/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 Sketch the graph of the function... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 105

Sketch the graph of the function. Identify any asymptotes. $$f(x)=\frac{7}{-x-1}$$

Short Answer

Expert verified
The graph of the function \(f(x) = \frac{7}{-x-1}\) has one vertical asymptote at \(x = -1\), and one horizontal asymptote at \(y = 0\). The graph approaches these asymptotes but never touches or crosses them.

Step by step solution

01

Identify the denominator and set it to zero

The denominator of the function is \(-x - 1\). By setting it to zero we get \(-x - 1 = 0\). Solving for \(x\), we get \(x = -1\). Our vertical asymptote is \(x = -1\).
02

Identify the Horizontal Asymptote

The horizontal asymptote is determined by comparing the degrees of the polynomial in the numerator and the denominator. They both have the degree of one. Hence, we have to just compare the ratio of the coefficients in front of \(x\) in both numerator and denominator. However, there is no \(x\) term in the numerator of our function, so the horizontal asymptote is \(y = 0\).
03

Graph the Function and Asymptotes

Now we have to graph the function. We know that the graph approaches the asymptotes but doesn't touch or cross them. The graph will approach and get closer to \(y = 0\) as \(x\) goes to positive or negative infinity, and the graph will break at \(x = -1\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vertical Asymptotes
When studying rational functions, the concept of vertical asymptotes is essential. A vertical asymptote is a line where the function's graph tends to infinity. Imagine it as a boundary your graph can approach, but never actually touch or cross.

Finding vertical asymptotes involves looking at the denominator of the rational function. If substituting a particular value for x makes the denominator zero, that value is where the vertical asymptote is located.
  • Identify the denominator of the function, for example, \(-x - 1\) in the function \(f(x) = \frac{7}{-x-1}\).
  • Set the denominator equal to zero and solve: \(-x - 1 = 0\).
  • Solving gives you \(x = -1\), which is your vertical asymptote.
As \(x\) approaches \(-1\), the function value grows significantly towards positive or negative infinity, forming a vertical line over x = -1.
Horizontal Asymptotes
The horizontal asymptote of a rational function indicates the behavior of the graph as \(x\) approaches positive or negative infinity. The graph will get closer and closer to this line, but just like the vertical asymptote, it will not cross it.

To determine the horizontal asymptote, you compare the degrees of the numerator and the denominator. Let鈥檚 look at some simple rules:
  • If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).
  • If the degrees are equal, then the horizontal asymptote is the ratio of their leading coefficients.
In our example, the numerator does not contain an \(x\) term, essentially making its degree 0, while the denominator's degree is 1. Thus, the horizontal asymptote is \(y = 0\), indicating no crossing over the x-axis no matter how large or small x becomes.
Denominator Zeroing
Understanding how the denominator affects your function is at the heart of analyzing rational equations. "Denominator zeroing" is the method to identify where the denominator equals zero, which is crucial in finding vertical asymptotes.

Consider the function \(f(x) = \frac{7}{-x-1}\). The first step is to set the denominator \(-x-1\) to zero, finding all the x-values where the function is undefined.

  • Set 鈭掟潙モ垝1=0 and solve for x. The solution, 饾懃 = -1, shows us a break in the function.
  • Where the function is undefined, it often leads to vertical asymptotes.
This crucial step of setting the denominator to zero gives insight into points where the graph acts distinctly, with values becoming extremely large or small but never reaching or encompassing these values.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(A) find the determinant of \(A,\) (b) find \(A^{-1},\) (c) find \(\operatorname{det}\left(A^{-1}\right),\) and (d) compare your results from parts (a) and (c). Make a conjecture based on your results. $$A=\left[\begin{array}{rr} 1 & 2 \\ -2 & 2 \end{array}\right]$$

Consider a company that specializes in potting soil. Each bag of potting soil for seedlings requires 2 units of sand, 1 unit of loam, and 1 unit of peat moss. Each bag of potting soil for general potting requires 1 unit of sand, 2 units of loam, and 1 unit of peat moss. Each bag of potting soil for hardwood plants requires 2 units of sand, 2 units of loam, and 2 units of peat moss. Find the numbers of bags of the three types of potting soil that the company can produce with the given amounts of raw materials. 500 units of sand 500 units of loam 400 units of peat moss

Find the domain of the function and identify any horizontal or vertical asymptotes. $$f(x)=\frac{x-4}{x^{2}+16}$$

Determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form. $$\left[\begin{array}{lllr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 2 \end{array}\right]$$

Evaluate the determinant, in which the entries are functions. Determinants of this type occur when changes of variables are made in calculus. $$\left|\begin{array}{cc} x & x \ln x \\ 1 & 1+\ln x \end{array}\right|$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.