Problem 28 - Precalculus Mathematics for Calculus

Question

Find all horizontal and vertical asymptotes (if any). $$s(x)=\frac{(2 x-1)(x+3)}{(3 x-1)(x-4)}$$

Step-by-Step Solution

Answer

Horizontal: \(y=\frac{2}{3}\); Vertical: \(x=\frac{1}{3}, x=4\).

Step 1: Understanding Horizontal Asymptotes

To find horizontal asymptotes, consider the degrees of the numerator and denominator polynomials. Here, both the numerator \((2x-1)(x+3)=2x^2+5x-3\) and the denominator \((3x-1)(x-4)=3x^2-13x+4\) are quadratic polynomials, i.e., their degree is 2.

Step 2: Calculating Horizontal Asymptotes

Since the degrees of the numerator and denominator are equal, the horizontal asymptote is given by the ratio of the leading coefficients of the numerator and the denominator. Therefore, the horizontal asymptote is \(y=\frac{2}{3}\).

Step 3: Understanding Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at those points. Determine the zeros of the denominator \((3x-1)(x-4)\) by setting it equal to zero.

Step 4: Calculating Vertical Asymptotes

Solve \(3x-1=0\) and \(x-4=0\) separately.\[3x-1=0 \rightarrow x=\frac{1}{3}\]and\[x-4=0 \rightarrow x=4\]Both provide potential vertical asymptotes as long as the numerator is not zero at these points.

Step 5: Checking Numerator for Corresponding Zeros

Ensure that the numerator \((2x-1)(x+3)\) is not zero at \(x=\frac{1}{3}\) and \(x=4\). If \(x=\frac{1}{3}\) or \(x=4\) do not zero the numerator, these points are confirmed as vertical asymptotes.

Step 6: Confirmation and Conclusion

At \(x=\frac{1}{3}\): \((2(\frac{1}{3})-1)((\frac{1}{3})+3)eq 0\). Similarly, at \(x=4\): \((2(4)-1)(4+3)eq 0\). Both calculations confirm non-zero results at both points. Therefore, \(x=\frac{1}{3}\) and \(x=4\) are vertical asymptotes.

Key Concepts

Horizontal Asymptotes

Horizontal asymptotes are crucial in understanding the end behavior of a function, particularly for rational functions, which are ratios of two polynomials. These asymptotes are horizontal lines that a graph approaches, but may never actually reach, as the input values approach positive or negative infinity. In our exercise, we inspect the function \( s(x) = \frac{(2x-1)(x+3)}{(3x-1)(x-4)} \). To identify any horizontal asymptotes, it's essential to compare the degrees of the polynomial in the numerator to that in the denominator.

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
• If they are equal, which is the case here with both being quadratic (degree 2), the horizontal asymptote is \( y = \frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}} \).
• When the degree of the numerator is greater than the denominator, no horizontal asymptote exists.

For our function, the leading coefficient of the numerator is 2, and for the denominator, it's 3. Hence, the horizontal asymptote is \( y = \frac{2}{3} \). This means as \( x \to \pm \infty \), \( s(x) \to \frac{2}{3} \). This simplifies predicting the behavior of the graph at extremes.

Vertical Asymptotes

Vertical asymptotes are the values of \( x \) where a rational function becomes undefined. This happens when the denominator of the function equals zero, provided that the numerator isn't zero at those same \( x \)-values. Understanding vertical asymptotes helps predict points on the graph where it will shoot up to positive or down to negative infinity, forming a vertical line. For our function, \( s(x) = \frac{(2x-1)(x+3)}{(3x-1)(x-4)} \), we must find values that make the denominator zero.

• Factor the denominator: \( (3x-1)(x-4) \).
• Set each factor to zero individually: \( 3x-1 = 0 \) leads to \( x = \frac{1}{3} \), and \( x-4 = 0 \) leads to \( x = 4 \).

Before confirming these as vertical asymptotes, check that the numerator does not zero out at these \( x \)-values:

• \( (2x-1)(x+3) eq 0 \) when \( x = \frac{1}{3} \) and \( x = 4 \),.
• Thus, both \( x = \frac{1}{3} \) and \( x = 4 \) are indeed vertical asymptotes.

These means the graph will exhibit steep lines at these \( x \)-values, indicating abrupt changes in the function's value.

Rational Functions

Rational functions are functions expressed as the ratio of two polynomials. They can be intimidating due to their complexity, but understanding their behavior and properties, such as asymptotes, simplifies them. In the case of \( s(x) = \frac{(2x-1)(x+3)}{(3x-1)(x-4)} \), we deal with polynomials of degree 2 in both the numerator and the denominator.

• The form is \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
• Key features to observe include horizontal asymptotes, which tell where the function stabilizes at extreme values; and vertical asymptotes, which tell at which points the function is not defined.
• Domains of rational functions exclude x-values that zero out the denominator.

To work with rational functions, note their structure and focus on identification of critical points such as asymptotes. This process demystifies their graphs, making it easier to predict and plot. Recognizing how these functions behave around asymptotes builds a solid foundation in both calculus and algebra, bridging understanding between different mathematical concepts.