3. (2002 National High School Competition Question) As shown in Figure $6-5$, given point $A(0,2)$ and two points $B, C$ on the parabola $y^{2}=x+4$, such that $A B \perp B C$. Find the range of the y-coordinate of point $C$.

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3. Let $B\left(y_{1}^{2}-4, y_{1}\right), C\left(y^{2}-4, y\right)$. Since $y_{1}^{2}-4 \neq 0$, we have $k_{A B}=\frac{y_{1}-2}{y_{1}^{2}-4}=\frac{1}{y_{1}+2}$.

Since $A B \perp B C$, we have $k_{B C}=-\left(y_{1}+2\right)$.
Thus, $\left\{\begin{array}{l}y-y_{1}=-\left(y_{1}+2\right)\left[x-\left(y_{1}^{2}-4\right)\right], \\ y^{2}=x+4 .\end{array}\right.$
Eliminating $x$, noting that $y \neq y_{1}$, we get
$$
\left(2+y_{1}\right)\left(y+y_{1}\right)+1=0 \Rightarrow y_{1}^{2}+(2+y) y_{1}+(2 y+1)=0 \text {. }
$$

By $\Delta \geqslant 0$, we solve to get $y \leqslant 0$ or $y \geqslant 4$.
When $y=0$, the coordinates of point $B$ are $(-3,-1)$; when $y=4$, the coordinates of point $B$ are $(5,-3)$, both satisfying the conditions. Therefore, the range of the y-coordinate of point $C$ is $y \leqslant 0$ or $y \geqslant 4$.