1. **Problem statement:** We have points $A, B, C$ with position vectors $\overrightarrow{OA} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}, \overrightarrow{OB} = \begin{bmatrix} 18 \\ 10 \end{bmatrix}, \overrightarrow{OC} = \begin{bmatrix} 13 \\ 1 \end{bmatrix}$ and a vector function $\vec{r}(t) = (1-t)^2 \overrightarrow{OA} + 2(1-t)t \overrightarrow{OB} + t^2 \overrightarrow{OC}$. We want to show the coordinate functions $x(t)$ and $y(t)$ are given by $x(t) = -21 t^2 + 32 t + 2$ and $y(t) = -17 t^2 + 16 t + 2$. 2. **Formula and rules:** The vector function is a quadratic Bezier curve formula. The coordinates are found by applying the formula component-wise: $$\vec{r}(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = (1-t)^2 \begin{bmatrix} 2 \\ 2 \end{bmatrix} + 2(1-t)t \begin{bmatrix} 18 \\ 10 \end{bmatrix} + t^2 \begin{bmatrix} 13 \\ 1 \end{bmatrix}.$$ 3. **Calculate $x(t)$:** $$x(t) = (1-t)^2 \cdot 2 + 2(1-t)t \cdot 18 + t^2 \cdot 13.$$ Expand $(1-t)^2 = 1 - 2t + t^2$: $$x(t) = 2(1 - 2t + t^2) + 36 t (1 - t) + 13 t^2.$$ Expand $36 t (1 - t) = 36 t - 36 t^2$: $$x(t) = 2 - 4 t + 2 t^2 + 36 t - 36 t^2 + 13 t^2.$$ Combine like terms: $$x(t) = 2 + (-4 t + 36 t) + (2 t^2 - 36 t^2 + 13 t^2) = 2 + 32 t - 21 t^2.$$ 4. **Calculate $y(t)$:** $$y(t) = (1-t)^2 \cdot 2 + 2(1-t)t \cdot 10 + t^2 \cdot 1.$$ Expand $(1-t)^2 = 1 - 2t + t^2$: $$y(t) = 2(1 - 2t + t^2) + 20 t (1 - t) + t^2.$$ Expand $20 t (1 - t) = 20 t - 20 t^2$: $$y(t) = 2 - 4 t + 2 t^2 + 20 t - 20 t^2 + t^2.$$ Combine like terms: $$y(t) = 2 + (-4 t + 20 t) + (2 t^2 - 20 t^2 + t^2) = 2 + 16 t - 17 t^2.$$ 5. **Conclusion:** We have shown that $$x(t) = -21 t^2 + 32 t + 2$$ $$y(t) = -17 t^2 + 16 t + 2,$$ which matches the given coordinate functions.