Answer :
Sure, let's break down the problem step by step to understand which piecewise definition matches the function [tex]\( y = |x + 5| - 2 \)[/tex].
The function [tex]\( y = |x + 5| - 2 \)[/tex] involves the absolute value expression [tex]\( |x + 5| \)[/tex]. The absolute value function can be split into two cases based on the value of [tex]\( x \)[/tex].
1. When [tex]\( x + 5 \geq 0 \)[/tex] (which simplifies to [tex]\( x \geq -5 \)[/tex]), the absolute value function [tex]\( |x + 5| \)[/tex] is simply [tex]\( x + 5 \)[/tex].
2. When [tex]\( x + 5 < 0 \)[/tex] (which simplifies to [tex]\( x < -5 \)[/tex]), the absolute value function [tex]\( |x + 5| \)[/tex] is [tex]\( -(x + 5) \)[/tex].
Let's define the function for these two cases:
### Case 1: [tex]\( x \geq -5 \)[/tex]
When [tex]\( x \geq -5 \)[/tex]:
[tex]\[ |x + 5| = x + 5 \][/tex]
So the function [tex]\( y \)[/tex] becomes:
[tex]\[ y = (x + 5) - 2 \][/tex]
[tex]\[ y = x + 3 \][/tex]
### Case 2: [tex]\( x < -5 \)[/tex]
When [tex]\( x < -5 \)[/tex]:
[tex]\[ |x + 5| = -(x + 5) = -x - 5 \][/tex]
So the function [tex]\( y \)[/tex] becomes:
[tex]\[ y = (-x - 5) - 2 \][/tex]
[tex]\[ y = -x - 7 \][/tex]
Now, combining both cases, we get the piecewise function:
[tex]\[ y = \begin{cases}
x + 3 & \text{if } x \geq -5 \\
-x - 7 & \text{if } x < -5
\end{cases} \][/tex]
So, the correct piecewise definition is:
[tex]\[ y = x + 3 \text{ for } x \geq -5 \text{ and } y = -x - 7 \text{ for } x < -5 \][/tex]
This matches option 3:
[tex]\[ y = x + 3 \text{ for } x \geq -5 \text{ and } y = -x - 7 \text{ for } x < -5 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{3} \][/tex]
The function [tex]\( y = |x + 5| - 2 \)[/tex] involves the absolute value expression [tex]\( |x + 5| \)[/tex]. The absolute value function can be split into two cases based on the value of [tex]\( x \)[/tex].
1. When [tex]\( x + 5 \geq 0 \)[/tex] (which simplifies to [tex]\( x \geq -5 \)[/tex]), the absolute value function [tex]\( |x + 5| \)[/tex] is simply [tex]\( x + 5 \)[/tex].
2. When [tex]\( x + 5 < 0 \)[/tex] (which simplifies to [tex]\( x < -5 \)[/tex]), the absolute value function [tex]\( |x + 5| \)[/tex] is [tex]\( -(x + 5) \)[/tex].
Let's define the function for these two cases:
### Case 1: [tex]\( x \geq -5 \)[/tex]
When [tex]\( x \geq -5 \)[/tex]:
[tex]\[ |x + 5| = x + 5 \][/tex]
So the function [tex]\( y \)[/tex] becomes:
[tex]\[ y = (x + 5) - 2 \][/tex]
[tex]\[ y = x + 3 \][/tex]
### Case 2: [tex]\( x < -5 \)[/tex]
When [tex]\( x < -5 \)[/tex]:
[tex]\[ |x + 5| = -(x + 5) = -x - 5 \][/tex]
So the function [tex]\( y \)[/tex] becomes:
[tex]\[ y = (-x - 5) - 2 \][/tex]
[tex]\[ y = -x - 7 \][/tex]
Now, combining both cases, we get the piecewise function:
[tex]\[ y = \begin{cases}
x + 3 & \text{if } x \geq -5 \\
-x - 7 & \text{if } x < -5
\end{cases} \][/tex]
So, the correct piecewise definition is:
[tex]\[ y = x + 3 \text{ for } x \geq -5 \text{ and } y = -x - 7 \text{ for } x < -5 \][/tex]
This matches option 3:
[tex]\[ y = x + 3 \text{ for } x \geq -5 \text{ and } y = -x - 7 \text{ for } x < -5 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{3} \][/tex]
