no that's not correct
I have this question: Is my answer correct? I came to this answer because I actually worked each out to find c1 and then graphed them. When I graphed the second one y became negative while x, (t), went to infinity. This is an electronic version of this graph: Oh, and I found that c1 for option 2 was (-8.46). The other options didn't turn out like this for their graph, but it could've been a mistake in a number of different steps. Is my answer right? Thanks!
If $k<0$, then as $t \to \infty$, $y \to a$.
If $k>0$, then as $t \to \infty$, $e^{kt} \to \infty$, so if you want $y \to -\infty$, you need $Ae^{kt} \to -\infty$, so $A<0$. Which of the choices offers $k>0$ and $A<0$? As the problem says, you must solve for $A$ in each case.