Problem 58 - Algebra and Trigonometry

Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find all asymptotes of the graph of \(f(x)=\frac{2 x^{2}-5}{x^{2}-2 x-15}\)

Step-by-Step Solution

Answer

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

Step 1 - Simplify the function

First, factor both the numerator and the denominator of the function. For the numerator, factorize if needed. The function is given as: \[ f(x) = \frac{2x^2 - 5}{x^2 - 2x - 15} \]

Step 2 - Factor the Denominator

Factorize the denominator: \( x^2 - 2x - 15 = (x - 5)(x + 3) \)So the function becomes: \[ f(x) = \frac{2x^2 - 5}{(x - 5)(x + 3)} \]

Step 3 - Vertical Asymptotes

Find the values of \( x \) that make the denominator zero, as these are the locations of the vertical asymptotes. Set the denominator equal to zero: \( (x - 5)(x + 3) = 0 \)Solve for \( x \): \( x = 5 \) and \( x = -3 \)These are the vertical asymptotes: \( x = 5 \) and \( x = -3 \)

Step 4 - Horizontal Asymptote

Determine the horizontal asymptote by comparing the degrees of the polynomial in the numerator and the denominator. Since both the numerator and denominator are of degree 2, the horizontal asymptote is found by dividing the leading coefficients: \( f(x) = \frac{2x^2 - 5}{x^2 - 2x - 15} \rightarrow \frac{2}{1} \)The horizontal asymptote is \( y = 2 \)

Key Concepts

factoring polynomials

When working with rational functions, factoring polynomials can simplify the equation and make it easier to identify asymptotes and intercepts. Polynomials can be factored by finding their roots or using methods like grouping. For example, let's look at the function: \[ f(x) = \frac{2x^2 - 5}{x^2 - 2x - 15} \] To factor the denominator, we see that: \[ x^2 - 2x - 15 = (x - 5)(x + 3) \] Thus, the function simplifies to: \[ f(x) = \frac{2x^2 - 5}{(x - 5)(x + 3)} \]

vertical asymptotes

Vertical asymptotes occur where the denominator of a rational function is zero, as the function approaches infinity or negative infinity at those points. For the function \[ f(x) = \frac{2x^2 - 5}{(x - 5)(x + 3)} \] We set the denominator equal to zero:\[ (x - 5)(x + 3) = 0 \] Solving for \( x \) gives: \[ x = 5 \] and \[ x = -3 \] Therefore, the vertical asymptotes are \( x = 5 \) and \( x = -3 \).

horizontal asymptotes

Horizontal asymptotes indicate the behavior of a function as \( x \) approaches infinity or negative infinity. For rational functions, horizontal asymptotes depend on the degrees of the polynomials in the numerator and the denominator. The given function is \[ f(x) = \frac{2x^2 - 5}{x^2 - 2x - 15} \] Since the degrees of both the numerator and the denominator are 2, we compare the leading coefficients. Here, the leading coefficients are 2 for the numerator and 1 for the denominator. Thus, the horizontal asymptote is found by: \[ y = \frac{2}{1} = 2 \] So, the horizontal asymptote is \( y = 2 \).

rational functions

A rational function is a quotient of two polynomials. Generally, it takes the form \[ f(x) = \frac{P(x)}{Q(x)} \] where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \) is not equal to zero. Rational functions can have asymptotic behavior and require careful analysis to identify vertical and horizontal asymptotes. Consider the function \[ f(x) = \frac{2x^2 - 5}{x^2 - 2x - 15} \] We can factor the denominator to identify vertical asymptotes and compare polynomial degrees to determine horizontal asymptotes. Such functions can model various real-world situations, providing insight into behaviors like growth, decay, and equilibrium.