Assuming $k_1$ is the slope in $x_1$ and $k_2$ is the slope in $x_2$.
Okay, say your polynomial is $y = a_3 x^3 + a_2 x^2 + a_1 x + a_0$. As shamovich already have stated you get a system of equations $Xa = y$ where $X$ is a matrix and $y$ and $a$ are vectors:
$$
\underbrace{
\begin{pmatrix}
x_1^3 & x_1^2 & x_1 & 1 \\
x_2^3 & x_2^2 & x_2 & 1 \\
3x_1^2 & 2x_1 & 1 & 0 \\
3x_2^2 & 2x_2 & 1 & 0
\end{pmatrix}
}_{=X}
\underbrace{
\begin{pmatrix}
a_3 \\ a_2 \\ a_1 \\ a_0
\end{pmatrix}
}_{=a}
=
\underbrace{
\begin{pmatrix}
y_1 \\ y_2 \\ k_1 \\ k_2
\end{pmatrix}
}_{=y}
$$
The first two rows are simply the equations $y_i = a_3 x_i^3 + a_2x_i^2 + a_1x_i + a_0$ and the last two rows are the equations $k_i = 3a_3 x_i^2 + 2a_2 x_i + a_1$.
I assume you know your points $(x_i, y_i)$ for $i = 1,2$ and $k_1, k_2$, so $X$ and $y$ are known. Solve for $x$ (you can use e.g. Matlab or Mathematica for this). As long as $x_1 \neq x_2$ there is a solution, since the determinant of the matrix is $-(x_1 - x_2)^4$.
You can derive the inverse $X^{-1}$. It has the form:
$$X^{-1} =
\begin{pmatrix}
\frac{2}{\left(-x_1+x_2\right){}^3} & \frac{2}{\left(x_1-x_2\right){}^3} & \frac{1}{\left(x_1-x_2\right){}^2} & \frac{1}{\left(x_1-x_2\right){}^2} \\
\frac{3 \left(x_1+x_2\right)}{\left(x_1-x_2\right){}^3} & -\frac{3 \left(x_1+x_2\right)}{\left(x_1-x_2\right){}^3} & -\frac{x_1+2 x_2}{\left(x_1-x_2\right){}^2} & -\frac{2 x_1+x_2}{\left(x_1-x_2\right){}^2} \\
-\frac{6 x_1 x_2}{\left(x_1-x_2\right){}^3} & \frac{6 x_1 x_2}{\left(x_1-x_2\right){}^3} & \frac{x_2 \left(2 x_1+x_2\right)}{\left(x_1-x_2\right){}^2} & \frac{x_1 \left(x_1+2 x_2\right)}{\left(x_1-x_2\right){}^2} \\
\frac{\left(3 x_1-x_2\right) x_2^2}{\left(x_1-x_2\right){}^3} & \frac{x_1^2 \left(x_1-3 x_2\right)}{\left(x_1-x_2\right){}^3} & -\frac{x_1 x_2^2}{\left(x_1-x_2\right){}^2} & -\frac{x_1^2 x_2}{\left(x_1-x_2\right){}^2}
\end{pmatrix}$$
So $a = X^{-1} y$ (but, as always, if you are using software, it is better to use built-in solvers than using the inverse explicitly).
Explicit formulas for the coefficients one by one is given by:
$$\begin{align}
a_3 &= \frac{\left(k_1+k_2\right) \left(x_1-x_2\right)-2 y_1+2 y_2}{\left(x_1-x_2\right){}^3} \\
a_2 &= \frac{-k_1 \left(x_1-x_2\right) \left(x_1+2 x_2\right)+k_2 \left(-2 x_1^2+x_1 x_2+x_2^2\right)+3 \left(x_1+x_2\right) \left(y_1-y_2\right)}{\left(x_1-x_2\right){}^3} \\
a_1 &= \frac{k_2 x_1 \left(x_1-x_2\right) \left(x_1+2 x_2\right)-x_2 \left(k_1 \left(-2 x_1^2+x_1 x_2+x_2^2\right)+6 x_1 \left(y_1-y_2\right)\right)}{\left(x_1-x_2\right){}^3} \\
a_0 &= \frac{x_2 \left(x_1 \left(-x_1+x_2\right) \left(k_2 x_1+k_1 x_2\right)-x_2 \left(-3 x_1+x_2\right) y_1\right)+x_1^2 \left(x_1-3 x_2\right) y_2}{\left(x_1-x_2\right){}^3}
\end{align}$$
If you know for a fact that your points lie on a polynomial of degree 5, you can use finite differences. This is especially easy when the $x$-values are in arithmetic progression, as in your case. Write the $y$-values:
175 125 100 125 0 -125 -100 -125 -175
Then write the difference between each number and the next:
-50 -25 25 -125 -125 25 -25 -50
Again:
25 50 -150 0 150 -50 -25
Again:
25 -200 150 150 -200 25
Again:
-225 350 0 -350 225
And once more:
575 -350 -350 575
Now if your points were really from a polynomial of degree 5, that last line would have been constant, but it's not, so they're not. But after all, you said they were estimated points - they still might be close to some polynomial of degree 5. To find the polynomial of degree 5 that comes closest to your points, there is a method called least squares, and there are many expositions of it on the web.
EDIT: Here's a start on least squares:
You want a polynomial $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ such that $p(800)=175$, $p(600)=125$, etc. We already know that's impossible, so we settle for making all the numbers $p(800)-175$, $p(600)-125$, etc., small. In fact, we form the quantity $$(p(800)-175)^2+(p(600)-125)^2+\cdots+(p(-800)-175)^2$$ and we try to minimize it. This quantity is a function of the 6 variables $a,b,c,d,e,f$, and there are standard calculus techniques for minimizing such a function. It gets very messy, but fortunately you don't actually have to do it; someone has done it for you, in the most general case, and found that you can write down a very simple matrix equation which you can solve for the unknowns $a,b,c,d,e,f$. And that's what you'll find if you search for least squares polynomial fit.
Best Answer
Lagrange Interpolation is the technique most commonly used for this purpose. Just take 5 points and plug them into Lagrange's formula (which I think is best articulated in the Examples section of the Wiki page). Assuming all of your points do indeed lie on a polynomial of degree 4, you will get that polynomial via Lagrange interpolation.