<選填>設點 \(A(-2, 2)\)、\(B(4, 8)\) 為坐標平面上兩點,且點 \(C\) 在二次函數 \(y = \frac{1}{2}x^2\) 的圖形上。當 \(C\) 的 \(x\) 坐標為 __________ 時,內積 \(\overrightarrow{AB} \cdot \overrightarrow{AC}\) 有最小值__________。
\[
\boxed{\text{設點法}}
\]
\[
\begin{aligned}
&\text{設 } C\left(t,\ \frac{1}{2}t^2\right) \\
&\overrightarrow{AB} = (4-(-2),\ 8-2) = (6,\ 6) \\
&\overrightarrow{AC} = \left(t-(-2),\ \frac{1}{2}t^2-2\right) = \left(t+2,\ \frac{1}{2}t^2-2\right)
\end{aligned}
\]
\[
\boxed{\text{內積計算}}
\]
\[
\begin{aligned}
\overrightarrow{AB} \cdot \overrightarrow{AC}
&= (6,\ 6) \cdot \left(t+2,\ \frac{1}{2}t^2-2\right) \\
&= 6(t+2) + 6\left(\frac{1}{2}t^2-2\right) \\
&= 6t + 12 + 3t^2 - 12 \\
&= 3t^2 + 6t
\end{aligned}
\]
\[
\boxed{\text{配方法求極值}}
\]
\[
\begin{aligned}
3t^2 + 6t &= 3(t^2 + 2t) \\
&= 3\left[(t+1)^2 - 1\right] \\
&= 3(t+1)^2 - 3 \\
&\geq -3 \quad (\text{當 } t+1=0 \text{ 時取等號})
\end{aligned}
\]
\[
\boxed{\text{結論}}
\]
\[
\begin{aligned}
&\text{當 } t = -1 \text{ 時，} \overrightarrow{AB} \cdot \overrightarrow{AC} \text{ 有最小值 } -3 \\
&C \text{ 點座標為 } (-1,\ \frac{1}{2})
\end{aligned}\]
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