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Consider the end behavior of the function [tex]$g(x)=4|x-2|-3$[/tex].

As [tex]$x$[/tex] approaches negative infinity, [tex][tex]$g(x)$[/tex][/tex] approaches [tex]\square[/tex] infinity. As [tex]$x$[/tex] approaches positive infinity, [tex]$g(x)$[/tex] approaches [tex]\square[/tex] infinity.

Sagot :

GinnyAnsweridn
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.
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