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To determine the end behavior of the function [tex]\( f(x) = 8|x-1| + 13 \)[/tex], let's analyze how the function behaves as [tex]\( x \)[/tex] approaches negative infinity and positive infinity step-by-step.
### Step-by-Step Analysis
1. Understanding the Function:
The function given is [tex]\( f(x) = 8|x-1| + 13 \)[/tex], which consists of an absolute value function [tex]\( |x-1| \)[/tex], multiplied by a constant (8), and then added by another constant (13).
2. Behavior of the Absolute Value Function:
Recall that the absolute value [tex]\( |x-1| \)[/tex] measures the distance of [tex]\( x \)[/tex] from 1, and it is always non-negative. The absolute value function splits into two cases:
- When [tex]\( x \geq 1 \)[/tex], [tex]\( |x-1| = x-1 \)[/tex].
- When [tex]\( x < 1 \)[/tex], [tex]\( |x-1| = 1-x \)[/tex].
3. As [tex]\( x \)[/tex] Approaches Negative Infinity:
- When [tex]\( x \)[/tex] is a very large negative number (i.e., [tex]\( x \)[/tex] is much less than 1), the term [tex]\( |x-1| \)[/tex] simplifies to [tex]\( 1-x \)[/tex] since [tex]\( x < 1 \)[/tex].
- As [tex]\( x \)[/tex] approaches negative infinity, the term [tex]\( 1-x \)[/tex] will grow larger and larger positively. Specifically, [tex]\( 1-x \rightarrow \infty \)[/tex] when [tex]\( x \rightarrow -\infty \)[/tex].
Therefore, [tex]\( 8|x-1| = 8(1-x) \)[/tex] will also tend to infinity since [tex]\( 8 \times \infty = \infty \)[/tex].
Thus, the whole function [tex]\( f(x) = 8|x-1| + 13 \)[/tex] will grow indefinitely because a positive constant added to an infinitely large number is still infinitely large.
Consequently, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] will also approach positive infinity.
4. As [tex]\( x \)[/tex] Approaches Positive Infinity:
- When [tex]\( x \)[/tex] is a very large positive number (i.e., [tex]\( x \)[/tex] is much greater than 1), the term [tex]\( |x-1| \)[/tex] simplifies to [tex]\( x-1 \)[/tex] since [tex]\( x \geq 1 \)[/tex].
- As [tex]\( x \)[/tex] approaches positive infinity, the term [tex]\( x-1 \)[/tex] will grow larger and positively. Specifically, [tex]\( x-1 \rightarrow \infty \)[/tex] when [tex]\( x \rightarrow \infty \)[/tex].
Therefore, [tex]\( 8|x-1| = 8(x-1) \)[/tex] will also tend to infinity since multiplying infinity by 8 still results in infinity.
Thus, the whole function [tex]\( f(x) = 8|x-1| + 13 \)[/tex] will also grow indefinitely as [tex]\( x \)[/tex] approaches positive infinity.
Consequently, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] will also approach positive infinity.
### Conclusion
The correct statement regarding the end behavior of the function [tex]\( f(x) = 8|x-1| + 13 \)[/tex] is:
B. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
So, the answer is:
B. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
### Step-by-Step Analysis
1. Understanding the Function:
The function given is [tex]\( f(x) = 8|x-1| + 13 \)[/tex], which consists of an absolute value function [tex]\( |x-1| \)[/tex], multiplied by a constant (8), and then added by another constant (13).
2. Behavior of the Absolute Value Function:
Recall that the absolute value [tex]\( |x-1| \)[/tex] measures the distance of [tex]\( x \)[/tex] from 1, and it is always non-negative. The absolute value function splits into two cases:
- When [tex]\( x \geq 1 \)[/tex], [tex]\( |x-1| = x-1 \)[/tex].
- When [tex]\( x < 1 \)[/tex], [tex]\( |x-1| = 1-x \)[/tex].
3. As [tex]\( x \)[/tex] Approaches Negative Infinity:
- When [tex]\( x \)[/tex] is a very large negative number (i.e., [tex]\( x \)[/tex] is much less than 1), the term [tex]\( |x-1| \)[/tex] simplifies to [tex]\( 1-x \)[/tex] since [tex]\( x < 1 \)[/tex].
- As [tex]\( x \)[/tex] approaches negative infinity, the term [tex]\( 1-x \)[/tex] will grow larger and larger positively. Specifically, [tex]\( 1-x \rightarrow \infty \)[/tex] when [tex]\( x \rightarrow -\infty \)[/tex].
Therefore, [tex]\( 8|x-1| = 8(1-x) \)[/tex] will also tend to infinity since [tex]\( 8 \times \infty = \infty \)[/tex].
Thus, the whole function [tex]\( f(x) = 8|x-1| + 13 \)[/tex] will grow indefinitely because a positive constant added to an infinitely large number is still infinitely large.
Consequently, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] will also approach positive infinity.
4. As [tex]\( x \)[/tex] Approaches Positive Infinity:
- When [tex]\( x \)[/tex] is a very large positive number (i.e., [tex]\( x \)[/tex] is much greater than 1), the term [tex]\( |x-1| \)[/tex] simplifies to [tex]\( x-1 \)[/tex] since [tex]\( x \geq 1 \)[/tex].
- As [tex]\( x \)[/tex] approaches positive infinity, the term [tex]\( x-1 \)[/tex] will grow larger and positively. Specifically, [tex]\( x-1 \rightarrow \infty \)[/tex] when [tex]\( x \rightarrow \infty \)[/tex].
Therefore, [tex]\( 8|x-1| = 8(x-1) \)[/tex] will also tend to infinity since multiplying infinity by 8 still results in infinity.
Thus, the whole function [tex]\( f(x) = 8|x-1| + 13 \)[/tex] will also grow indefinitely as [tex]\( x \)[/tex] approaches positive infinity.
Consequently, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] will also approach positive infinity.
### Conclusion
The correct statement regarding the end behavior of the function [tex]\( f(x) = 8|x-1| + 13 \)[/tex] is:
B. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
So, the answer is:
B. As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
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