Answer :
Consider the function [tex]\( g(x) = 4|x - 2| - 3 \)[/tex].
We need to observe the behavior of [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches negative infinity and positive infinity.
1. As [tex]\( x \)[/tex] approaches negative infinity:
- When [tex]\( x \)[/tex] is a very large negative number, [tex]\( x - 2 \)[/tex] is also very large in the negative direction.
- Therefore, [tex]\( |x - 2| = -(x - 2) \)[/tex], since the absolute value converts it to a positive number.
- So, [tex]\( g(x) = 4|x - 2| - 3 \)[/tex] becomes [tex]\( g(x) = 4(-(x - 2)) - 3 = 4(-x + 2) - 3 = -4x + 8 - 3 = -4x + 5 \)[/tex].
Given that [tex]\( x \)[/tex] is approaching negative infinity, [tex]\( -4x \)[/tex] will become a very large positive number.
Combining this result, [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches negative infinity will indeed go to positive infinity. Therefore:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
2. As [tex]\( x \)[/tex] approaches positive infinity:
- When [tex]\( x \)[/tex] is a very large positive number, [tex]\( x - 2 \)[/tex] is also very large in the positive direction.
- In this case, [tex]\( |x - 2| = x - 2 \)[/tex] because it's already a positive value.
- Thus, [tex]\( g(x) = 4|x - 2| - 3 \)[/tex] becomes [tex]\( g(x) = 4(x - 2) - 3 = 4x - 8 - 3 = 4x - 11 \)[/tex].
Given that [tex]\( x \)[/tex] is approaching positive infinity, [tex]\( 4x \)[/tex] will become a very large positive number.
Combining this result, [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive infinity will indeed go to positive infinity. Therefore:
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
Putting it all together, we complete the statement:
Consider the end behavior of the function [tex]\( g(x) = 4|x - 2| - 3 \)[/tex].
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
We need to observe the behavior of [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches negative infinity and positive infinity.
1. As [tex]\( x \)[/tex] approaches negative infinity:
- When [tex]\( x \)[/tex] is a very large negative number, [tex]\( x - 2 \)[/tex] is also very large in the negative direction.
- Therefore, [tex]\( |x - 2| = -(x - 2) \)[/tex], since the absolute value converts it to a positive number.
- So, [tex]\( g(x) = 4|x - 2| - 3 \)[/tex] becomes [tex]\( g(x) = 4(-(x - 2)) - 3 = 4(-x + 2) - 3 = -4x + 8 - 3 = -4x + 5 \)[/tex].
Given that [tex]\( x \)[/tex] is approaching negative infinity, [tex]\( -4x \)[/tex] will become a very large positive number.
Combining this result, [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches negative infinity will indeed go to positive infinity. Therefore:
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
2. As [tex]\( x \)[/tex] approaches positive infinity:
- When [tex]\( x \)[/tex] is a very large positive number, [tex]\( x - 2 \)[/tex] is also very large in the positive direction.
- In this case, [tex]\( |x - 2| = x - 2 \)[/tex] because it's already a positive value.
- Thus, [tex]\( g(x) = 4|x - 2| - 3 \)[/tex] becomes [tex]\( g(x) = 4(x - 2) - 3 = 4x - 8 - 3 = 4x - 11 \)[/tex].
Given that [tex]\( x \)[/tex] is approaching positive infinity, [tex]\( 4x \)[/tex] will become a very large positive number.
Combining this result, [tex]\( g(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive infinity will indeed go to positive infinity. Therefore:
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
Putting it all together, we complete the statement:
Consider the end behavior of the function [tex]\( g(x) = 4|x - 2| - 3 \)[/tex].
As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.