Example 14 (1) Let the parabola be $y=2 x^{2}$. If it is shifted $p$ units to the right, or $q$ units downward, it can make the parabola intersect the line $y=x-4$ at exactly one point. Find the values of $p, q$;
(2) If the parabola $y=2 x^{2}$ is shifted $p$ units to the left and $q$ units upward, the resulting parabola passes through the points $(1,3)$ and $(4,9)$. Find the values of $p, q$.
(3) If the parabola $y=a x^{2}+b x+c$ is shifted 3 units to the left and 2 units downward, the resulting graph is the parabola $y=a x^{2}$ passing through the point $\left(-1,-\frac{1}{2}\right)$. Find the original quadratic function.
(1) The parabola $y=2 x^{2}$, after being shifted to the right by $p$ units, becomes the parabola $y=2(x-p)^{2}$. Thus, the equation $2(x-p)^{2}=x-4$ has two identical roots, which means the equation
$$
2 x^{2}-(4 p+1) x+2 p^{2}+4=0
$$

has a discriminant
$$
\Delta=(4 p+1)^{2}-4 \cdot 2 \cdot\left(2 p^{2}+4\right)=0 .
$$

Therefore, $p=\frac{31}{8}$. At this point, the intersection point is $\left(\frac{33}{8}, \frac{1}{8}\right)$.
Shifting the parabola $y=2 x^{2}$ downward by $q$ units, we get the parabola $y=2 x^{2}-q$. Thus, the equation $2 x^{2}-q=x-4$ has two identical roots, which means
$$
\Delta=1-4 \cdot 2(4-q)=0 .
$$

Therefore, $q=\frac{31}{8}$. At this point, the intersection point is $\left(\frac{1}{4},-\frac{15}{4}\right)$.
(2) Shifting the parabola $y=2 x^{2}$ to the left by $p$ units and upward by $q$ units, we get the parabola $y=2(x+p)^{2}+q$. Thus, by the given conditions, we have
$$
\left\{\begin{array}{l}
3=2(1+p)^{2}+q \\
9=2(4+p)^{2}+q
\end{array}\right.
$$

Solving the system of equations, we get $p=-2, q=1$, which means the parabola was shifted to the right by 2 units and upward by 1 unit.
(3) First, the parabola $y=a x^{2}$ passes through the point $\left(-1,-\frac{1}{2}\right)$, from which we can find $a=-\frac{1}{2}$. Let the original quadratic function be $y=-\frac{1}{2}(x-h)^{2}+k$. By the given conditions, we have
$$
\left\{\begin{array}{l}
-h+3=0, \\
k-2=0 .
\end{array}\right.
$$

Solving the system of equations, we get $h=3, k=2$. Therefore, the original quadratic function is
$$
y=-\frac{1}{2}(x-3)^{2}+2 \text {. }
$$