Determine the end behavior for f(x) = = As x →→∞, f(x) → −1 As x → ∞, f(x) → −1 As x →→∞, f(x) → 0 As x → ∞, f(x) → 0 - As x→→∞, f(x) → 2 As x →∞, f(x) → 2 As x → −∞, ƒ(x) → 1 As x →∞, f(x) → 1 2x²+8 x²-x-6
Determine the end behavior for f(x) = = As x →→∞, f(x) → −1 As x → ∞, f(x) → −1 As x →→∞, f(x) → 0 As x → ∞, f(x) → 0 - As x→→∞, f(x) → 2 As x →∞, f(x) → 2 As x → −∞, ƒ(x) → 1 As x →∞, f(x) → 1 2x²+8 x²-x-6
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Title:** Determine the End Behavior for \( f(x) = \frac{2x^2 + 8}{x^2 - x - 6} \)
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**Question:**
Determine the end behavior for \( f(x) = \frac{2x^2 + 8}{x^2 - x - 6} \)
---
**Options:**
1. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow -1 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow -1 \)
2. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow 0 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow 0 \)
3. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow 2 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow 2 \)
4. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow 1 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow 1 \)
---
**Explanation:**
To determine the end behavior of the function \( f(x) = \frac{2x^2 + 8}{x^2 - x - 6} \), we need to look at the degrees of the polynomials in the numerator and the denominator.
- The degree of the numerator polynomial \( 2x^2 + 8 \) is 2.
- The degree of the denominator polynomial \( x^2 - x - 6 \) is also 2.
When the degrees of the numerator and the denominator are the same, the end behavior of the function is determined by the ratio of the leading coefficients:
\[
\text{End behavior} = \frac{\text{Leading coefficient of numerator}}{\text{Leading coefficient of denominator}} = \frac{2}{1} = 2
\]
Thus, as \( x \rightarrow \infty \) and as \( x \rightarrow -\infty \), the function \( f(x) \) approaches 2. Therefore, the correct answer is:
\( \bigcirc \) As](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f34e597-e324-4f60-b87e-69201cdf9380%2Ff162e80e-d12c-4744-8b6b-4530bb282a40%2Fijk6new_processed.png&w=3840&q=75)
Transcribed Image Text:**Title:** Determine the End Behavior for \( f(x) = \frac{2x^2 + 8}{x^2 - x - 6} \)
---
**Question:**
Determine the end behavior for \( f(x) = \frac{2x^2 + 8}{x^2 - x - 6} \)
---
**Options:**
1. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow -1 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow -1 \)
2. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow 0 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow 0 \)
3. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow 2 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow 2 \)
4. \( \bigcirc \) As \( x \rightarrow -\infty \), \( f(x) \rightarrow 1 \)
As \( x \rightarrow \infty \), \( f(x) \rightarrow 1 \)
---
**Explanation:**
To determine the end behavior of the function \( f(x) = \frac{2x^2 + 8}{x^2 - x - 6} \), we need to look at the degrees of the polynomials in the numerator and the denominator.
- The degree of the numerator polynomial \( 2x^2 + 8 \) is 2.
- The degree of the denominator polynomial \( x^2 - x - 6 \) is also 2.
When the degrees of the numerator and the denominator are the same, the end behavior of the function is determined by the ratio of the leading coefficients:
\[
\text{End behavior} = \frac{\text{Leading coefficient of numerator}}{\text{Leading coefficient of denominator}} = \frac{2}{1} = 2
\]
Thus, as \( x \rightarrow \infty \) and as \( x \rightarrow -\infty \), the function \( f(x) \) approaches 2. Therefore, the correct answer is:
\( \bigcirc \) As
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