Question

Identify the horizontal and vertical asymptotes, if any, of the given function. $$ f(x)=\frac{x^{2}-9}{9 x-9} $$

Short answer

Vertical asymptote at \( x = 1 \); no horizontal asymptote.

Step-by-step solution

Simplify the Function

The given function is \( f(x) = \frac{x^2 - 9}{9x - 9} \). Start by simplifying the expression. We can factor both the numerator and the denominator. The numerator \( x^2 - 9 \) can be factored as \((x + 3)(x - 3)\) because it is a difference of squares. The denominator \( 9x - 9 \) can be factored as \( 9(x - 1) \). So, the function becomes \( f(x) = \frac{(x + 3)(x - 3)}{9(x - 1)} \).

Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is non-zero. Set the denominator \( 9(x-1) = 0 \) to find the value of \( x \). Solving \( 9(x-1) = 0 \) gives \( x = 1 \). Since the numerator does not become zero at \( x = 1 \), there is a vertical asymptote at \( x = 1 \).

Identify Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the numerator and the denominator. The degree of the numerator \((x^2 - 9)\) is 2 and the degree of the denominator \((9x - 9)\) is 1. Since the degree of the numerator (2) is greater than the degree of the denominator (1), the function does not have a horizontal asymptote.

Key concepts

Horizontal Asymptotes

Horizontal asymptotes tell us how the function behaves as the variable tends towards infinity or negative infinity. For rational functions, we often examine the degrees of the polynomial in both the numerator and the denominator. Here's how it works:

• If the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is at the line \( y = 0 \).
• If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients of the numerator and the denominator.
• If the degree of the numerator is greater than that of the denominator, no horizontal asymptote exists.

In the original problem, the numerator's degree \((x^2 - 9)\) is 2 and the denominator's degree \((9x - 9)\) is 1. Since the numerator's degree is greater, no horizontal asymptote exists for the function \( f(x)=\frac{x^2 - 9}{9x - 9} \). This is key whenever solving similar problems.

Vertical Asymptotes

Vertical asymptotes occur where a function shoots off to infinity. This typically happens where the denominator equals zero, causing the function to be undefined, but only when the numerator doesn't also become zero at that point. To find vertical asymptotes:

• Set the denominator of the function equal to zero.
• Solve the equation for \( x \).
• Check that these \( x \) values don't make the numerator zero as well.

For example, let's revisit the exercise's function \( f(x)=\frac{(x+3)(x-3)}{9(x-1)} \):
1. Set \( 9(x-1) = 0 \). Solving, we get \( x = 1 \).
2. Check \( (x+3)(x-3) eq 0 \) at \( x = 1 \).
The value \( x = 1 \) doesn't zero the numerator; thus, there is a vertical asymptote at \( x = 1 \). This method is applicable for almost any rational function!

Rational Functions

Rational functions are quotients of polynomial functions. They appear as the division of one polynomial by another, and they can have interesting features like asymptotes, which graphically show where the function might go towards infinity or a specific line. Understanding rational functions:

• They can be expressed in the form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.
• The behavior of the function, especially around asymptotes, depends on these polynomials' degrees and their factorization.
• Rational functions can model real-life situations, such as rates and ratios.

In our exercise, \( f(x)=\frac{(x+3)(x-3)}{9(x-1)} \) represents a rational function. By factoring, simplifying, and understanding the degrees of \( P(x) \) and \( Q(x) \), we can determine both vertical and horizontal asymptotes. Grasping these features will significantly aid in graphing functions and analyzing their long-term behaviors.