/*! 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 22 Explain why it is possible for c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Explain why it is possible for curves to intersect horizontal and oblique asymptotes but not to intersect vertical asymptotes.

Short Answer

Expert verified
Curves can intersect horizontal/oblique asymptotes but not vertical ones due to redefined behavior at vertical asymptotes.

Step by step solution

01

Understand Asymptotes

First, understand what asymptotes are. An asymptote is a line that a curve approaches as it heads towards infinity. We have three kinds of asymptotes: horizontal, vertical, and oblique (or slant).
02

Analyze Horizontal Asymptotes

Horizontal asymptotes occur when the curve approaches a horizontal line as the x-values go to positive or negative infinity. A curve can cross its horizontal asymptote because asymptotes describe end behavior, not the function value over the entire domain.
03

Analyze Vertical Asymptotes

Vertical asymptotes occur when the function goes to infinity as the x-value approaches a certain point. A curve cannot cross a vertical asymptote because the function is undefined at that point, often due to division by zero.
04

Analyze Oblique Asymptotes

Oblique asymptotes arise when the relationship between the numerator and denominator degrees in a rational function is such that the curve behaves like a linear function as x approaches infinity. Like horizontal, a curve can intersect its oblique asymptote; these intersections often happen closer to the origin.
05

Conclusion

Horizontal and oblique asymptotes describe the behavior as x approaches infinity, allowing crossing points. Vertical asymptotes represent undefined behavior, preventing intersection.

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 91Ó°ÊÓ!

Key Concepts

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

Horizontal Asymptotes
Horizontal asymptotes are lines that a graph approaches as the value of \(x\) tends to infinity. These asymptotes suggest how the graph behaves at extreme values, specifically when \(x\) becomes very large or very small. However, unlike barriers etched in stone, horizontal asymptotes are more like tendencies. They describe what happens in the distant realm of extremes, not necessarily what's going on in the world where \(x\) values are finite.

This means a curve can indeed cross a horizontal asymptote and, in fact, may do so multiple times within its domain. The most crucial element is that as \(x\) approaches infinity or negative infinity, the curve will eventually level off and align with the asymptote. To visualize this concept, imagine a rocket launched horizontally that bounces above and below a guiding laser line—eventually, it stabilizes along the laser's path.
  • The fact that curves can intersect horizontal asymptotes is due to these lines describing asymptotic behavior, but not dictating intersection restrictions.
  • Horizontal asymptotes provide a picture of end behavior and aren't necessarily a border for the graph within finite intervals.
Vertical Asymptotes
Vertical asymptotes differ significantly from their horizontal cousins. They arise in functions where the curve skydives or ascends without ever touching down—imagine a line going straight up at a certain \(x\) value, leaving the curve in turmoil as it attempts to touch this vertical boundary. To recognize them, look for places where the function becomes undefined, often as the denominator of a rational function approaches zero.

The idea behind vertical asymptotes is that as \(x\) approaches a certain value, the function shoots towards infinity or negative infinity. Because at these points the values of the function become undefined, the curve cannot cross such asymptotes. Vertical asymptotes, therefore, act like invisible walls that a curve cannot cross, standing as the places where the function fails to be defined:
  • Vertical asymptotes mark locations in the graph where the function's value is not only large but mathematically undefined.
  • These asymptotes highlight the inability of the curve to traverse beyond these certain \(x\) values, creating natural boundaries within the graph.
  • If you see a steep climb or drop around a certain value on the x-axis, you’re likely near a vertical asymptote!
Oblique Asymptotes
Oblique asymptotes are unique as they only appear in certain rational functions where the degree of the polynomial in the numerator is one higher than the degree of the denominator. These asymptotes resemble a slanted line, giving curves a tilt as they sail off into the infinite. Much like horizontal asymptotes, oblique ones primarily describe behavior at infinity.

Because they describe behavior at infinity, oblique asymptotes can indeed be crossed by their curves. This is particularly common near the origin, where the extra components of the polynomial dominate the function's shape. As \(x\) grows large, however, the graph settles into the linear glide path set by the oblique asymptote:
  • Oblique asymptotes imply a linear pattern that the function adopts as \(x\) goes to infinity.
  • Curves are free to meet and cross these slanting guides within smaller intervals, but will eventually align as \(x\) extends to greatness.
  • The spectacle of curves crossing these asymptotes showcases their flexible nature in capturing the curve's end game.

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 rectangular piece of cardboard is 13 inches long and 9 inches wide. From each corner, a square piece is cut out, and then the flaps are turned up to form an open box. Determine the length of a side of the square pieces so that the volume of the box is as large as possible.

For each of the following functions, find the \(x\) intercepts and find the turning points. Express your answers to the nearest tenth. (a) \(f(x)=x^{3}+2 x^{2}-3 x+4\) (b) \(f(x)=42 x^{3}-x^{2}-246 x-35\) (c) \(f(x)=x^{4}-4 x^{2}-4\)

A polynomial function with real coefficients is continuous everywhere; that is, its graph has no holes or breaks. This is the basis for the following property: If \(f(x)\) is a polynomial with real coefficients, and if \(f(a)\) and \(f(b)\) are of opposite sign, then there is at least one real zero between \(a\) and \(b\). This property, along with our knowledge of polynomial functions, provides the basis for locating and approximating irrational solutions of a polynomial equation. Consider the equation \(x^{3}+2 x-4=0\). Applying Descartes' rule of signs, we can determine that this equation has one positive real solution and two nonreal complex solutions. (You may want to confirm this!) The rational root theorem indicates that the only possible rational solutions are 1,2 , and 4 . Using a little more compact format for synthetic division, we obtain the following results when testing for 1 and 2 as possible solutions: $$ \begin{array}{r|rrrr} & 1 & 0 & 2 & -4 \\ 1 & 1 & 1 & 3 & -1 \\ 2 & 1 & 2 & 6 & 8 \end{array} $$ Because \(f(1)=-1\) and \(f(2)=8\), there must be an irrational solution between 1 and 2 . Furthermore, \(-1\) is closer to 0 than is 8 , so our guess is that the solution is closer to 1 than to 2 . Let's start looking at \(1.0,1.1,1.2\), and so on, until we can place the solution between two numbers. Because \(f(1.1)=-0.469\) and \(f(1.2)=0.128\), the irrational solution must be between \(1.1\) and 1.2. Furthermore, because \(0.128\) is closer to 0 than is \(-0.469\), our guess is that the solution is closer to \(1.2\) than to \(1.1\). Let's start looking at \(1.15,1.16\), and so on. $$ \begin{array}{l|rrrrr} & 1 & 0 & 2 & -4 \\ \ 1.15 & 1 & 1.15 & 3.3225 & -0.179 \\ 1.16 & 1 & 1.16 & 3.3456 & -0.119 \\ 1.17 & 1 & 1.17 & 3.3689 & -0.058 \\ 1.18 & 1 & 1.18 & 3.3924 & 0.003 \end{array} $$ Because \(f(1.17)=-0.058\) and \(f(1.18)=0.003\), the irrational solution must be between \(1.17\) and \(1.18\). Therefore we can use \(1.2\) as a rational approximation to the nearest tenth. For each of the following equations, (a) verify that the equation has exactly one irrational solution, and (b) find an approximation, to the nearest tenth, of that solution. (a) \(x^{3}+x-6=0 \) (b) \(x^{3}-6 x-6=0 \) (c) \(x^{3}-27 x-60=0 \) (d) \(x^{3}-x^{2}-x-1=0 \) (e) \(x^{3}-2 x-10=0\) (f) \(x^{3}-5 x^{2}-1=0 \)

Graph each of the following rational functions: $$ f(x)=\frac{3}{(x+2)(x-4)} $$

Graph \(f(x)=x^{3}\). Now predict the graphs for \(f(x)=\) \(x^{3}+2, f(x)=-x^{3}+2\), and \(f(x)=-x^{3}-2\). Graph these three functions on the same set of axes with the graph of \(f(x)=x^{3}\).

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.