinite (either positive or negative). This means we need to evaluate the following limits:

  1. lim x→1- f(x)
  2. lim x→1+ f(x)
  3. lim x→-1- f(x)
  4. lim x→-1+ f(x)

Let's start with x = 1. As x approaches 1 from the left (x→1-), x is slightly less than 1. Thus, (x-1) is a small negative number, and (x+1) is close to 2 (positive). The numerator, (x-4)(x+9), approaches (1-4)(1+9) = (-3)(10) = -30, which is negative. Therefore, the function f(x) approaches a negative number divided by a small negative number, resulting in positive infinity. This can be written as:

lim x→1- f(x) = +∞

Now, let's consider x approaching 1 from the right (x→1+). In this case, x is slightly greater than 1, so (x-1) is a small positive number, and (x+1) is still close to 2 (positive). The numerator remains negative (-30). Thus, the function f(x) approaches a negative number divided by a small positive number, resulting in negative infinity:

lim x→1+ f(x) = -∞

Since the limits as x approaches 1 from the left and right are infinite, we confirm that there is a vertical asymptote at x = 1. We can perform a similar analysis for x = -1. As x approaches -1 from the left (x→-1-), x is slightly less than -1, making (x+1) a small negative number, and (x-1) is close to -2 (negative). The numerator, (x-4)(x+9), approaches (-1-4)(-1+9) = (-5)(8) = -40, which is negative. Thus, the function f(x) approaches a negative number divided by a small positive number, resulting in negative infinity:

lim x→-1- f(x) = -∞

Finally, as x approaches -1 from the right (x→-1+), x is slightly greater than -1, so (x+1) is a small positive number, and (x-1) remains close to -2 (negative). The numerator is still negative (-40). Therefore, the function f(x) approaches a negative number divided by a small negative number, resulting in positive infinity:

lim x→-1+ f(x) = +∞

Again, the infinite limits as x approaches -1 from both sides confirm the existence of a vertical asymptote at x = -1. Through this detailed limit analysis, we have definitively shown that the function f(x) = (x-4)(x+9) / (x^2-1) has vertical asymptotes at x = 1 and x = -1. This rigorous approach ensures that our conclusions are mathematically sound and provides a comprehensive understanding of the function's behavior near these asymptotes.

Conclusion and Final Answer

In summary, our comprehensive analysis of the function f(x) = (x-4)(x+9) / (x^2-1) has led us to a clear understanding of its vertical asymptotes. By factoring the denominator, identifying potential asymptotes, and rigorously confirming their existence through limit analysis, we have demonstrated a robust method for handling such problems. We first factored the denominator as (x-1)(x+1), revealing potential vertical asymptotes at x = 1 and x = -1. We then performed limit analysis, examining the behavior of the function as x approached these values from both the left and the right. The infinite limits obtained in each case conclusively confirmed the presence of vertical asymptotes at x = 1 and x = -1. Therefore, the final answer is that the function f(x) has vertical asymptotes at x = 1 and x = -1. This process not only provides the correct answer but also reinforces the underlying principles of calculus and rational functions. The ability to identify and analyze vertical asymptotes is a crucial skill in mathematics, with applications in graphing functions, understanding their behavior, and solving more complex problems in calculus and beyond. Through this guide, we hope to have illuminated the path to mastering this skill.

Final Answer: The final answer is \boxed{x=1, x=-1}