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I would like to model the second order rate equation using the following criteria:
begin{align}
A_text{initial}&=8,500,000,\
A_text{final} &=1,200,000,\
t &=38.
end{align}

I'd like to create a plot that looks like the 2nd order chart shown in the following image:

How can I use my criteria to extrapolate the data in between $A_text{initial}$ and $A_text{final}$?

The second order differential rate law is $-frac{d[A]}{dt} = k[A]^2$

First separate the two varaibles: $[A]$ and $t$ by putting them on different sides of the equation

$-frac{d[A]}{[A]^2} = k dt$

Now integrate both sides

$int -frac{d[A]}{[A]^2} = int k dt$

Doing the integration gives the formula describing the concentration as a function of time:

$frac{1}{[A]} = kt+c$

The value $c$ is an unknown constant from the integration. There is also the unknown constant $k$, we will have to use your data to find these two constants.

First, at the starting time ($t=0$) you know that $[A]=8.5 times 10^6$. Putting this into the formula we got from integrating gives:

$frac{1}{8.5 times 10^6} = k(0)+c$

So clearly $c=frac{1}{8.5 times 10^6}$. Now put that information into the formula that relates concentration and time

$frac{1}{[A]} = kt+frac{1}{8.5 times 10^6}$

Now use your data for the final time. When $t=38$ then $[A] = 1.2 times 10^6$

$frac{1}{1.2 times 10^6} = k(38)+frac{1}{8.5 times 10^6}$

This gives us the value of $k$:

$k = frac{1}{38}(frac{1}{1.2 times 10^6} - frac{1}{8.5 times 10^6}) approx 1.88 times 10^{-8}$

Finally, putting this value of $k$ into the formula for concentration gives:

$frac{1}{[A]} = (1.88 times 10^{-8})t+frac{1}{8.5 times 10^6}$

You can use this formula to predict the concentration for any time $t$

Assuming $A:,mathbb Rtomathbb R$ sends time to some value and you have two values: $$A(t_0)=A_text{initial}tag1label{eq1}$$ and $$A(t_1)=A_text{final},tag2label{eq2}$$
where $t_0=0$ is the time at which you have $A_text{initial}=8,500,000$ and $t_1=38$ is the time at which you have $A_text{final}=1,200,000$,

and you want $A$ to satisfy the following differential equation:
$$-frac{mathrm dA(t)}{mathrm dt}=kA(t)^2$$
(which I'm not sure is exactly what the notation in the picture means),

divide it by $-A(t)^2$ to have
$$frac{mathrm dA(t)}{A(t)^2mathrm dt}=-k$$
and integrate with respect to $t$:
begin{align}
&intfrac{mathrm dA(t)}{A(t)^2mathrm dt},mathrm dt=-int k,mathrm dt\
iff&intfrac{mathrm dA(t)}{A(t)^2}=-kt+c_1\
iff&-frac1{A(t)}+c_2=-kt+c_1;
end{align}
of course, the use of two integration constants in this case is superfluous, so define $c_3=c_2-c_1$ and subtract $c_2$ from both sides:
$$-frac1{A(t)}=-kt+c_1-c_2=-kt-(c_2-c_1)=-kt-c_3;$$
now just invert and multiply both sides by $-1$:
$$A(t)=frac1{kt+c_3};tag3label{eq3}$$
now recall you have the first two equations $eqref{eq1}$ and $eqref{eq2}$ and observe the form $eqref{eq3}$ has two free variables $k$ and $c_3$, in which we can write the system
begin{cases}
A_text{initial}=A(t_0)=dfrac1{kt_0+c_3},\
A_text{final}=A(t_1)=dfrac1{kt_1+c_3};
end{cases}
now recall $t_0=0$, so we trivially have $c_3=1/A(0)$, which leaves us with one equation:
$$A_text{final}=A(t_1)=dfrac1{kt_1+1/A_text{initial}};$$
invert it:
$$frac1{A_text{final}}=kt_1+frac1{A_text{initial}};$$
subtract $1/A_text{initial}$ from both sides:
$$kt_1=frac1{A_text{final}}-frac1{A_text{initial}};$$
divide both sides by $t_1$:
$$k=frac1{t_1A_text{final}}-frac1{t_1A_text{initial}};$$
now substitute it all back into $eqref{eq3}$:
$$A(t)=frac1{kt+c_3}=1Big/left(left(frac1{t_1A_text{final}}-frac1{t_1A_text{initial}}right)t+frac1{A_text{initial}}right);$$
now just substitute your original numerical values into the last expression and evaluate it at whatever point $t$ you want.