2nd degree quadratic fit between 2 points with known slopes

[mwg83]

If you know 2 the points (x1,y1) with slope m1, and (x2,y2) with slope m2. And both slopes are positive, it is possible to fit a quadratic between them?

How would you derive A, B, and C for the form y=Ax^2+Bx+C

 

[MarkFL]

We have:

[MATH]y(x)=Ax^2+Bx+C[/MATH]
Thus:

[MATH]y'(x)=2Ax+B[/MATH]
Now, from the two given points, we have:

[MATH]y_1=Ax_1^2+Bx_1+C[/MATH]
[MATH]y_2=Ax_2^2+Bx_2+C[/MATH]
And from the two given slopes, we have:

[MATH]m_1=2Ax_1+B[/MATH]
[MATH]m_2=2Ax_2+B[/MATH]
Using these last 2 equations, if we subtract the former from the latter, we will eliminate \(B\):

[MATH]m_2-m_1=2A\left(x_2-x_1\right)[/MATH]
Solve this for \(A\), and then substitute for \(A\) into either of those equations whose difference we found to get \(B\). Then using either of the two equations that arose from the points on the quadratic, substitute for \(A\) and \(B\) to get \(C\).

 

[ksdhart2]

Sorry, but I don't understand what you're asking. From what I can gather, you have two points... but how can a point have a slope?

 

[Otis]

… (x1,y1) with slope m1 …

… (x2,y2) with slope m2 …

how can a point have a slope?

Hi ksdhart2. I did a double-take on that wording, too.

I'm thinking it's a sloppy description about slopes of lines tangent to the parabola, at points (x₁,y₁) and (x₂,y₂).

?

 

[MarkFL]

I took the wording to mean that at the point \(\left(x_i,y_i\right)\) on the quadratic, the quadratic will have an instantaneous slope of \(m_i\).

 

[HallsofIvy]

A general quadratic in x and y can be written Ax^2+ Bxy+ Cy^2+ D= 0. Given two points that curve passes through, and the slopes of the curve at those points give 4 equations to solve for A, B, C, and D. However, If we were to divide through by any one of those coefficients we would get an equation for exactly the same curve with only 3 coefficients.