Example 4. For what value of $a$ does the curve $y=\sqrt{x+1}$ and the line $y=x+a$ have two intersection points?
Solve as shown in Figure 4: $\because y=\sqrt{x+1}, \therefore y \geqslant 0$ squaring both sides gives $y^{2}=x+1(y \geqslant 0)$,
$\therefore$ the curve of $y=\sqrt{x}+1$ is the part of the parabola $y^{2}=x+1$ above the x-axis (including the point on the x-axis).
Obviously, the intersection points should satisfy:
$$
\left\{\begin{array}{l}
y^{2}=x+1, \\
y=x+a, \\
y \geqslant 0 .
\end{array}\right.
$$

From (1) and (2), eliminating $x$, we get
$$
\begin{array}{c}
y^{2}-y+a-1=0 .(4) \\
\Delta=1-4(a-1) \\
=-4 a+5 \text {. }
\end{array}
$$

When $a0$. Let the y-coordinates of the two intersection points of the line and the parabola $y^{2}=x+1$ be $y_{1}$ and $y_{2}$, then

When $1 \leqslant a<\frac{5}{4}$, $y_{1} y_{2}=a-1 \geqslant 0$, and $y_{1} + y_{2}=1$, so $y_{1} \geqslant 0, y_{2} \geqslant 0$. At this time, the line $y=x+a$ intersects the curve $y=\sqrt{x+1}$ at two points.

When $a<1$, $y_{1} y_{2}=a-1<0$, the line $y=x+a$ intersects the parabola $y^{2}=x+1$ at two points, but only one of them is above the x-axis. Therefore, at this time, the line intersects the curve $y=\sqrt{x+1}$ at only one point.

In summary, when $1 \leqslant a<\frac{5}{4}$, the line $y=x+a$ and the curve $y=\sqrt{x+1}$ intersect at two points.