Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
How do I evalute the horizontal and vertical asymptotes of this equation?
![The given equation is:
\[ y = \frac{x^2 - 1}{2x^2 - 2x} \]
This equation represents a rational function where \( y \) is expressed as the ratio of two polynomials in \( x \). The numerator of the fraction is \( x^2 - 1 \) and the denominator is \( 2x^2 - 2x \).
To further understand this function, we can analyze its components:
1. **Numerator**: \( x^2 - 1 \)
- This is a quadratic expression that can be factored as \( (x + 1)(x - 1) \).
2. **Denominator**: \( 2x^2 - 2x \)
- This can also be factored by taking out a common factor of 2, resulting in \( 2(x^2 - x) = 2x(x - 1) \).
With these factorizations, the function can be rewritten as:
\[ y = \frac{(x + 1)(x - 1)}{2x(x - 1)} \]
We can simplify further by canceling the common factor \((x - 1)\) from the numerator and the denominator, given \( x \neq 1 \):
\[ y = \frac{x + 1}{2x} \quad \text{for} \quad x \neq 1 \]
Thus, the simplified form of the function is:
\[ y = \frac{x + 1}{2x} \]
In this simplified form, it is easier to analyze the behavior of \( y \) in terms of \( x \). It's important to note the excluded value \( x \neq 1 \) because at \( x = 1 \), the original denominator becomes zero, which makes the function undefined at that point.
### Graphical Representation:
If provided, the plot of this function will show:
- A vertical asymptote at \( x = 1 \) due to the excluded value.
- Analyzing \( y = \frac{x + 1}{2x} \):
- Horizontal asymptote as \( x \) approaches \( \infty \) will be \( y = \frac{1}{2} \).
- The function will approach \( y = 0 \) as \( x \) approaches negative or positive infinity.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4b94795-d280-48fd-aba6-be8468932522%2F781f6287-64c8-419c-b3bf-497ffd648d9f%2Fbasjh7.png&w=3840&q=75)
Transcribed Image Text:The given equation is:
\[ y = \frac{x^2 - 1}{2x^2 - 2x} \]
This equation represents a rational function where \( y \) is expressed as the ratio of two polynomials in \( x \). The numerator of the fraction is \( x^2 - 1 \) and the denominator is \( 2x^2 - 2x \).
To further understand this function, we can analyze its components:
1. **Numerator**: \( x^2 - 1 \)
- This is a quadratic expression that can be factored as \( (x + 1)(x - 1) \).
2. **Denominator**: \( 2x^2 - 2x \)
- This can also be factored by taking out a common factor of 2, resulting in \( 2(x^2 - x) = 2x(x - 1) \).
With these factorizations, the function can be rewritten as:
\[ y = \frac{(x + 1)(x - 1)}{2x(x - 1)} \]
We can simplify further by canceling the common factor \((x - 1)\) from the numerator and the denominator, given \( x \neq 1 \):
\[ y = \frac{x + 1}{2x} \quad \text{for} \quad x \neq 1 \]
Thus, the simplified form of the function is:
\[ y = \frac{x + 1}{2x} \]
In this simplified form, it is easier to analyze the behavior of \( y \) in terms of \( x \). It's important to note the excluded value \( x \neq 1 \) because at \( x = 1 \), the original denominator becomes zero, which makes the function undefined at that point.
### Graphical Representation:
If provided, the plot of this function will show:
- A vertical asymptote at \( x = 1 \) due to the excluded value.
- Analyzing \( y = \frac{x + 1}{2x} \):
- Horizontal asymptote as \( x \) approaches \( \infty \) will be \( y = \frac{1}{2} \).
- The function will approach \( y = 0 \) as \( x \) approaches negative or positive infinity.
Expert Solution
Step 1
Evaluate the horizontal and vertical asymptotes of the equation.
Given:
Find the horizontal asymptotes
both the numerator and denominator have same degree,
therefore balance section. and the horizontal asymptotes is ratio of coefficient of balance degree of x.
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