1. **State the problem:** We are given parametric equations $x(t) = t^2 - 2t$ and $y(t) = t + 1$ for $-2 \leq t \leq 4$. We need to sketch the curve, indicate the direction, and find the Cartesian equation. 2. **Find the Cartesian equation:** From $y(t) = t + 1$, solve for $t$: $$t = y - 1.$$ Substitute into $x(t)$: $$x = (y - 1)^2 - 2(y - 1) = (y - 1)^2 - 2y + 2.$$ Expand and simplify: $$x = y^2 - 2y + 1 - 2y + 2 = y^2 - 4y + 3.$$ So the Cartesian equation is: $$x = y^2 - 4y + 3.$$ 3. **Rewrite the Cartesian equation in standard form:** Complete the square for $y$: $$x = (y^2 - 4y + 4) - 4 + 3 = (y - 2)^2 - 1.$$ This is a parabola opening to the right with vertex at $(x,y) = (-1, 2)$. 4. **Indicate direction:** As $t$ increases from $-2$ to $4$, find points: - At $t = -2$: $x = (-2)^2 - 2(-2) = 4 + 4 = 8$, $y = -2 + 1 = -1$. - At $t = 4$: $x = 16 - 8 = 8$, $y = 5$. The curve starts at $(8, -1)$ and moves to $(8, 5)$ passing through the vertex at $(-1, 2)$. 5. **Summary:** The curve is a right-opening parabola $x = (y - 2)^2 - 1$ with direction from $(8, -1)$ to $(8, 5)$ as $t$ goes from $-2$ to $4$. Final answer: $$x = (y - 2)^2 - 1,$$ with direction indicated by increasing $t$ from $-2$ to $4$.