/*! 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 10 Give the equations of any vertic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. State the domain of \(f .\) $$f(x)=\frac{-6}{x+9}$$

Short Answer

Expert verified
Vertical asymptote at \( x=-9 \), horizontal asymptote at \( y=0 \); domain: all real numbers except \( x = -9 \).

Step by step solution

01

Identify Asymptotes

The given function is \( f(x) = \frac{-6}{x+9} \). The function is rational, written as \( \frac{N(x)}{D(x)} \) with \( N(x) = -6 \) and \( D(x) = x+9 \).For vertical asymptotes, set \( D(x) = 0 \). Therefore, solve \( x+9=0 \).Hence, the vertical asymptote is at \( x = -9 \).For horizontal asymptotes, compare the degrees of the numerator and the denominator. The degree of the numerator is 0 and the degree of the denominator is 1.Therefore, the horizontal asymptote is at \( y = 0 \).There are no oblique asymptotes as the numerator's degree is not one less than the denominator's.
02

Determine Domain

The function \( f(x) = \frac{-6}{x+9} \) is undefined where the denominator is zero, i.e., \( x = -9 \).Thus, the domain of \( f(x) \) includes all real numbers except \( x = -9 \). This can be written as \( \{x \in \mathbb{R} \mid x eq -9\} \).

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.

Vertical Asymptotes
When dealing with rational functions, vertical asymptotes occur where the denominator equals zero, and the function cannot be evaluated. Essentially, these are the places where the function shoots off to infinity. In our specific example, the function given is \( f(x) = \frac{-6}{x+9} \). To find the vertical asymptote, we set the denominator \( x+9 \) equal to zero and solve for \( x \). This gives us:
  • \( x + 9 = 0 \)
  • Thus, \( x = -9 \)
Therefore, there is a vertical asymptote at \( x = -9 \). This means that as \( x \) approaches \( -9 \), the value of \( f(x) \) either increases or decreases without bound. Recognizing vertical asymptotes is crucial for understanding where the function is not defined.
Horizontal Asymptotes
Horizontal asymptotes in rational functions indicate the behavior of the function as the input \( x \) becomes very large, either positively or negatively. These asymptotes tell us what the value of \( f(x) \) will approach as \( x \) reaches infinity. The function \( f(x) = \frac{-6}{x+9} \) has a numerator with a degree of 0 and a denominator with a degree of 1.
  • If the degree of the numerator is less than the degree of the denominator, like here, the horizontal asymptote is always \( y = 0 \).
  • If the degrees are the same, the horizontal asymptote will be the ratio of the leading coefficients.
  • If the numerator's degree is greater, no horizontal asymptote exists. Instead, there might be an oblique asymptote.
In this case, the horizontal asymptote is at \( y = 0 \), meaning the function approaches \( y = 0 \) when \( x \) is very large or very small.
Domain of a Function
To fully comprehend a rational function, understanding its domain is crucial. The domain of a function is essentially all the possible values of \( x \) that you can plug into the function without causing any mathematical errors, such as division by zero. For the function \( f(x) = \frac{-6}{x+9} \), it is important to exclude any \( x \) values that make the denominator zero.
  • Find the values where the denominator is zero: \( x+9 = 0 \).
  • From this, we know that \( x = -9 \) is not in the domain because it would make the function undefined.
Thus, the domain of \( f(x) \) includes all real numbers except \( x = -9 \).This can be formally written as:
  • \( \{x \in \mathbb{R} \mid x eq -9\} \)
Understanding the domain helps avoid undefined points in the function, ensuring you work within the valid range of the function's definition.

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

Solve each problem. Body Mass Index The federal government has devel. oped the body mass index (BMI) to determine ideal weights. A person's BMI is directly proportional to his or her weight in pounds and inversely proportional to the square of his or her height in inches. (A BMI of 19 to 25 corresponds to a healthy weight.) A 6-foot-tall person weighing 177 pounds has a BMI of \(24 .\) Find the BMI (to the nearest whole number) of a person whose weight is 130 pounds and whose height is 66 inches.

If we are given the graph of \(y=f(x)\), we can obtain the graph of \(y=-f(x)\) by reflecting across the \(x\)-axis, and we can obtain the graph of \(y=f(-x)\) by reflecting across the y-axis.You are given the graph of a rational function \(y=f(x)\). Draw \(a\) sketch by hand of the graph of (a) \(y=-f(x)\) and (b) \(y=f(-x)\) (Check your book to see graph)

Solve each problem. Gravity The weight of an object varies inversely with the square of its distance from the center of Earth. The radius of Earth is approximately 4000 miles. If a person weighs 160 pounds on Earth's surface, what would this individual weigh 8000 miles above the surface of Earth?

Solve each equation and inequality. (These types of equations and inequalities occur in calculus.) (a) \(\frac{(x-1)(2 x)-\left(x^{2}\right)(1)}{(x-1)^{2}}=0\) (b) \(\frac{(x-1)(2 x)-\left(x^{2}\right)(1)}{(x-1)^{2}}>0\)

Solve each equation and inequality. (These types of equations and inequalities occur in calculus.) (a) \(\frac{\left(x^{2}+1\right)(2 x)-\left(x^{2}-1\right)(2 x)}{\left(x^{2}+1\right)^{2}}=0\) (b) \(\frac{\left(x^{2}+1\right)(2 x)-\left(x^{2}-1\right)(2 x)}{\left(x^{2}+1\right)^{2}} \geq 0\)

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.