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Chapter 4: Problem 25

Find all horizontal and vertical asymptotes (if any). $$ s(x)=\frac{6 x^{2}+1}{2 x^{2}+x-1} $$

Short Answer

Expert verified
The horizontal asymptote is \( y = 3 \); vertical asymptotes are \( x = \frac{1}{2} \) and \( x = -1 \).

Step by step solution

01

Determine Horizontal Asymptotes

To find horizontal asymptotes, compare the degrees of the polynomial in the numerator and the polynomial in the denominator. Both numerator and denominator are degree 2 polynomials. For rational functions where the degrees are equal, the horizontal asymptote is found by taking the ratio of the leading coefficients. The leading coefficient of the numerator is 6, and the leading coefficient of the denominator is 2, which gives:\[y = \frac{6}{2} = 3\]Thus, the horizontal asymptote is \( y = 3 \).
02

Determine Vertical Asymptotes

Finding vertical asymptotes requires us to set the denominator equal to zero and solve for \( x \). Solve for \( x \) using the equation:\[ 2x^2 + x - 1 = 0 \]Factor the quadratic:\( 2x^2 + x - 1 = 0 \)Factoring, \((2x - 1)(x + 1) = 0 \)Set each factor to zero and solve for \( x \):\[2x - 1 = 0 \rightarrow x = \frac{1}{2} \\]\[x + 1 = 0 \rightarrow x = -1 \\]Therefore, the vertical asymptotes are \( x = \frac{1}{2} \) and \( x = -1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Functions
A rational function is a fraction where both the numerator and the denominator are polynomials. It's important to note that the denominator of a rational function can't be zero since division by zero is undefined. This type of function is expressed in the form:
  • \( f(x) = \frac{P(x)}{Q(x)} \)
where \( P(x) \) and \( Q(x) \) are polynomials. The properties of rational functions are intriguing, especially when investigating asymptotes, which are lines that the graph approaches as it extends towards infinity. The two main types of asymptotes we'll discuss are horizontal and vertical asymptotes.
Polynomial Degree Comparison
When working with rational functions, understanding the degree of the polynomials involved is crucial. The degree of a polynomial is the highest power of the variable in the expression. In the given problem, the function:
  • \( s(x)=\frac{6x^{2}+1}{2x^{2}+x-1} \)
has a numerator and a denominator, each of degree 2. This implies that they have the same degree. Comparing the degrees of the numerator and the denominator helps identify the type and location of horizontal asymptotes.
When both polynomial degrees are equal, the horizontal asymptote can be determined from the leading coefficients.
Leading Coefficients
The leading coefficient of a polynomial is the coefficient of the term with the highest degree. Identifying the leading coefficients is essential in finding the horizontal asymptotes of a rational function. In our example, the leading coefficients are:
  • Numerator (6)
  • Denominator (2)
To find the horizontal asymptote for rational functions where the degrees are equal, calculate the ratio of the leading coefficients:
\( y = \frac{6}{2} = 3 \).
Thus, the horizontal asymptote for the given function is \( y = 3 \). Understanding and using leading coefficients makes complex polynomial relationships approachable.
Factoring Quadratics
Factoring is a method used to simplify quadratic expressions, usually aiming to find the roots or solve equations. It is particularly useful for determining vertical asymptotes in rational functions. To find the vertical asymptotes, we focus on the denominator and set it to zero:
  • \( 2x^2 + x - 1 = 0 \)
Factoring this quadratic, we express it as:
  • \((2x - 1)(x + 1) = 0\)
We solve for each factor to find where the denominator becomes zero:
\( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \), and \( x + 1 = 0 \) gives \( x = -1 \).
So, the vertical asymptotes are \( x = \frac{1}{2} \) and \( x = -1 \). Factoring quadratics effectively opens up the underlying math, allowing us to explore the behavior of rational functions around roots.

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Most popular questions from this chapter

Transformations of \(y=1 / x^{2}\) In Example 2 we saw that some simple rational functions can be graphed by shifting, stretching, or reflecting the graph of \(y=1 / x\) . In this exercise we consider rational functions that can be graphed by transforming the graph of \(y=1 / x^{2},\) shown on the following page. (a) Graph the function $$ r(x)=\frac{1}{(x-2)^{2}} $$ by transforming the graph of \(y=1 / x^{2}\) . (b) Use long division and factoring to show that the function $$ s(x)=\frac{2 x^{2}+4 x+5}{x^{2}+2 x+1} $$ can be written as $$ s(x)=2+\frac{3}{(x+1)^{2}} $$ Then graph \(s\) by transforming the graph of \(y=1 / x^{2} .\) (c) One of the following functions can be graphed by trans- forming the graph of \(y=1 / x^{2} ;\) the other cannot. Use transformations to graph the one that can be, and explain why this method doesn't work for the other one. $$ p(x)=\frac{2-3 x^{2}}{x^{2}-4 x+4} \quad q(x)=\frac{12 x-3 x^{2}}{x^{2}-4 x+4} $$

The Doppler Effect As a train moves toward an observer (see the figure), the pitch of its whistle sounds higher to the ob- server than it would if the train were at rest, because the crests of the sound waves are compressed closer together. This phenomenon is called the Doppler effect. The observed pitch \(P\) is a function of the speed \(v\) of the train and is given by $$ P(v)=P_{0}\left(\frac{s_{0}}{s_{0}-v}\right) $$ where \(P_{0}\) is the actual pitch of the whistle at the source and \(S_{0}=332 \mathrm{m} / \mathrm{s}\) is the speed of sound in air. Suppose that a train has a whistle pitched at \(P_{0}=440 \mathrm{Hz}\) . Graph the func- tion \(y=P(v)\) using a graphing device. How can the vertical asymptote of this function be interpreted physically?

Graphs with Holes In this chapter we adopted the convention that in rational functions, the numerator and denominator don't share a common factor. In this exercise we consider the graph of a rational function that does not satisfy this rule. (a) Show that the graph of $$ r(x)=\frac{3 x^{2}-3 x-6}{x-2} $$ is the line \(y=3 x+3\) with the point \((2,9)\) removed. [Hint. Factor. What is the domain of \(r ? ]\) (b) Graph the rational functions: $$ \begin{aligned} s(x) &=\frac{x^{2}+x-20}{x+5} \\ t(x) &=\frac{2 x^{2}-x-1}{x-1} \\ u(x) &=\frac{x-2}{x^{2}-2 x} \end{aligned} $$

Use Descartes鈥� Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros. $$ P(x)=x^{5}+4 x^{3}-x^{2}+6 x $$

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. $$ r(x)=\frac{2 x^{3}+2 x}{x^{2}-1} $$
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