Finding polynomials with their values at points Is there any way I can find a polynomial given any $2$ points (with $x$ coordinate OF MY CHOICE): Let's say there's some polynomial I don't know $(p(x)=2x^3+x^2+3)$, but my machine will give me an output. I give one $x$ value of my choice, and it returns $p(x)$, where $p(x)$ is the polynomial function. I give another value of my choice., $x+h$, and get the output $p(x+h)$. Given these outputs, I have to find $p(x)$ as a polynomial.
What I've done is plugged in $0$, which gives me the final term of the polynomial that is not multiplied by any power of $x$. Then I plug in $1$, getting another output. When I find the "slope" of the two points, I get the sum of all the coefficients of all the terms that are powers of $x$. If I do this for the given $p(x)$, I get $3$, which is the sum of $2$ and $1$. However, I can't figure out what powers of $x$ there are and what specific coefficients there are. Does anyone know how to solve this?
@GerryMyerson and @Shash said I can find the polynomial given the bound of the coefficients. I am confused as to what that means. There is only one number that is the sum of the coefficients. How is there a bound? Also, how do I find this sum of coefficients with just one value? I need to use one more value, M+1, as Shash said, so I can't use 2 values to find the max/sum, as I won't be able to ask for a value that is M+1. Can anyone help? Thanks.
EDIT: Non-negative integer coefficients are assumed.
 A: With two points you can only uniquely determine $p(x)$ if it has degree one. In general if $p(x)$ has degree $n$, you will need the computer give $(n+1)$ outputs, each for a different input.
For example if $p(x)$ has degree 2 and you input $0$ and $1$ and get outputs $a$ and $b$ respectively, there are an infinite number of parabolas which pass through these points. Here are some:
$p(x) = (b-a)x^2+a$
$p(x) = bx^2-ax+a$
$p(x) = 2bx^2-(b+a)x+a$
$p(x) = -3ax^2+(b+2a)x+a$
A: No. You would a minimum of $n$ points, where $n$ is the dimension of the polynomial. For instance, let's try to find a cubic polynomial $p$ where $p(0)=0$ and $p(1)=1$. Notice that $p_1(x)=x^3-x^2+1$ and $p_2(x)=x^3-x+1$ both satisfy the given criteria.
A: A polynomial of degree $n$ has $n+1$ coefficients.  If you know the degree but don't know anything about those coefficients, you'll need  $n+1$ values of the polynomial to determine them.  If you don't know the degree, no amount of values of the polynomial will suffice.
A: If all the coefficients are non-negative integers and if we know the max of all the coefficients, then it is easy to find the coefficients with just ${one\  point}$ of choice. Let's say the max is $M$. Then evaluate the polynomial at $M+1$. The output will be equal to the decimal number system representation of a number whose base $M+1$ number has digits as the coefficients of the polynomial. So, given the output, just get the base $M+1$ representation.
So, even if have just an upper bound (can be very lose upper bound), just set that as $M$ and proceed as above.
As Gerry Myerson points out in the comments, if you do not have an upper bound, you can use $M = p(1)$.  Then only two evaluations will be required to determine the coefficients.
