(3) Let the function $f(x)=x^{2}+x+m\left(m \in \mathbf{R}^{+}\right)$, if $f(t)<0$, then your judgment on the existence of zeros of the function $y=f(x)$ in the interval $(t, t+1)$ is $\qquad$ .
(3) There is at least one zero point. Hint: Since $f(t)0$, then $f(x)=$ $x^{2}+x+m$ intersects the $x$-axis at two points $A$ and $B$, with the x-coordinates being $x_{1}$ and $x_{2}$ respectively, and $x_{1}0 .
\end{array}\right.
$$

Therefore, $-10$, thus
$$
f(t+1)=(t+1)^{2}+(t+1)+m>0,
$$

which means $f(t) f(t+1)<0$, so the function $y=f(x)$ has at least one zero point in the interval $(t, t+1)$.