Example 7 For what real number $a$ does the function $f(x)=x^{2}+x+a\left(x \geqslant-\frac{1}{2}\right)$ have two distinct intersection points with its inverse function image.
Solution: According to the problem, the original problem is equivalent to
$$
\left\{\begin{array}{l}
y=x^{2}+x+a \\
x=y^{2}+y+a \\
x \geqslant-\frac{1}{2}, y \geqslant-\frac{1}{2}
\end{array}\right.
$$

having two distinct real roots.
From (1)-(2) we get $y-x=(x-y)(x+y+1)$. Since $x \geqslant-\frac{1}{2}, y \geqslant-\frac{1}{2}$, then $x+y+1>0$, hence $x-y=0$. Substituting it into equation (1) yields $x^{2}+a=0$.

When $a=0$, $x=0, y=0$, at this time there is only one solution, which does not meet the requirement. Therefore, $a<0$, and $x=\sqrt{-a}$ or $x=-\sqrt{-a}$.
From $\sqrt{-a} \geqslant-\frac{1}{2},-\sqrt{-a} \geqslant-\frac{1}{2}$ we get $a \leqslant-\frac{1}{4}$.
Thus, when and only when $a \in\left[-\frac{1}{4}, 0\right)$, the graph of the function and its inverse function have two different intersection points.