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Chapter 2: Problem 34

Find the slant asymptote of each rational function. $$F(x)=\frac{x^{3}-1}{x^{2}}$$

Short Answer

Expert verified
The slant asymptote of the function \(F(x) = \frac{x^{3}-1}{x^{2}}\) is \(y = x\)

Step by step solution

01

Identify the Degrees of the Numerator and the Denominator

Here, the degree of the numerator, which is the polynomial \(x^{3} - 1\), is 3, and the degree of the denominator, which is the polynomial \(x^{2}\), is 2. As the degree of the numerator is greater than the degree of the denominator by one, a slant asymptote exists.
02

Perform Polynomial Division

Perform polynomial long division to obtain the equation of the slant asymptote. Divide the numerator \(x^{3}-1\) by the denominator \(x^{2}\) using long division of polynomials. It yields \(x- \frac{1}{x^{2}}\).
03

Identify the Slant Asymptote

Exclude the remainder derived from the polynomial division to get the equation for slant asymptote. Here there is a fraction -\(\frac{1}{x^{2}}\) as a remainder which tends to 0 as x tends to infinity. So the slant asymptote of the function \(F(x)\) is \(y=x\).

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Key Concepts

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

Polynomial Long Division
Understanding polynomial long division is essential when dealing with rational functions, especially when you need to find slant asymptotes. It's quite similar to the long division process you learned in elementary school, but instead of numbers, you're dividing polynomials.

To perform polynomial long division, you'll want to:
  • Divide the first term of the numerator by the first term of the denominator and write the result above the division bar.
  • Multiply the entire denominator by that result and subtract from the numerator.
  • Repeat the process with the new (lower degree) polynomial that's now your numerator, until the degree of the remainder is less than the degree of the denominator.

It's a systematic process that sometimes involves adding zero terms to 'fill in the gaps' in the polynomial. For example, in our exercise, dividing \(x^3\) by \(x^2\) gives a quotient of \(x\) and a remainder of \(x^3 - x^2\cdot x\), which simplifies to \( -1 \). This step is critical for identifying the equation of the slant asymptote when the degree of the numerator is exactly one more than the degree of the denominator.
Rational Functions
A rational function is a fraction where both the numerator and the denominator are polynomials. The fascinating aspect of rational functions is their behavior as x approaches large values or infinity. This behavior often relates to asymptotes鈥攍ines which the graph of the function approaches but never actually touches.

There are three types of asymptotes to consider:
  • Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator.
  • Vertical asymptotes occur at values of x that make the denominator zero (assuming these x-values do not also zero out the numerator).
  • Slant (or oblique) asymptotes, which are the topic of our exercise, occur when the degree of the numerator is exactly one more than the degree of the denominator.

In our example, \( F(x) = \frac{x^3 - 1}{x^2} \), the numerator's degree is one higher than the denominator's, indicating a slant asymptote. By performing polynomial long division, we deduce the equation of this asymptote, which informs us how the function behaves as x becomes large.
Degrees of Polynomials
The degree of a polynomial represents the highest power of x in its expression. In simple terms, it's the largest exponent you'll see when the polynomial is written in standard form. The degree helps us predict a polynomial's end behavior and plays a significant role in identifying asymptotic behavior in rational functions.

For instance, in rational functions, when:
  • The degrees of the numerator and denominator are equal, the function has a horizontal asymptote at the ratio of their leading coefficients.
  • The degree of the numerator is less than the degree of the denominator, the x-axis (y=0) is a horizontal asymptote.
  • The degree of the numerator is one more than the degree of the denominator, as in the exercise \( F(x) = \frac{x^3 - 1}{x^2} \), a slant asymptote exists, found by polynomial long division.

In our example, the numerator (degree 3) is one degree higher than the denominator (degree 2). This information alone lets us predict the existence of a slant asymptote even before performing the actual division. Recognizing these patterns makes analyzing rational functions more intuitive.

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Most popular questions from this chapter

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=6 x^{4}+23 x^{3}+19 x^{2}-8 x-4$$

Find the zeros of each polynomial function. If a zero is a multiple zero, state its multiplicity. $$P(x)=2 x^{3}+9 x^{2}-2 x-9$$

The property that the product of conjugates of the form \((a+b i)(a-b i)\) is equal to \(a^{2}+b^{2}\) can be used to factor the sum of two perfect squares over the set of complex numbers. For example, \(x^{2}+y^{2}=(x+y i)(x-y i) .\) In Exercises 71 to \(74,\) factor the binomial over the set of complex numbers. $$x^{2}+16$$

Determine the degree of the numerator and the degree of the denominator of \(\frac{x^{3}+3 x^{2}-5}{x^{2}-4} \cdot[\mathrm{A} .2]\)

DIGITS OF PI In 1999, Professor Yasumasa Kanada of the University of Tokyo used a supercomputer to compute 206,158,430,000 digits of pi ( \(\pi\) ). (Source: Guinness World Records 2001, Bantam Books, p. \(252 .\) ) Computer scientists often try to find mathematical models that approximate the time a computer program takes to complete a calculation or mathematical procedure. Procedures for which the completion time can be closely modeled by a polynomial are called polynomial time procedures. Here is an example. A student finds that the time, in seconds, required to compute \(n \times 10,000\) digits of pi on a personal computer using the mathematical program MAPLE is closely approximated by $$\begin{aligned} T(n)=& 0.23245 n^{3}+0.53797 n^{2} \\ &+7.88932 n-8.53299 \end{aligned}$$ a. Evaluate \(T(n)\) to estimate how long, to the nearest second, the computer takes to compute 50,000 digits of \(\mathrm{pi}\). b. About how many digits of pi can the computer compute in 5 minutes? Round to the nearest thousand digits.
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