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Chapter 1: Problem 37

Determine all vertical and slant asymptotes. $$y=\frac{x^{3}}{x^{2}+x-4}$$

Short Answer

Expert verified
The vertical asymptotes are at \(x=2\) and \(x=-2\). The slant asymptote is \(y=x\).

Step by step solution

01

Find the Vertical Asymptotes

To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\):\n\n\(x^{2}+x-4=0\)\n\nThis can be factored to:\n\n\( (x-2)(x+2) = 0\) \n\nSo, \(x =2\) and \(x=-2\) are your vertical asymptotes.
02

Check for Slant Asymptotes

A slant asymptote occurs when the degree of the numerator is one greater than the degree of the denominator. Since the degree of the numerator is 3 and the degree of the denominator is 2, there is a slant asymptote.
03

Calculate Slant Asymptote

To calculate the slant asymptote, perform a long division by dividing the denominator into the numerator. \nThe quotient y=x gives you the equation for the slant asymptote.

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Key Concepts

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

Vertical Asymptotes
A vertical asymptote is a line that a graph approaches but never touches or crosses. This typically occurs in rational functions where the denominator can be zero, leading to undefined values for the function. To find vertical asymptotes, you'll need to:
  • Set the denominator of the rational function equal to zero.

  • Solve for the variable to find the values of the vertical asymptotes.
In the exercise given, the function is \[y = \frac{x^3}{x^2 + x - 4}\]Here, to find vertical asymptotes, we set:\[x^2 + x - 4 = 0\]Factoring gives us \[(x - 2)(x + 2) = 0\]Therefore, the vertical asymptotes are at \[x = 2\] and \[x = -2\]Notice how the graph approaches these lines but never actually reaches or intersects them.
Slant Asymptotes
Slant asymptotes, or oblique asymptotes, appear when the degree of the polynomial in the numerator is exactly one higher than the degree of the polynomial in the denominator. They provide a straight line that the graph of the function approaches as it extends towards infinity.If you recognize that your function might have a slant asymptote, you will:
  • Perform polynomial long division with the numerator divided by the denominator.
For our function \(y = \frac{x^3}{x^2 + x - 4}\), the numerator is degree 3 and the denominator is degree 2, suggesting a slant asymptote exists.Conducting long division, we find that yields the quotient:\[y = x\]Thus, the slant asymptote is the line \[y = x\], as the remainder does not affect the asymptote.
Rational Functions
Rational functions are fractions composed of two polynomials 鈥� a numerator and a denominator. These functions can exhibit various asymptotic behaviors because they can become undefined when the denominator equals zero. Key characteristics of rational functions include:
  • Numerator and Denominator: Both are polynomials, which means they have powers of variables.

  • Asymptotes: Vertical asymptotes occur where the denominator is zero and there's no common factor with the numerator. Slant asymptotes appear when the degree of the numerator is one higher than the degree of the denominator.

  • Behavior at Infinity: Rational functions often approach a particular line (asymptote) as the independent variable grows large positively or negatively.
Understanding how these elements interact helps you consistently determine asymptotic behavior and plot rational functions accurately.

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Most popular questions from this chapter

Symbolically find the largest \(\delta\) corresponding to \(\varepsilon=0.1\) in the definition of \(\lim _{x \rightarrow 1^{-}} 1 / x=1 .\) Symbolically find the largest 8 corresponding to \(\varepsilon=0.1\) in the definition of \(\lim _{x \rightarrow+1} 1 / x=1\) Which \(\delta\) could be used in the definition of \(\lim _{x \rightarrow 1} 1 / x=1 ?\) Briefly explain. Then prove that \(\lim _{x \rightarrow 1} 1 / x=1\)

Use numerical and graphical evidence to conjecture values for each limit. $$\lim _{x \rightarrow 0} \frac{x^{2}+x}{\sin x}$$

In exercise \(85,\) you needed to find an example indicating that the following statement is not (necessarily) true: if \(h(a)=0\) then \(f(x)=\frac{g(x)}{h(x)}\) has a vertical asymptote at \(x=a .\) This is not true, but perhaps its converse is true: if \(f(x)=\frac{g(x)}{h(x)}\) has a vertical asymptote at \(x=a,\) then \(h(a)=0 .\) Is this statement true? What if \(g\) and \(h\) are polynomials?

Consider the following arguments concerning \(\lim _{x \rightarrow 0^{+}} \sin \frac{\pi}{x}\) First, as \(x>0\) approaches \(0, \frac{\pi}{x}\) increases without bound; since \(\sin t\) oscillates for increasing \(t\), the limit does not exist. Second: taking \(x=1,0.1,0.01\) and so on, we compute \(\sin \pi=\sin 10 \pi=\sin 100 \pi=\cdots=0 ;\) therefore the limit equals \(0 .\) Which argument sounds better to you? Explain. Explore the limit and determine which answer is correct.

As we see in Chapter 2, the velocity of an object that has traveled \(\sqrt{x}\) miles in \(x\) hours at the \(x=1\) hour mark is given by \(v=\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1} .\) Estimate this limit.
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