Predicting the graph of the given reaction 
I have no idea on how to answer this question. Please help.

I'm unsure why this was migrated from chemistry.
The rates of chemical reactions are governed by rate equations, for example (Where $\mathrm{A}$ is the reactant, $k$ is the rate constant, and $n$ is the reaction order which usually must be experimentally determined:
$$rate=-\dfrac{dA}{dt}=k[\mathrm{A}]^n$$
We can solve the differential equation to get a generalized equation for $\mathrm{A}$ as a function of time.
$$-\dfrac{dA}{[\mathrm{A}]^n}=kdt$$
$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]^n}=\int_0^tkdt=kt$$
If $n=0$ (zeroth order), then the left integral evaluates to:
$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]^0}=[\mathrm{A}]_0-[\mathrm{A}]_t=kt$$
This equation is linear. Since your graph is not linear, your data is not zerothorder.
If $n=1$ (first order):
$$-\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{dA}{[\mathrm{A}]}=\ln{[\mathrm{A}]_t}-\ln{[\mathrm{A}]_0}=\ln{\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}=-kt$$
Solving for $[\mathrm{A}]_t$ gives you:
$$\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=e^{-kt}$$
$${[\mathrm{A}]_t}=[\mathrm{A}]_0e^{-kt}$$
Your graph looks like an exponential decay, so you can find the slope of the tangent line at time $t=0$ by differentiating, except you need to know $k$ to evaluate at $t=0$.
$$\dfrac{d}{dt}[\mathrm{A}]_t=\dfrac{d}{dt}[\mathrm{A}]_0e^{-kt}=-k[\mathrm{A}]_0e^{-kt}$$
However, since the half-life of a first order reaction is given by 
$$t_{1/2}=\dfrac{\ln{(2)}}{k}$$
you can solve for $k$ since the half-life is marked on the graph. 
