1. The problem is to graph the piecewise function $$f(x) = \begin{cases} 2x + 8 & 0 \leq x \leq 2 \\ (x - 4)^2 + 8 & 2 < x < 7 \\ 17 & 7 \leq x \leq 10 \end{cases}$$ over the domain $0 \leq x \leq 10$. 2. We will analyze each piece separately and find key points to plot. 3. For the first piece, $f(x) = 2x + 8$ for $0 \leq x \leq 2$: - At $x=0$, $f(0) = 2(0) + 8 = 8$. - At $x=2$, $f(2) = 2(2) + 8 = 12$. This is a straight line segment from $(0,8)$ to $(2,12)$. 4. For the second piece, $f(x) = (x - 4)^2 + 8$ for $2 < x < 7$: - This is a parabola opening upwards with vertex at $x=4$. - At $x=4$, $f(4) = (4-4)^2 + 8 = 8$ (minimum point). - At $x=2$, $f(2) = (2-4)^2 + 8 = 4 + 8 = 12$ (matches the end of first piece). - At $x=7$, $f(7) = (7-4)^2 + 8 = 9 + 8 = 17$. 5. For the third piece, $f(x) = 17$ for $7 \leq x \leq 10$: - This is a constant function, a horizontal line at $y=17$ from $x=7$ to $x=10$. 6. To graph the entire function: - Plot the line segment from $(0,8)$ to $(2,12)$. - Plot the parabola from just above $x=2$ to just before $x=7$, passing through $(4,8)$ and $(7,17)$. - Plot the horizontal line from $(7,17)$ to $(10,17)$. 7. The function is continuous at $x=2$ and $x=7$ because the values match at these points. Final answer: The graph consists of a line segment from $(0,8)$ to $(2,12)$, a parabola from $x=2$ to $x=7$ with vertex at $(4,8)$, and a horizontal line at $y=17$ from $x=7$ to $x=10$.