1. **State the problem:** We have a piecewise function $g(x)$ defined on several intervals with some missing parts. We need to: - a. Complete the rule for $g(x)$. - b. Find $g(1)$, $g(-12)$, and $g(15)$. - c. Find the $x$-values where $g(x) = -6$. 2. **Given function:** $$ g(x) = \begin{cases} -10, & -15 \leq x < -10 \\ -6, & -10 \leq x < -8 \\ \_, & \_ \leq x < -1 \\ \_, & -1 \leq x < 1 \\ 4, & \_ \leq x < 10 \\ 8, & 10 \leq x < 15 \end{cases} $$ 3. **Complete the missing intervals and values:** From the graph description and intervals, the missing intervals are: - For $-8 \leq x < -1$, $g(x) = -2$ (assuming from graph horizontal segment between $-8$ and $-1$). - For $-1 \leq x < 1$, $g(x) = 0$ (assuming from graph horizontal segment between $-1$ and $1$). - For $1 \leq x < 10$, $g(x) = 4$ (given the next interval after $1$ up to $10$). So the completed function is: $$ g(x) = \begin{cases} -10, & -15 \leq x < -10 \\ -6, & -10 \leq x < -8 \\ -2, & -8 \leq x < -1 \\ 0, & -1 \leq x < 1 \\ 4, & 1 \leq x < 10 \\ 8, & 10 \leq x < 15 \end{cases} $$ 4. **Evaluate $g(1)$, $g(-12)$, and $g(15)$:** - $g(1)$: Since $1 \leq x < 10$ corresponds to $g(x) = 4$, so $g(1) = 4$. - $g(-12)$: Since $-15 \leq x < -10$ corresponds to $g(x) = -10$, so $g(-12) = -10$. - $g(15)$: The function is defined only up to $x < 15$, so $g(15)$ is undefined. 5. **Find $x$-values where $g(x) = -6$:** From the piecewise definition, $g(x) = -6$ on the interval $-10 \leq x < -8$. **Final answers:** - a. Completed rule as above. - b. $g(1) = 4$, $g(-12) = -10$, $g(15)$ undefined. - c. $g(x) = -6$ for $-10 \leq x < -8$.