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Sagot :
Let’s analyze the end behavior of the function [tex]\( g(x) = \frac{x^2 + 6x}{30} \)[/tex].
Step-by-Step Analysis:
1. Understanding the Function: We have [tex]\( g(x) = \frac{x^2 + 6x}{30} \)[/tex].
2. Leading Term Dominance: For large values of [tex]\( x \)[/tex] (either very large positive or very large negative), the [tex]\( x^2 \)[/tex] term in the numerator will dominate over the [tex]\( 6x \)[/tex] term since [tex]\( x^2 \)[/tex] grows much faster than [tex]\( x \)[/tex].
3. End Behavior as [tex]\( x \to \infty \)[/tex]:
- As [tex]\( x \)[/tex] becomes very large, [tex]\( x^2 \)[/tex] will dominate, and [tex]\( g(x) \)[/tex] will behave approximately like [tex]\( \frac{x^2}{30} \)[/tex].
- Thus, as [tex]\( x \to \infty \)[/tex], [tex]\( \frac{x^2}{30} \)[/tex] increases without bound.
- Therefore, [tex]\( g(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex].
4. End Behavior as [tex]\( x \to -\infty \)[/tex]:
- Similarly, for very large negative values of [tex]\( x \)[/tex], [tex]\( x^2 \)[/tex] still dominates because [tex]\( (-x)^2 = x^2 \)[/tex].
- So, [tex]\( g(x) \)[/tex] will again behave like [tex]\( \frac{x^2}{30} \)[/tex].
- Therefore, as [tex]\( x \to -\infty \)[/tex], [tex]\( \frac{x^2}{30} \)[/tex] increases without bound.
- Thus, [tex]\( g(x) \to \infty \)[/tex] as [tex]\( x \to -\infty \)[/tex].
From this analysis, we can conclude that as [tex]\( x \)[/tex] approaches either [tex]\( -\infty \)[/tex] or [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
Selecting Correct Statement:
Given these conclusions, the correct statement is:
- As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex]; and as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
However, this exact match is not provided in the options. The closest option, aligning to our conclusion, is the interpretation of the positive bound behavior:
The closest correct option seems to be:
- As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] or [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
Which aligns with one intermediate result as:
So, reflecting on the best available correct behavior:
Correct Answer: As [tex]\( x \)[/tex] approaches [tex]\( -\infty, g(x) \)[/tex] approaches [tex]\( \infty \)[/tex]; and as [tex]\( x \)[/tex] approaches [tex]\( \infty, g(x) approaches \infty \)[/tex].
Step-by-Step Analysis:
1. Understanding the Function: We have [tex]\( g(x) = \frac{x^2 + 6x}{30} \)[/tex].
2. Leading Term Dominance: For large values of [tex]\( x \)[/tex] (either very large positive or very large negative), the [tex]\( x^2 \)[/tex] term in the numerator will dominate over the [tex]\( 6x \)[/tex] term since [tex]\( x^2 \)[/tex] grows much faster than [tex]\( x \)[/tex].
3. End Behavior as [tex]\( x \to \infty \)[/tex]:
- As [tex]\( x \)[/tex] becomes very large, [tex]\( x^2 \)[/tex] will dominate, and [tex]\( g(x) \)[/tex] will behave approximately like [tex]\( \frac{x^2}{30} \)[/tex].
- Thus, as [tex]\( x \to \infty \)[/tex], [tex]\( \frac{x^2}{30} \)[/tex] increases without bound.
- Therefore, [tex]\( g(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex].
4. End Behavior as [tex]\( x \to -\infty \)[/tex]:
- Similarly, for very large negative values of [tex]\( x \)[/tex], [tex]\( x^2 \)[/tex] still dominates because [tex]\( (-x)^2 = x^2 \)[/tex].
- So, [tex]\( g(x) \)[/tex] will again behave like [tex]\( \frac{x^2}{30} \)[/tex].
- Therefore, as [tex]\( x \to -\infty \)[/tex], [tex]\( \frac{x^2}{30} \)[/tex] increases without bound.
- Thus, [tex]\( g(x) \to \infty \)[/tex] as [tex]\( x \to -\infty \)[/tex].
From this analysis, we can conclude that as [tex]\( x \)[/tex] approaches either [tex]\( -\infty \)[/tex] or [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
Selecting Correct Statement:
Given these conclusions, the correct statement is:
- As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex]; and as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
However, this exact match is not provided in the options. The closest option, aligning to our conclusion, is the interpretation of the positive bound behavior:
The closest correct option seems to be:
- As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] or [tex]\( \infty \)[/tex], [tex]\( g(x) \)[/tex] approaches [tex]\( \infty \)[/tex].
Which aligns with one intermediate result as:
So, reflecting on the best available correct behavior:
Correct Answer: As [tex]\( x \)[/tex] approaches [tex]\( -\infty, g(x) \)[/tex] approaches [tex]\( \infty \)[/tex]; and as [tex]\( x \)[/tex] approaches [tex]\( \infty, g(x) approaches \infty \)[/tex].
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