Answer:
From top to bottom:
![0, 1, 2, -8, -8](https://tex.z-dn.net/?f=0%2C%201%2C%202%2C%20-8%2C%20-8)
Step-by-step explanation:
We are given the piecewise function:
![g(x) = \left\{ \begin{array}{ll} 0.5x+1 & \text{if } x \leq 2 \\ -8 & \text{if } x > 2 \end{array} \right.](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Cleft%5C%7B%20%20%20%20%20%20%20%20%5Cbegin%7Barray%7D%7Bll%7D%20%20%20%20%20%20%20%20%20%20%20%200.5x%2B1%20%26%20%5Ctext%7Bif%20%7D%20x%20%5Cleq%202%20%5C%5C%20%20%20%20%20%20%20%20%20%20%20%20-8%20%26%20%5Ctext%7Bif%20%7D%20x%20%3E%202%20%20%20%20%20%20%20%20%5Cend%7Barray%7D%20%20%20%20%5Cright.)
Row 1:
We want to find g(-2).
Since -2 is less than (or equal to) 2, we will use the first equation. Thus:
![g(-2)=0.5(-2)+1=-1+1=0](https://tex.z-dn.net/?f=g%28-2%29%3D0.5%28-2%29%2B1%3D-1%2B1%3D0)
Row 2:
Likewise, 0 is less than or equal to 2. We will continue to use the first equation:
![g(0)=0.5(0)+1=1](https://tex.z-dn.net/?f=g%280%29%3D0.5%280%29%2B1%3D1)
Row 3:
2 is not less than 2 but it is equal to 2. So we will continue to use the first equation:
![g(2)=0.5(2)+1=1+1=2](https://tex.z-dn.net/?f=g%282%29%3D0.5%282%29%2B1%3D1%2B1%3D2)
Row 4 and 5:
Both 4 and 6 are greater than 2. Thus, we will use the second equation. Therefore:
![g(4)=-8\text{ and } g(6)=-8](https://tex.z-dn.net/?f=g%284%29%3D-8%5Ctext%7B%20and%20%7D%20g%286%29%3D-8)