How to get a parabola from 2 points and a slope I need to get the equation of a parabola, i have 2 points and a slope
The 2 points are as follows:


*

*A = (8, 2.912)

*B = (16, 2.912)

*Slope = 0.364


The slope starts at (0,0) and goes through A
Technically i have an additional slope, but that is just mirrored for B.
How do i get the parabola connecting these two points, where the slope is a tangent?
 A: I’m assuming that when you say that the ‘slope’ starts at $(0,0)$ and passes through $A$, you mean that the tangent line to the parabola at $A$ goes through the origin. Since $\frac{2.912}8=0.364$, that line does in fact have a slope of $0.364$, but you should not confuse the line with its slope, any more than you would confuse yourself with your height or weight.
You know that the axis of the parabola lies midway between $A$ and $B$, so it’s the line $x=12$. This means that the equation of the parabola has the form $$y=a(x-12)^2+b\tag{1}$$ for some constants $a$ and $b$. You know that $(8,2.912)$ lies on the curve; substituting that into $(1)$ gives you one equation involving $a$ and $b$. Differentiating $(1)$ and substituting for the slope of the tangent at that point gives you a second, and you should then be able to solve the resulting system for $a$ and $b$.
A: Assuming the parabola is of the form $y = a(x-b)^2 + c$, the fact that $y(8) = y(16)$ tells you that the vertex of the parabola is halfway between $x=8$ and $x=16$; that is, at $x=12$. So $b=12$. 
The slope of the parabola at the point $x=k$ is given by $2a(k-b)$ (this follows by taking the derivative: $y' = 2a(x-b)$, so evaluating at $x=k$ gives the quantity I just gave you). If you know $k$, then since we now know $b$ we can obtain the value of $a$. That will tells us the value of $a$ and $b$, and we only need to figure out the value of $c$. Since we know the value of $y$ at specific points, we can plug them into the formula and solve for $c$. 
Once you know $a$, $b$, and $c$, you know the parabola.
