Problem 6 - College Algebra and Calculus: An Applied Approach

Question

Find the domain of the function and identify any horizontal and vertical asymptotes. $$f(x)=\frac{3 x^{2}+x-5}{x^{2}+1}$$

Step-by-Step Solution

Answer

The domain of the function \(f(x)=\frac{3 x^{2}+x-5}{x^{2}+1}\) is all real numbers. The function has no vertical asymptotes and has a horizontal asymptote at \(y = 3\).

Step 1: Find the Domain

We cannot have the denominator of a fraction equal to zero because divisibility by zero is undefined in mathematics. So, the denominator of the function \(x^{2}+1 = 0\) should be solved for \(x\) to define the domain. However, \(x^{2} cannot be -1, so there are no values of \(x\) that make the denominator zero. Hence, the domain of the function is all real numbers.

Step 2: Identify the Vertical Asymptote(s)

Vertical asymptotes are the values of \(x\) for which the function is undefined. This is where the denominator of the function equals zero. As we established in the previous step, there are no such \(x\)-values. Therefore, the function does not have any vertical asymptotes.

Step 3: Identify the Horizontal Asymptote(s)

The degree of the polynomial in the numerator is 2 and in the denominator it's also 2. Therefore, as \(x \rightarrow \pm\infty\), the function behaves like \(\frac{3x^{2}}{x^{2}}\), which simplifies to \(3\). Hence, the function has a horizontal asymptote at \(y = 3\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible values of the independent variable, usually represented by \(x\), for which the function is defined. In simpler terms, it tells us what \(x\) values can be plugged into a function without causing any mathematical mishaps, like dividing by zero.

For the function \(f(x) = \frac{3x^2 + x - 5}{x^2 + 1}\), determining the domain involves checking the denominator, \(x^2 + 1\). A crucial detail in rational functions is that the denominator must not be zero because division by zero is undefined. Here, when solving \(x^2 + 1 = 0\), we realize \(x^2 = -1\) has no real solution, as squares of real numbers are never negative.

This means \(x\) can indeed be any real number because there are no restrictions from the denominator. Therefore, the domain of this function is all real numbers, denoted by \( (-\infty, \infty) \).

• Ensure the denominator never equals zero.
• The domain is the set of all permissible \(x\) values.

Vertical Asymptotes

Vertical asymptotes occur in rational functions at \(x\) values where the function is undefined, specifically where the denominator equals zero. They appear as imaginary vertical lines on the graph towards which the function rises or falls indefinitely.

For our function \(f(x) = \frac{3x^2 + x - 5}{x^2 + 1}\), once you've determined that the denominator is never zero for real values of \(x\) (since \(x^2 = -1\) has no real solution), you discover there are no vertical asymptotes.

• Vertical asymptotes occur where the denominator equals zero.
• In this function, no \(x\) values make the denominator zero, so there are no vertical asymptotes.

Understanding vertical asymptotes is critical when sketching the graph of a rational function or analyzing its behavior near these points.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a rational function as \(x\) approaches positive or negative infinity. It is a horizontal line that the graph of the function approaches closely, but does not necessarily touch or cross.

In the function \(f(x) = \frac{3x^2 + x - 5}{x^2 + 1}\), to identify horizontal asymptotes, we look at the degrees of the polynomials in the numerator and the denominator. The degrees of both polynomials are 2, indicating they are equal. In such a case, the horizontal asymptote is the ratio of the leading coefficients of these polynomials.

• Numerator and denominator both have degree 2.
• Leading coefficient of numerator: 3.
• Leading coefficient of denominator: 1.

The horizontal asymptote is \(y = \frac{3}{1} = 3\). Thus, as \(x\) approaches infinity or negative infinity, the function approaches the line \(y = 3\). This tells us about the long-term behavior of the function and is vital for graph sketching and analysis.