Example 1. The vertex of the parabola $y=f(x)$ is $(-2,3)$, and the difference between the two roots of $\mathrm{f}(\mathrm{x})=0$ is 2. Find $\mathrm{f}(\mathrm{x})$.
$$
\begin{array}{l}
\text { Solve for } f(x)=a x^{2}+bx+c, \text { then from the given, we have } \\
-\frac{b}{2 a}=-2, \\
\left\{\begin{array} { l } 
{ \frac { 4 a c - b ^ { 2 } } { 4 a } = 3 , } \\
{ \sqrt { \Delta } = 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-3, \\
b=-12, \\
c=-9
\end{array}\right.\right. \\
\frac{\sqrt{\Delta}}{(a)}=2 \\
\therefore f(x)=-3 x^{2}-12 x-9. \\
\end{array}
$$

Method: Since the vertex is known, we can let
$$
f(x)=a(x+2)+3.
$$

Also, since the difference between the two roots of $f(x)=0$ is 2,
$$
\therefore 2 \sqrt{-\frac{3}{a}}=2.
$$

Solving for $a$ gives $a=-3$.
$$
\therefore f(x)=-3 x^{2}-12 x-9.
$$

Note: When finding the expression of a quadratic function, generally set $f(x)=ax^2+bx+c$. If the given conditions include the vertex, extremum, or axis of symmetry, setting $f(x)=a(x+m)^2+n$ is more convenient.
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