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Chapter 2: Problem 38

Graph the rational functions in Exercises \(27-38 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{x^{3}+1}{x^{2}} $$

Short Answer

Expert verified
Vertical asymptote at \(x=0\), oblique asymptote at \(y=x\).

Step by step solution

01

Factor the Function

The given rational function is \( y = \frac{x^3 + 1}{x^2} \). To analyze this function, first factor the numerator if possible. Here, \( x^3 + 1 \) can be factored using the formula for the sum of cubes: \( x^3 + 1 = (x + 1)(x^2 - x + 1) \). Therefore, the function becomes \( y = \frac{(x + 1)(x^2 - x + 1)}{x^2} \).
02

Identify the Vertical Asymptotes

Vertical asymptotes occur where the denominator of the rational function is zero, and the numerator is not zero simultaneously. Since \( x^2 = 0 \) solves to \( x = 0 \), check if the numerator is zero when \( x = 0 \): for \( x = 0 \), \( x+1 \) and \( x^2 - x + 1 \) are not zero. Hence, \( x = 0 \) is a vertical asymptote.
03

Identify the Horizontal Asymptote (if any)

Horizontal asymptotes are determined by comparing the degrees of the polynomial in the numerator and the denominator. The numerator is of degree 3 and the denominator is of degree 2. When the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. However, there is an oblique asymptote, determined by polynomial long division.
04

Find the Oblique Asymptote

Since the degree of the numerator is higher than the degree of the denominator, we perform long division of \( x^3 + 1 \) by \( x^2 \). Dividing \( x^3 + 0x^2 + 0x + 1 \) by \( x^2 \) gives \( x + 0 \) as the result with a remainder. Thus, the oblique asymptote is \( y = x \).
05

Sketch the Graph

To sketch the graph, plot the vertical asymptote (a dashed line at \( x = 0 \)), and the oblique asymptote (a dashed line following \( y = x \)). Since the rational function simplifies as \( y = x + \frac{-x + 1}{x^2} \), plot points around the asymptotes to determine the graph's behavior. Use additional points to ensure the graph transitions smoothly around its asymptotes and intercepts.

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Key Concepts

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

Vertical Asymptote
In rational functions, vertical asymptotes occur at values of x that make the denominator zero, given that the numerator does not equal zero at those points. This creates a vertical line on a graph where the function's value approaches infinity or negative infinity as it nears the asymptote.

For the function \( y = \frac{x^3 + 1}{x^2} \), identify the vertical asymptote by setting the denominator equal to zero. Here, since \( x^2 = 0 \) results in \( x = 0 \), you need to confirm that the numerator is not zero for \( x = 0 \). Substituting \( x = 0 \) into the numerator, which is \( (x + 1)(x^2 - x + 1) \), produces non-zero terms, establishing a vertical asymptote at \( x = 0 \).
  • The denominator \( x^2 = 0 \) gives \( x = 0 \).
  • The numerator \( (x + 1)(x^2 - x + 1) eq 0 \) at \( x = 0 \).
  • Therefore, a vertical asymptote exists at \( x = 0 \).
Oblique Asymptote
An oblique (or slant) asymptote appears when the degree of the numerator is exactly one more than the degree of the denominator in a rational function. In such cases, there won't be a horizontal asymptote but instead an oblique line that the graph approaches as \( x \) gets large in magnitude.

To find this oblique asymptote for \( y = \frac{x^3 + 1}{x^2} \), perform polynomial long division. Dividing \( x^3 + 1 \) by \( x^2 \) simplifies the leading term down by one degree, yielding the linear equation \( y = x \). The remainder of this division does not affect the asymptote but helps in sketching the curve's exact path.
  • The numerator is \( x^3 + 1 \).
  • The denominator is \( x^2 \).
  • Dividing yields \( y = x \) with a remainder that diminishes as \( x \) becomes larger.
  • The oblique asymptote is \( y = x \).
Polynomial Long Division
Polynomial long division is a method used to divide one polynomial by another of lower degree. It's much like arithmetic long division but applies to polynomials.

To divide the numerator \( x^3 + 1 \) by the denominator \( x^2 \) in the function \( y = \frac{x^3 + 1}{x^2} \), follow these steps:
  • Divide the leading term of the numerator (\( x^3 \)) by the leading term of the denominator (\( x^2 \)) to get \( x \).
  • Multiply \( x \) by \( x^2 \) and subtract from the original numerator to find the new polynomial.
  • Proceed dividing again until the degree of the remainder is less than the degree of the denominator.
The final result is \( y = x \) with a remainder, which indicates the division outcome. This remainder will factor into the rational function's precise behavior near its oblique asymptote. Remember that this remainder becomes negligible for very large \( |x| \), reinforcing that \( y = x \) is the line the function nears.

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Most popular questions from this chapter

Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises \(45-48 .\) a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$ q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h} $$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$ f(x)=\cos x+4 \sin (2 x), \quad x_{0}=\pi $$

In Exercises \(5-10\) , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$ y=2 \sqrt{x}, \quad(1,2) $$

In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ h(t)=t^{3}, \quad(2,8) $$

A continuous function \(y=f(x)\) is known to be negative at \(x=0\) and positive at \(x=1 .\) Why does the equation \(f(x)=0\) have at least one solution between \(x=0\) and \(x=1 ?\) Illustrate with a sketch.

In Exercises \(41-44,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0 .\) If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=\frac{10^{|x|}-1}{x} $$
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