/*! 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 56 Rational Function Does the graph... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 56

Rational Function Does the graph of every rational function have a vertical asymptote? Explain.

Short Answer

Expert verified
No, not all rational functions have vertical asymptotes. This depends on the specific form of the function, particularly whether the denominator polynomial becomes zero at a value of x where the numerator polynomial does not.

Step by step solution

01

Understand vertical asymptotes

A vertical asymptote of a function is a vertical line x = a where the function tends to either positive infinity or negative infinity as x approaches a. Typically in a rational function (a function of the form f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials), we would find vertical asymptotes at values of x for which the denominator Q(x) is zero (as division by zero is undefined), while the numerator P(x) is not zero.
02

Understand exceptions to vertical asymptotes

However, there is an exception to this rule. If the denominator and the numerator both become zero at the same value of x, it can result in one of the two circumstances. If the order of the zero in the denominator is greater, then we still get a vertical asymptote. However, if the zeroes are of the same order, then that x value is a hole in the function rather than an asymptote.
03

Conclusion

So, not all rational functions have vertical asymptotes. Whether or not a given rational function has a vertical asymptote depends on the specific form of the function, particularly the properties of the denominator polynomial in relation to the numerator polynomial.

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!

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

Using the Intermediate Value Theorem In Exercises \(95-98\) , verify that the Intermediate Value Theorem applies to the indicated interval and find the value of \(c\) guaranteed by the theorem. $$ f(x)=\frac{x^{2}+x}{x-1}, \quad\left[\frac{5}{2}, 4\right], \quad f(c)=6 $$

Continuity of a Composite Function In Exercises \(67-72\) discuss the continuity of the composite function \(h(x)=f(g(x))\) $$ \begin{array}{l}{f(x)=\tan x} \\ {g(x)=\frac{x}{2}}\end{array} $$

Using the Intermediate Value Theorem In Exercises \(91-94,\) use the Intermediate Value Theorem and a graphing utility to approximate the zero of the function in the interval \([0,1] .\) Repeatedly "zoom in" on the graph of the function to approximate the zero accurate to two decimal places. Use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. $$ h(\theta)=\tan \theta+3 \theta-4 $$

Testing for Continuity In Exercises \(77-84\) , describe the interval(s) on which the function is continuous. $$ f(x)=\frac{x}{x^{2}+x+2} $$

Prove or disprove: If \(x\) and \(y\) are real numbers with \(y \geq 0\) and \(y(y+1) \leq(x+1)^{2},\) then \(y(y-1) \leq x^{2}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Calculus

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.