# Polynomial passing through two points with specific tangents

How can I calculate a Polynomial that passes through the origin and a given point (P) having given tangents at the origin (Ot) and the given point (Op)?

The given point will always be to the right of the origin (Px > 0). The polynomial should be a function of x (i.e. it moves from left to right).

I believe a 3rd degree polynomial will always fit these constraints, but I'm not certain.

I am looking for a function f( x, Px, Py, Ot, Pt ) = ? which plots a polynomial for x, satisfying the four given values.

When x = 0,   f()   = 0
When x = Px,  f()   = Py
When x = 0,   f'()  = Ot
When x = Px,  f'()  = Pt

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The slope of a tangent line is given by the derivative, so you want to solve a system of the form $$f(0)=0, f(p)=q,$$ $$f'(0)=a, f'(p)=b.$$ If you write out the polynomial $f$ in terms of its coefficients and then compute what the above equations should be, you'll have a linear system in terms of the coefficients. You'll have four equations, so to ensure a solution you'll want four unknowns i.e. a third-degree polynomial. –  anon Aug 10 '11 at 23:35
Like anon said. But please start from f(x)=ax+cx^2+dx^3, to solve only two equations for two unknowns (c and d). –  Did Aug 10 '11 at 23:52
@anon: You could put that in the answer box. I'd vote for it –  Ross Millikan Aug 11 '11 at 0:21

The slope of a tangent line is the derivative, so you want to solve a system of the form ($p\ne 0$)

$$f(0)=0, f(p)=q,$$ $$f'(0)=a, f'(p)=b.$$

The coefficients of $f$ are the unknowns, and we have four equations making a linear system for the coefficients. To ensure a solution, we need also four unknowns i.e. a degree $3$ polynomial. From the left two conditions we know that $f(x)=dx^3+cx^2+ax$. The right two conditions then are

$$\begin{pmatrix}p^3 & p^2\\3p^2 & 2p\end{pmatrix} \begin{pmatrix}d\\c\end{pmatrix}=\begin{pmatrix}q-ap\\b-a\end{pmatrix}.$$

Solving,

$$\begin{pmatrix}d\\c\end{pmatrix}=-\frac{1}{p^4}\begin{pmatrix}2p & -p^2\\-3p^2 & p^3\end{pmatrix}\begin{pmatrix}q-ap\\b-a\end{pmatrix}.$$

Thus we have

$$f(x,p,q,a,b)=-\frac{2p(q-ap)-p^2(b-a)}{p^4}x^3-\frac{-3p^2(q-ap)+p^3(b-a)}{p^4}x^2+ax.$$

EDIT: Also, the polynomial can be quadratic or linear if and only if the corresponding coefficients above vanish appropriately. To see this, solve the quadratic case to get the condition

$$\frac{q-ap}{p^2}=\frac{b-a}{2p},$$

which is equivalent to $d=0$. By geometry, the linear case occurs when $a=b=p/q$, which can also be shown as logically equivalent to $d=c=0$.

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In general one needs to solve a (confluent) Vandermonde system to do Hermite interpolation, but as can be seen in anon's answer, specially arranged data can allow for shortcuts. –  Ｊ. Ｍ. Aug 11 '11 at 2:25