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Suppose you have the points $(x_0=0,y_0=0) \quad (x_1=0,y_1=0)$ and the derivative at $x_1$ equal to $0$. How can I find a polynomial of degree 3 that would fit these criterias? I was under the impression that for $P_n$ you needed $n+1$ points, I can imagine graphically how such a polynomial would exist but I am stuck trying to find an algebraic solution.

Points to relevant lit. would be appreciated as well.

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  • $\begingroup$ Informally, for a polynomial of degree $n$ you need $n+1$ independent facts (i.e. you shouldn't be able to derive any of the facts from any of the other facts). A point on the graph is one fact, but so is a derivative at a point, or the $x$-value of an extremal or inflection point. $\endgroup$ – Arthur Feb 18 '15 at 18:17
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You can find "a" polynomial by setting up a system of equations. You have too few independent equations to find a unique polynomial with these characteristics. Suppose the polynomial you find is $P(x)=ax^3 + bx^2 + cx + d$. The derivative is $P^\prime(x) = 3ax^2 + 2bx + c$. You then plug in the x and y values you have (to either the poly or its derivative) then solve the system. As I said, it will not be unique with so few points. But any solution to the system is a polynomial desired.

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  • $\begingroup$ The system I can set up is p(0) = 0, p(1) = 1, p'(1) = 0. These are 3 equations and I have four variables, I tried setting the fourth variable to a given value (1) but the solution is inconsistent with the points. $\endgroup$ – arynaq Feb 18 '15 at 18:31

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