It is a logistic function. this means that it will first cause and exponential increase, which will slow down and eventually level off. I would look something like this: 
The equation of this is $P(t)=\frac{KP_0 e^{rt}}{K+P_0(e^{rt}-1)}$
$P_0 =$ population at time $t = 0$
$K$ = final population after some (long) time, also called the “carrying capacity”, which limits growth
$r$ = initial growth rate
Initial: $1$ person Total amount of people: $10,000$ $50$ people after $5$ days $x$ people after $10$ days
$t$ is measured in days
$P(t)=\frac{KP_0 e^{rt}}{K+P_0(e^{rt}-1)}$
$50=\frac{10000*1*e^{5r}}{10,000+1(e^{5r}-1)}$
$50=\frac{10000e^r5}{9,999+e^{5r}}$
$50(9999)+50(e^{5r})=10,000e^{5r}$
$499950+50e^{5r}=10,000e^{5r}$
$499950=10,000e^{5r}-50e^{5r}$
$499950=9950e^{5r}$
$499950=9950e^{5r}$
$50.2462=e^{5r}$
$\ln(50.2462)=\ln(e^{5r})$
$\ln(50.2462)=5r$
$\frac{\ln(50.2462)}{5}=r$
This then needs to be added to the initial equation
$P(t)=\frac{KP_0 e^{\frac{\ln(50.2462)}{5} t}}{K+P_0 (e^((ln(50.2462))/5 t)-1)}$
${t = 10}$
$P(10)=\frac{10,000*1e^((\ln(50.2462))/5(10))}{10,000+1(e^((ln(50.2462))/5(10))-1)}$
$P(10)=\frac{10,000*e^((\ln(50.2462))/5(10))}{9,999+e^((ln(50.2462))/5(10))}$
$P(10)=\frac{25246837.4700}{12523.6838}$
$P(10) = 2015.9274$
It will take after $10$ days, $2016$ people will be infected.
I got my information (the equation) from this website: http://www.intmath.com/blog/h1n1-and-the-logistic-equation/3498