What can we say of the vector field of $\frac{dy}{dt}=f(t,y),$ when $f(t,3)=-1 \, \forall t$?

Question

Let $f(t,y),$ be such that $f(t,3)=-1 $ is continuous for all $t$.

1. What information can we get from this fact for the direction field of the differential equation $\frac{dy}{dt}=f(t,y)?$

2. If $y(0) < 3$, can $y(t) \rightarrow \infty$ as $t\rightarrow \infty$?

My question:

1. As far as I know, the only fact that I can deduce from the hypothesis is that the horizontal line $y=3$ is filled with points with constant slope ($\frac{dy}{dt}(t,3)=-1$) for all $t$.

Can something else be deduced from the fact that $f(t,3)$ is continuous?

2. I believe $y(t) \not \rightarrow \infty,$ when $t \rightarrow \infty$ if $y(0) <3,$ because if that where so, since $y(0) <3$, by continuity, there would be a $t_0>0,$ such that $\frac{dy}{dt}(t_0,3) > 0,$ and since $\frac{dy}{dt}(t_0,3)=f(t, 3) = -1$ for all $t$, that cannot happen. Hence $y(t) \not \rightarrow \infty.$

Is this argument correct?

Thanks in advance for your help.