/*! 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 71 Make a hand-drawn graph. Be sure... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 71

Make a hand-drawn graph. Be sure to label all the asymptotes. List the domain and the \(x\) -intercepts and the \(y\) -intercepts. Check your work using \(a\) graphing calculator. $$f(x)=\frac{x^{3}+1}{x}$$

Short Answer

Expert verified
Vertical asymptote at x=0. No y-intercept. x-intercept at -1. Domain: (-鈭�,0) 鈭� (0,鈭�).

Step by step solution

01

- Identify the function

The function given is \( f(x) = \frac{x^3 + 1}{x} \). Let's rewrite it to simplify: \( f(x) = x^2 + \frac{1}{x} \).
02

- Find the asymptotes

To find the vertical asymptotes, set the denominator equal to zero: \( x = 0 \). There is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator. However, there is an oblique asymptote: divide \( x^3 + 1 \) by \( x \), which yields \( x^2 \). So, the oblique asymptote is \( y = x^2 \).
03

- Determine the intercepts

For the y-intercept, set \( x = 0 \). Because \( \frac{1}{0} \) is undefined, there is no y-intercept. For the x-intercepts, set the numerator equal to zero: \( x^3 + 1 = 0 \). This factors to \( (x+1)(x^2-x+1) = 0 \). Thus, the x-intercept is \( x = -1 \).
04

- Define the domain

The domain is all real numbers except where the function is undefined. Since \( x \) cannot be zero: \( \text{Domain} = (-\infty, 0) \cup (0, \infty) \).
05

- Draw the graph

Draw the graph of \( f(x) = x^2 + \frac{1}{x} \) by hand. Mark the vertical asymptote at \( x = 0 \) and the oblique asymptote along \( y = x^2 \). Label the x-intercept at \( (-1, 0) \).
06

- Verify with a graphing calculator

Use a graphing calculator to plot \( f(x) = x^2 + \frac{1}{x} \) and check that your hand-drawn graph matches the calculator's output. Ensure all asymptotes and intercepts match.

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Key Concepts

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

vertical asymptote
Understanding vertical asymptotes is crucial in precalculus graphing. A vertical asymptote occurs where the function approaches infinity as the input approaches a specific value. It happens where the denominator of a rational function equals zero, making the function undefined.

For the function given, \( f(x) = \frac{x^3 + 1}{x} \), the denominator is \( x \). Setting \( x \) to zero indicates the vertical asymptote at \( x = 0 \). Graphically, this means the function has a value approaching positive or negative infinity as \( x \) approaches zero from either side.

Always remember:
  • Vertical asymptotes indicate undefined points.
  • They're a key part of sketching rational functions.
oblique asymptote
An oblique asymptote, also known as a slant asymptote, appears when the degree of the polynomial in the numerator is exactly one greater than the degree of the polynomial in the denominator.

In our function \( f(x) = \frac{x^3 + 1}{x} \), the degree of the numerator (3) is one more than the degree of the denominator (1). To find the oblique asymptote, we perform polynomial division:

\[ \frac{x^3 + 1}{x} = x^2 + \frac{1}{x} \]

This yields \[ y = x^2 \], representing the oblique asymptote. This indicates that as \( x \) approaches positive or negative infinity, the function \( f(x) \) behaves like \( y = x^2 \).

Key points:
  • Oblique asymptotes occur only with specific polynomials.
  • They aid in understanding the end behavior of graphs.
domain and intercepts
Domains and intercepts provide valuable insight into the behavior of functions.

For the given function \( f(x) = \frac{x^3 + 1}{x}\), the domain excludes values that make the denominator zero. Therefore, the domain is:

\(\text{Domain} = (-\infty, 0) \cup (0, \infty) \)\br>
Intercepts are where the function crosses the axes:
  • Y-intercept: Set \( x = 0 \). But \( \frac{1}{0} \) is undefined, so there's no y-intercept.
  • X-intercepts: Set the numerator to zero: \( x^3 + 1 = 0\). This factors to \((x + 1)(x^2 - x + 1) = 0 \), yielding \( x = -1 \) as the x-intercept.
Always remember:
  • Domain indicates possible input values.
  • X and y-intercepts show how functions cross the axes.
graphing calculator
Using a graphing calculator can verify hand-drawn graphs and help visualize functions more accurately.

Here are steps to use it for our function \( f(x) = x^2 + \frac{1}{x} \):
  • Enter the function: Input \( f(x) \) into the calculator.
  • View the graph: Observe the plotted graph on the screen.
  • Check asymptotes and intercepts: Confirm the vertical asymptote at \( x = 0 \), the oblique asymptote \( y = x^2 \), and the x-intercept at \( x = -1 \).
A graphing calculator:
  • Provides a visual aid for complex functions.
  • Helps in verifying manual calculations.
Whether learning or teaching, it is a powerful tool to ensure understanding and accuracy in precalculus.

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