Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Can the graph of a polynomial have vertical or horizontal asymptotes? Explain.

Short Answer

Expert verified
If so, under what conditions? Answer: The graph of a polynomial function can have a horizontal asymptote only if it is a constant or linear function. Polynomial functions cannot have vertical asymptotes.

Step by step solution

01

Definition of Asymptote:

An asymptote is a line that the graph of a function approaches but never touches as the input (x) approaches infinity or negative infinity. There are two types of asymptotes: horizontal and vertical. A horizontal asymptote is a horizontal line that the graph of the function approaches as the input (x) goes to infinity or negative infinity. A vertical asymptote is a vertical line that the graph of the function approaches but never touches or crosses.
02

Properties of Polynomial Functions:

Polynomial functions have the form P(x) = a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0, where a_n, a_(n-1), ..., a_1, a_0 are constants and n is a non-negative integer. Polynomial functions are continuous on the entire real line, which means that they do not have any breaks or jumps in their graphs. Also, they are smooth on the entire real line, meaning that they do not have any sharp corners or cusps. #tagName# Vertical Asymptote of Polynomial Functions:
03

A vertical asymptote occurs when the denominator of a rational function approaches zero but the numerator does not. However, polynomial functions do not have denominators, as they are not rational functions. Therefore, polynomial functions cannot have vertical asymptotes. #tagName# Horizontal Asymptote of Polynomial Functions:

A function has a horizontal asymptote if its graph approaches a horizontal line as the input (x) goes to infinity or negative infinity. Polynomial functions of degree 0 or 1 (constant or linear functions) have horizontal asymptotes because they approach a constant value as x goes to infinity or negative infinity. Higher-degree polynomial functions do not have horizontal asymptotes because their graphs continuously increase or decrease as x goes to infinity or negative infinity, depending on their leading term. In conclusion, the graph of a polynomial function can have a horizontal asymptote only if it is a constant or linear function, and it cannot have a vertical asymptote.

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 Vaia!

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

Show that any exponential function \(b^{x}\), for \(b>1,\) grows faster than \(x^{p},\) for \(p>0\)

Find the solution of the following initial value problems. $$f^{\prime}(u)=4(\cos u-\sin 2 u) ; f(\pi / 6)=0$$

Modified Newton's method The function \(f\) has a root of multiplicity 2 at \(r\) if \(f(r)=f^{\prime}(r)=0\) and \(f^{\prime \prime}(r) \neq 0 .\) In this case, a slight modification of Newton's method, known as the modified (or accelerated) Newton's method, is given by the formula $$x_{n+1}=x_{n}-\frac{2 f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}, \quad \text { for } n=0,1,2, \ldots$$ This modified form generally increases the rate of convergence. a. Verify that 0 is a root of multiplicity 2 of the function \(f(x)=e^{2 \sin x}-2 x-1\) b. Apply Newton's method and the modified Newton's method using \(x_{0}=0.1\) to find the value of \(x_{3}\) in each case. Compare the accuracy of each value of \(x_{3}\) c. Consider the function \(f(x)=\frac{8 x^{2}}{3 x^{2}+1}\) given in Example 4. Use the modified Newton's method to find the value of \(x_{3}\) using \(x_{0}=0.15 .\) Compare this value to the value of \(x_{3}\) found in Example 4 with \(x_{0}=0.15\)

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=e^{x}\left(x^{2}-7 x-12\right)$$

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=x^{3}-3 x^{2}$$

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free