1. **State the problem:** We have a piecewise function for position $x(t)$ over time $t$ defined as: $$x(t) = \begin{cases} 10t & 0 \leq t \leq 1 \\ -\frac{4}{3}t^2 + \frac{38}{3}t - \frac{4}{3} & 1 < t \leq 4 \\ 2t + 20 & 4 < t \leq 5 \end{cases}$$ We want to graph this function showing the position in meters over time in seconds. 2. **Explain the segments:** - From $0$ to $1$ second, the position increases linearly with slope $10$. - From $1$ to $4$ seconds, the position follows a quadratic curve. - From $4$ to $5$ seconds, the position again changes linearly but with slope $2$ and intercept $20$. 3. **Plotting approach:** - Plot each segment on its respective interval. - Ensure continuity at the boundaries $t=1$ and $t=4$. 4. **Check continuity at $t=1$:** - Left limit: $x(1) = 10 \times 1 = 10$ - Right limit: $x(1) = -\frac{4}{3}(1)^2 + \frac{38}{3}(1) - \frac{4}{3} = -\frac{4}{3} + \frac{38}{3} - \frac{4}{3} = \frac{30}{3} = 10$ 5. **Check continuity at $t=4$:** - Left limit: $x(4) = -\frac{4}{3}(16) + \frac{38}{3}(4) - \frac{4}{3} = -\frac{64}{3} + \frac{152}{3} - \frac{4}{3} = \frac{84}{3} = 28$ - Right limit: $x(4) = 2 \times 4 + 20 = 8 + 20 = 28$ 6. **Final graph function:** $$ y = \begin{cases} 10t & 0 \leq t \leq 1 \\ -\frac{4}{3}t^2 + \frac{38}{3}t - \frac{4}{3} & 1 < t \leq 4 \\ 2t + 20 & 4 < t \leq 5 \end{cases} $$ This piecewise function can be graphed to show the position over time with smooth transitions at $t=1$ and $t=4$. **Answer:** The graph consists of three segments as defined above, continuous at the joining points.