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
#12 can you show me how to do this right? I’ve attached my work.
![**Problem Statement:**
Find all horizontal and vertical asymptotes (if any). (If an answer does not exist, enter DNE.)
\[
s(x) = \frac{16x^2 + 1}{4x^2 + 2x - 6}
\]
**Answers Provided:**
- Vertical asymptote: \( x = 1 \) (smaller value) ❌
- Vertical asymptote: \( x = -\frac{6}{4} \) (larger value) ❌
- Horizontal asymptote: \( y = 16 \) ❌
**Note:**
The problem requires identifying the correct horizontal and vertical asymptotes of the given rational function. The incorrect answers are marked with red crosses (❌).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f440a4a-a87c-43e8-9c30-e4890b6a6760%2Feae68dfd-725f-4f63-9bdf-c53bab29e88a%2F1ba3npj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find all horizontal and vertical asymptotes (if any). (If an answer does not exist, enter DNE.)
\[
s(x) = \frac{16x^2 + 1}{4x^2 + 2x - 6}
\]
**Answers Provided:**
- Vertical asymptote: \( x = 1 \) (smaller value) ❌
- Vertical asymptote: \( x = -\frac{6}{4} \) (larger value) ❌
- Horizontal asymptote: \( y = 16 \) ❌
**Note:**
The problem requires identifying the correct horizontal and vertical asymptotes of the given rational function. The incorrect answers are marked with red crosses (❌).
![The problem presented is:
\[ s(x) = \frac{16x^2 + 1}{4x^2 + 2x - 6} \]
**Solution Steps:**
1. **Identify and Factor the Denominator:**
- The task is to simplify the expression, starting by factoring the denominator: \( 4x^2 + 2x - 6 \).
2. **Factorization Steps:**
- Decompose into simpler expressions using cross multiplication indications:
- Write down the product \( ac = 24 \) (from \( a = 4, c = -6 \)).
- Find the factors of 24 which sum up to \( b = 2 \). These are \(-4\) and \(6\).
3. **Intermediate Steps:**
- Break the middle term using these factors:
\[
4x^2 - 4x + 6x - 6
\]
- Factor by grouping:
\[
4x(x-1) + 6(x-1)
\]
- Combine common factors:
\[
(x-1)(4x+6)
\]
4. **Vertical Asymptote (VA) Identification:**
- Set the denominator equal to zero to find the vertical asymptotes.
- \( x - 1 = 0 \) results in a vertical asymptote at \( x = 1 \).
5. **Simplifying by Solving Remainder Equation:**
- Solve the equation \( 4x + 6 = 0 \):
\[
4x = -6
\]
- Simplify to find:
\[
x = -\frac{6}{4} = -\frac{3}{2}
\]
**Conclusion:**
The function has a vertical asymptote at \( x = 1 \). Further simplification might include reducing or canceling terms, if possible, depending on the form of \( s(x) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f440a4a-a87c-43e8-9c30-e4890b6a6760%2Feae68dfd-725f-4f63-9bdf-c53bab29e88a%2F4dsc04r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The problem presented is:
\[ s(x) = \frac{16x^2 + 1}{4x^2 + 2x - 6} \]
**Solution Steps:**
1. **Identify and Factor the Denominator:**
- The task is to simplify the expression, starting by factoring the denominator: \( 4x^2 + 2x - 6 \).
2. **Factorization Steps:**
- Decompose into simpler expressions using cross multiplication indications:
- Write down the product \( ac = 24 \) (from \( a = 4, c = -6 \)).
- Find the factors of 24 which sum up to \( b = 2 \). These are \(-4\) and \(6\).
3. **Intermediate Steps:**
- Break the middle term using these factors:
\[
4x^2 - 4x + 6x - 6
\]
- Factor by grouping:
\[
4x(x-1) + 6(x-1)
\]
- Combine common factors:
\[
(x-1)(4x+6)
\]
4. **Vertical Asymptote (VA) Identification:**
- Set the denominator equal to zero to find the vertical asymptotes.
- \( x - 1 = 0 \) results in a vertical asymptote at \( x = 1 \).
5. **Simplifying by Solving Remainder Equation:**
- Solve the equation \( 4x + 6 = 0 \):
\[
4x = -6
\]
- Simplify to find:
\[
x = -\frac{6}{4} = -\frac{3}{2}
\]
**Conclusion:**
The function has a vertical asymptote at \( x = 1 \). Further simplification might include reducing or canceling terms, if possible, depending on the form of \( s(x) \).
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