Problem 18 - Precalculus : Building Concepts and Connections

Question

Find the domain and the vertical and horizontal asymptotes (if any). $$f(x)=\frac{3 x+5}{x^{2}-x-2}$$

Step-by-Step Solution

Answer

The domain of the function is \( x \in \mathbb{R} - \{-1,2\} \), the vertical asymptotes are at \( x=-1 \) and \( x=2 \), and the horizontal asymptote is at \( y=0 \).

Step 1: Find the Domain

Set the denominator of the function equal to zero and solve for \( x \). \[ x^{2}-x-2=0 \] The solutions to this equation will provide the values not included in the domain. The quadratic equation can be factored: \( x^{2}-x-2=(x-2)(x+1)=0 \), so \( x=2, -1 \). Therefore, the domain of the function is \( x \in \mathbb{R} - \{-1,2\} \).

Step 2: Find the Vertical Asymptotes

The solutions for x from the previous step will now represent the vertical asymptotes. Therefore, the vertical asymptotes are at \( x=-1 \) and \( x=2 \).

Step 3: Find the Horizontal Asymptotes

In order to find the horizontal asymptote of a rational function, we must look at the degree of the polynomials in the numerator and the denominator. If the degree in the denominator is higher (which is the case here), then the orizontal asymptote is at \( y=0 \). Also we can analyze the behavior of function as \( x \) approaches to positive and negative infinity.

Key Concepts

Domain of a Function

Understanding the domain of a rational function is crucial because it defines where the function is actually defined and free of any undefined expressions. A rational function, like \( f(x) = \frac{3x+5}{x^2-x-2} \), can have values in its denominator that make the function undefined because division by zero isn't allowed.
To find these problematic values, we set the denominator equal to zero and solve for \( x \):

• \( x^2-x-2 = 0 \) turns into \( (x-2)(x+1) = 0 \).
• This results in \( x = 2 \) and \( x = -1 \).

Therefore, the domain of this function is all real numbers except \( x = 2 \) and \( x = -1 \). We express it as \( x \in \mathbb{R} - \{-1,2\} \), which means all real numbers except these points.

Vertical Asymptotes

Vertical asymptotes occur in rational functions where the denominator vanishes while the numerator is non-zero, causing the function to blow up to infinity. In simpler terms, vertical asymptotes represent the positions where the function becomes undefined and tends to tire towards infinity.
When you find the values that make the denominator zero, these same values often represent the vertical asymptotes. This is the case for our function \( f(x) = \frac{3x+5}{x^2-x-2} \), which had problems at \( x = 2 \) and \( x = -1 \).
Thus, for this function, these x-values can be thought of as walls or barriers that the function will never quite reach and they are precisely the vertical asymptotes:

• The vertical asymptotes are at \( x = -1 \) and \( x = 2 \).

Testing these further: as \( x \) approaches these values from either left or right, \( f(x) \) will either increase or decrease without bound.

Horizontal Asymptotes

Horizontal asymptotes in rational functions tell us about the behavior of the function as \( x \) heads towards positive or negative infinity. They provide insights into the long-term behavior of the graph of the function.
To find horizontal asymptotes, compare the degrees of the polynomial in the numerator and the denominator. Here, for the function \( f(x) = \frac{3x+5}{x^2-x-2} \):

• The numerator has a degree of 1 (since it's \( 3x+5 \)) and the denominator has a degree of 2 (since it's \( x^2-x-2 \)).
• Since the degree of the denominator is greater, the horizontal asymptote is at \( y = 0 \).

This means as \( x \) approaches infinity or negative infinity, \( f(x) \) is getting closer and closer to the line \( y = 0 \), though this horizontal line is never actually reached.