Problem 27 - Algebra and Trigonometry

Question

Find all horizontal and vertical asymptotes (if any). $$ s(x)=\frac{(5 x-1)(x+1)}{(3 x-1)(x+2)} $$

Step-by-Step Solution

Answer

Vertical asymptotes: \(x = \frac{1}{3}\) and \(x = -2\). Horizontal asymptote: \(y = \frac{5}{3}\).

Step 1: Simplify the Function

Start by expanding both the numerator and the denominator of the given function \( s(x) = \frac{(5x-1)(x+1)}{(3x-1)(x+2)} \). This is useful to identify the behavior of the function when finding horizontal asymptotes. Expand the products:- Numerator: \((5x - 1)(x + 1) = 5x^2 + 5x - x - 1 = 5x^2 + 4x - 1\)- Denominator: \((3x - 1)(x + 2) = 3x^2 + 6x - x - 2 = 3x^2 + 5x - 2\)Thus, the function is simplified to \( s(x) = \frac{5x^2 + 4x - 1}{3x^2 + 5x - 2} \).

Step 2: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator is equal to zero, provided the numerator is non-zero at those points.First, find the roots of the denominator: \(3x^2 + 5x - 2 = 0\).Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 5\), \(c = -2\).Calculate:\[x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-2)}}{6}\]\[x = \frac{-5 \pm \sqrt{25 + 24}}{6}\]\[x = \frac{-5 \pm \sqrt{49}}{6}\]\[x = \frac{-5 \pm 7}{6}\]Thus, the solutions are \(x = \frac{-5 + 7}{6} = \frac{1}{3}\) and \(x = \frac{-5 - 7}{6} = -2\).Verify if the numerator is zero at these points. Since neither makes the numerator zero, the vertical asymptotes are at \(x = \frac{1}{3}\) and \(x = -2\).

Step 3: Identify Horizontal Asymptotes

Horizontal asymptotes depend on the degrees of the numerator and denominator:1. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).2. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is \(y = \frac{a}{b}\), where \(a\) and \(b\) are the leading coefficients.Here, both the numerator \(5x^2 + 4x - 1\) and the denominator \(3x^2 + 5x - 2\) are quadratic (degree 2), so compare their leading coefficients:\(y = \frac{5}{3}\).The horizontal asymptote is \(y = \frac{5}{3}\).

Key Concepts

Horizontal Asymptotes

Horizontal asymptotes help us understand the end behavior of a rational function. In simple terms, they show us the value that the function approaches as the input, or x-values, become extremely large or small.

For rational functions, the rule of thumb is to compare the degrees of the polynomials in the numerator and the denominator:

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at the line \(y = 0\).
• If the degree of the numerator is equal to the degree of the denominator, you find the horizontal asymptote by dividing the leading coefficients of the numerator and the denominator.
• If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In our function \(s(x) = \frac{5x^2 + 4x - 1}{3x^2 + 5x - 2}\), both the numerator and the denominator have a degree of 2. Thus, the horizontal asymptote is found by dividing the leading coefficient of the numerator, 5, by the leading coefficient of the denominator, 3, giving us \(y = \frac{5}{3}\). This tells us that as \(x\) becomes very large or very small, the value of \(s(x)\) approaches \(\frac{5}{3}\).

Vertical Asymptotes

Vertical asymptotes illustrate where a function tends towards infinity; they occur when the function's denominator is zero and the numerator is not zero at the same x-value.

To find where these asymptotes occur for our rational function \(s(x) = \frac{5x^2 + 4x - 1}{3x^2 + 5x - 2}\), we set the denominator equal to zero and solve the equation:
\(3x^2 + 5x - 2 = 0\).

Applying the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 3\), \(b = 5\), and \(c = -2\), reveals solutions:

• \(x = \frac{1}{3}\)
• \(x = -2\)

These solutions point to the x-values at which vertical asymptotes occur, because the numerator is not zero at these points. Therefore, the vertical asymptotes are \(x = \frac{1}{3}\) and \(x = -2\). These are where \(s(x)\) approaches positive or negative infinity.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. These functions can have unique graphical features such as asymptotes and holes.

Asymptotes—whether horizontal or vertical—help define the behavior of the function as input values become very large, very small, or approach certain points. The rational function in our exercise, \(s(x) = \frac{(5x-1)(x+1)}{(3x-1)(x+2)}\), once expanded to \(s(x) = \frac{5x^2 + 4x - 1}{3x^2 + 5x - 2}\), exhibits both kinds of asymptotes:

• A horizontal asymptote at \(y = \frac{5}{3}\), where the function levels off as \(x\) tends to infinity or negative infinity.
• Vertical asymptotes at \(x = \frac{1}{3}\) and \(x = -2\), where the function diverges, exhibiting breaks in the graph.

Understanding these properties allows us to sketch an accurate graph of the function and anticipate its behavior across different regions of its domain.