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Chapter 4: Problem 16

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

Short Answer

Expert verified
Vertical asymptote: \(x = 1\). Horizontal asymptote: \(y = 2\).

Step by step solution

01

Identify the Vertical Asymptote

Vertical asymptotes occur where the denominator of the function is zero, and the numerator is not zero at the same point. Identify the vertical asymptote by setting the denominator equal to zero: \(x - 1 = 0\). Solving this equation gives \(x = 1\). Therefore, there is a vertical asymptote at \(x = 1\).
02

Identify the Horizontal Asymptote

Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator. In \(s(x) = \frac{2x + 3}{x - 1}\), both the numerator and the denominator are linear polynomials (degree 1). The horizontal asymptote, when the degrees are equal, is found using the leading coefficients: \(\frac{2}{1} = 2\). Thus, the horizontal asymptote is \(y = 2\).

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Key Concepts

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

Vertical Asymptotes
Vertical asymptotes in rational functions are crucial points where the function ceases to exist. This happens because, at these points, we're dividing by zero - a mathematical no-no! For example, consider the function \[s(x) = \frac{2x + 3}{x - 1}\]Here, the denominator is \(x - 1\). To find vertical asymptotes, set the denominator equal to zero: \[x - 1 = 0\]Solving this equation gives us \(x = 1\). At \(x = 1\), the function doesn鈥檛 have a value. Therefore, the vertical asymptote is a vertical line on the graph at \(x = 1\). It's like a marker, indicating where the function "blows up". Remember:
  • Vertical asymptotes occur where the denominator is zero.
  • They represent values that the function cannot take.
Horizontal Asymptotes
Horizontal asymptotes tell us about the behavior of a function as it stretches towards infinity in either direction. With the function \[s(x) = \frac{2x + 3}{x - 1}\]we look at the degrees of both the numerator and the denominator. Both components are linear, which means they have the same degree of 1. When the numerator and the denominator share the same degree, we focus on their leading coefficients to find the horizontal asymptote:\[\frac{2}{1} = 2\]Hence, we have a horizontal asymptote at \(y = 2\). This means, as \(x\) goes to infinity or negative infinity, the function approaches the line \(y = 2\) but never truly touches it. Key things to note:
  • Horizontal asymptotes show end behavior of rational functions.
  • When degrees match, use leading coefficients for \(y\)-value.
Rational Functions
Rational functions are like the interesting twin of simpler algebraic functions. They are any function that can be written as a fraction of two polynomials. Our given function \[s(x) = \frac{2x + 3}{x - 1}\]is a stellar example. All rational functions can have interesting features like vertical and horizontal asymptotes. This function has a linear polynomial in the numerator, \(2x + 3\), and another in the denominator, \(x - 1\). Some things to remember about rational functions are:
  • They can be undefined for certain \(x\)-values where the denominator is zero.
  • They often feature asymptotes, showcasing behavior and limits.
  • Their graphs can appear highly varied - not just lines or curves!
Understanding rational functions is like unlocking doors to explore complex behaviors and patterns in algebra, making them fascinating to study!

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Most popular questions from this chapter

Volume of a Silo A grain silo consists of a cylindrical main section and a hemispherical roof. If the total volume of the silo (including the part inside the roof section) is \(15,000 \mathrm{ft}^{3}\) and the cylindrical part is 30 \(\mathrm{ft}\) tall, what is the radius of the silo, correct to the nearest tenth of a foot?

Show that the given values for \(a\) and \(b\) are lower and upper bounds for the real zeros of the polynomial. $$ P(x)=3 x^{4}-17 x^{3}+24 x^{2}-9 x+1 ; \quad a=0, b=6 $$

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. $$P(x)=2 x^{6}+5 x^{4}-x^{3}-5 x-1$$

Polynomials of Odd Degree The Conjugate Zeros Theorem says that the complex zeros of a polynomial with real coefficients occur in complex conjugate pairs. Explain how this fact proves that a polynomial with real coefficients and odd degree has at least one real zero.

Use a graphing device to find all real solutions of the equation, correct to two decimal places. $$ 4.00 x^{4}+4.00 x^{3}-10.96 x^{2}-5.88 x+9.09=0 $$
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