Let's analyze the end behavior of the function [tex]\( g(x) = 4|x-2| - 3 \)[/tex].
### As [tex]\( x \)[/tex] approaches negative infinity:
1. Consider [tex]\( |x-2| \)[/tex] for [tex]\( x \)[/tex] approaching a very large negative value.
2. When [tex]\( x \)[/tex] is very negative, [tex]\( x-2 \)[/tex] is also very negative. Thus, [tex]\( |x-2| \)[/tex] becomes the positive equivalent of a very large negative number.
3. Therefore, [tex]\( |x-2| \)[/tex] will grow without bound.
4. Multiplying this by 4, we get [tex]\( 4|x-2| \)[/tex] which will also grow without bound.
5. Subtracting 3 from a very large number will still result in a very large number.
6. Hence, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
### As [tex]\( x \)[/tex] approaches positive infinity:
1. Consider [tex]\( |x-2| \)[/tex] for [tex]\( x \)[/tex] approaching a very large positive value.
2. When [tex]\( x \)[/tex] is very positive, [tex]\( x-2 \)[/tex] is also very positive. Thus, [tex]\( |x-2| \)[/tex] remains very large.
3. Therefore, [tex]\( |x-2| \)[/tex] will grow without bound.
4. Multiplying this by 4, we get [tex]\( 4|x-2| \)[/tex] which will also grow without bound.
5. Subtracting 3 from a very large number will still result in a very large number.
6. Hence, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( g(x) \)[/tex] approaches positive infinity.
So, we fill in the blanks as follows:
- 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.
