Growth Rate of a Population I think I know the answer but I just wan to confirm my suspicions. The question is:
Assume that the growth rate of a certain species is constant, but negative. Sketch the population if at $t=0$, $N=N_0$. What happens as $t \rightarrow \infty$
So I believe the graph would be an exponential function with a y-intercept at $N_0$. Essential it is $N(t) = t^2 + N_0$ with only positive values of $t$. And as $t \rightarrow \infty$ the population would also go towards $\infty$.
 A: It would not be an exponential graph
If the rate of change of a population of a certain species is constant, but negative then
$$\frac{\mathrm{d}N}{\mathrm{d}t} = -k $$
where $k>0$ is constant. Separating the variables gives:
$$\int \mathrm{d}N = -k\int\mathrm{d}t$$
It follows that $N = -kt+c$, where $c$ is constant. If $N=N_0$ when $t=0$ then
$N_0=c$ and hence
$$N(t) = N_0 - kt$$
You will be able to find $k$ when you have more information, e.g. $N=1000$ when $t=5$. 
In this case, the limit $t \to \infty$ makes no sense; you can't have a negative population. The model $N(t) = N_0-kt$ only makes sense for $N\ge 0$, i.e. $N_0-kt \ge 0$, i.e. $t \le \frac{1}{k}N_0$.
When would it be an exponential graph?
When the growth/decay rate is directly proportional to the current population then we get an exponential. If the rate is proportional to the current population then it is not consonant. 
In such a case, we have
$$\frac{\mathrm{d}N}{\mathrm{d}t} = kN$$
Separating variables gives
$$\int \frac{1}{N}\mathrm{d}N =k\int \mathrm{d}t$$
It follows that $\ln|N| = kt+c$ and hence $|N| = \alpha\mathrm{e}^{kt}$, where $\alpha = \mathrm{e}^c > 0$. Dropping the absolute value gives $N = \alpha\mathrm{e}^{kt}$ where $\alpha \in \mathbb R$.
If the population grows then your initial conditions will give $k>0$. If your population shrinks then your initial conditions will give $k<0$.
