Problem 15 - College Algebra with Corequisite Support

Question

For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. $$ f(x)=\frac{3 x-4}{x^{3}-16 x} $$

Step-by-Step Solution

Answer

Domain: all real numbers except -4, 0, 4. Vertical asymptotes at x = -4, 0, 4. Horizontal asymptote: y = 0.

Step 1: Finding the Domain

The domain of a function is all the values of \(x\) for which the function is defined. For rational functions, this means we must consider the values of \(x\) where the denominator is zero and exclude them from the domain. Set the denominator equal to zero: \[x^3 - 16x = 0\]Factor the equation:\[x(x^2 - 16) = 0\]This gives us \(x = 0\) or \(x^2 = 16\), leading to \(x = \pm 4\). Therefore, the function is undefined at \(x = 0, 4, -4\). The domain is all real numbers except \(-4, 0, 4\).

Step 2: Identifying Vertical Asymptotes

Vertical asymptotes occur at the values of \(x\) where the denominator is zero and the numerator is non-zero. From Step 1, we know the potential vertical asymptotes occur at \(x = -4, 0, 4\).Substitute these values into the numerator:For \(x = -4\): \(3(-4) - 4 = -12 - 4 = -16 eq 0\)For \(x = 0\): \(3(0) - 4 = -4 eq 0\)For \(x = 4\): \(3(4) - 4 = 12 - 4 = 8 eq 0\)Thus, vertical asymptotes are at \(x = -4, 0, 4\).

Step 3: Determining the Horizontal Asymptote

Horizontal asymptotes are determined by looking at the degrees of the polynomials in the numerator and denominator. In this function:The degree of the numerator (\(3x - 4\)) is 1.The degree of the denominator (\(x^3 - 16x\)) is 3.Because the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).

Key Concepts

Domain of a function

The domain of a function refers to all the possible values that the variable, often represented as \( x \), can take for which the function is well-defined. In simpler terms, it's the set of all \( x \) values where the function doesn't "break" or become undefined. For rational functions like \( f(x)=\frac{3x-4}{x^3-16x} \), the function breaks wherever the denominator equals zero since division by zero is undefined.

To find these points, set the denominator equal to zero: \[ x^3 - 16x = 0 \] Factor the equation: \[ x(x^2 - 16) = 0 \] This gives us solutions: \( x = 0 \) or solving \( x^2 = 16 \), which gives \( x = 4 \) and \( x = -4 \).

Therefore, the function is undefined at \( x = -4, 0, \) and \( 4 \). Hence, the domain of \( f(x) \) consists of all real numbers except these three points. When writing the domain, we would say:

• Domain: All real numbers except \(-4, 0, \) and \( 4 \)

Vertical Asymptotes

Vertical asymptotes occur in a rational function where the function approaches infinity or negative infinity because the denominator is zero and the numerator is not zero. These are essentially the points in the graph where it spikes upwards or downwards and do not intersect the line.

From our domain analysis, we know the values where the denominator is zero are \( x = -4, 0, 4 \). To confirm these are indeed vertical asymptotes, we check the numerator to ensure it is not zero at these points.

• At \( x = -4 \), the numerator is \( 3(-4) - 4 = -16 \), which is non-zero.
• At \( x = 0 \), the numerator is \( 3(0) - 4 = -4 \), which is non-zero.
• At \( x = 4 \), the numerator is \( 3(4) - 4 = 8 \), which is non-zero.

Since none of these reduce the numerator to zero, all three values lead to vertical asymptotes. Thus, the vertical asymptotes are located at \( x = -4, 0, 4 \). This means the graph of the function will tend towards infinity around these points.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a graph as \( x \) approaches infinity or negative infinity. They give us a sense of what value the function's output (\( y \)) approaches as inputs become very large or very small.

To determine horizontal asymptotes for rational functions, compare the degrees of the numerator and denominator polynomials.

• The degree of the numerator "\( 3x - 4 \)" is 1.
• The degree of the denominator "\( x^3 - 16x \)" is 3.

If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is always \( y = 0 \). This indicates that as \( x \) grows positively or negatively towards infinity, the output \( y \) will approach zero. Thus, for the given function \( f(x)=\frac{3x-4}{x^3-16x} \), the horizontal asymptote is \( y = 0 \). This profound relationship assures that regardless of how large or small \( x \) goes, the function levels out towards \( y = 0 \).