What is the minimum degree for a polynomial to pass through points with defined slopes

Question

I'm having some difficulty solving this problem. The information I have is the following:

What is the minimum degree for a polynomial for it to pass through points $A(x_1,y_1)$ and $B(x_2,y_2)$ with predefined slopes $S_1$ and $S_2$.

Thanks in advance!

Answer 1

The general case will require a polynomial of degree 3 or less because there are four constraints.

Depending on the specific points and slopes, the degree could be $0,1,2,$ or $3$.

I presume you mean the maximum required degree, not the maximum possible degree. You can always find polynomials of arbitrarily high degree that do the job because there are only four constraints, not enough to determine uniquely any polynomial of degree higher than three.

You can assume the form $$p(x)=A+ Bx+Cx^2+Dx^3$$ and the constraints $$p(a_1)=b_1,p(a_2)=b_2,p'(a_1)=m_1,p'(a_2)=m_2$$ where the $a_i, b_i,$ and $m_i$ are specified constants. This produces a system of four equations in the four unknowns $A,B,C,$ and $D$.

Answer 2

I interpret your question as the following:

We fix real numbers $x_1,x_2,y_1,y_2,S_1,S_2$ seek a polynomial function $f$ such that

$$ f(x_1)=y_1\\ f(x_2)=y_2\\ f'(x_1)=S_1\\ f'(x_2)=S_2\\ $$

We can interpret this algebraically by subtracting linear functions from the polynomial $f$:

Consider the function $$ g_1(x)=f(x_1-x)+S_1x-y_1 $$ Then $g_1(0)=f(x_1)-y_1=y_1-y_1=0$. So $g_1$ has a root at $0$.

But we can also see that $$g_1'(x)=-f'(x_1-x)+S_1$$ So $g_1'(0)=-f_1'(0)+S_1=-S_1+S_1=0$. Therefore, $g_1$ has at least a double root at $0$. So we must have $$g_1(x)=x^2h_1(x)$$ for some non-zero polynomial $h_1$.

Similarly, if $$ g_2(x)=f(x_2-x)+S_2x-y_2 $$ then $$ g_2(x)=x^2h_2(x) $$ for some non-zero polynomial $h_2$.

Now we can express $f$ in terms of $g_1,g_2$: $$ f(x)=g_1(x_1-x)-S_1(x_1-x)+y_1\\ f(x)=g_2(x_2-x)-S_2(x_2-x)+y_2 $$ and therefore we can express it in terms of $h_1$,$h_2$: $$ f(x)=(x_1-x)^2h_1(x_1-x)-S_1(x_1-x)+y_1\\ f(x)=(x_2-x)^2h_2(x_2-x)-S_2(x_2-x)+y_2 $$ Therefore $f$ is certainly at least quadratic (though that's obvious, since a polynomial of degree $\le1$ has constant slope). We now want to choose $h_1,h_2$ so that $$ (x_1-x)^2h_1(x_1-x)-S_1(x_1-x)+y_1=(x_2-x)^2h_2(x_2-x)-S_2(x_2-x)+y_2 $$ To make things easier, we might as well consider $h_1,h_2$ to be polynomials in $x$. So write $\bar{h}_1(x)=h_1(x_1-x)$ and $\bar{h}_2(x)=h_1(x_2-x)$. Then we are trying to find $\bar h_1,\bar h_2$ such that: $$ (x_1-x)^2\bar{h}_1(x)-S_1(x_1-x)+y_1=(x_2-x)^2\bar{h}_2(x)-S_2(x_2-x)+y_2 $$

At this point, we see that choosing $\bar h_1,\bar h_2$ to be constant polynomials is never going to work. But we can choose them to be linear, degree $1$ polynomials, and the problem then comes down to solving simultaneous equations.