Thread: Finding the equation of a polynomial without knowing any zeros

I am having trouble with finding the equation of a polynomial function, particularly because I do not know what any of the solutions (zeros) are.

However, I do know the following things:
1) The degree of the function
2) Coordinates for all local extrema and points of inflection
3) The end-behavior (in this case it is a positive even function, where both ends point up)
4) The Y-intercept (there is no X-intercept as there are no real solutions)
5) A number of other points

Note: There are no real solutions (zeros). I said earlier that I do not know what any of the zeros are (and I don't), I just know that there are no real ones.

I don't need an exact answer or an exact process, I just need to know how to approach this problem.
The frustrating thing is that every website I've seen either says a) how to find zeros given an equation, or b) how to find an equation given the zeros.

If you know n + 1 points for an nth degree polynomial, then you can use a system of equations.

If you know extrema points, then you know the value of the derivative there is zero, likewise if you know points of inflection, then you know the second derivative is zero there. All of this information will given you a system of linear equations.

here is how (in detail) you would do it for a polynomial of degree 2:

suppose we know that f(4) = 2. suppose further than f(1) = 6, and f(5) = 8 (in case you're wondering, i'm just making these numbers up).

we know that f(x) = ax² + bx + c. we're going to regard a,b and c as "independent variables". when x = 4, since f(4) = 2, we get:

(x²,x,1) = (16,4,1). so f(4) can be regarded as the dot-product (16,4,1).(a,b,c) = 2, or if you like:

16a + 4b + c = 2

in the same way, we get:

a + b + c = 6

25a + 5b + c = 8

this is the same as the matrix equation:



which one can solve by row-reduction, or cramer's method, for example.

in this case, the system is small enough that we can see that:

15a + 3b = -4
9a + b = 6

using b = 6 - 9a in the first equation:

15a + 3(6 - 9a) = -4
15a + 18 - 27a = -4
22 = 12a
a= 11/6, so b = -21/2, and thus c = 44/3.

as a check, we verify f(x) = (11/6)x² - (21/2)x + 44/3 fits the bill.

f(1) = 11/6 - 21/2 + 44/3 = 11/6 - 63/6 + 88/6 = 36/6 = 6.

f(4) = (11/6)(16) - (21/2)(4) + 44/3 = (11/3)(8) - 42 + 11/6 = 88/3 - 42 + 44/3 = 176/6 - 252/6 + 88/6 = 12/6 = 2.

f(5) = (11/6)(25) - (21/2)(5) + 44/3 = 275/6 - 105/2 + 44/3 = 275/6 - 315/6 + 88/6 = 48/6 = 8.

Thanks, guys, that's exactly what I needed.

(Although this is going to be a lot more complicated than your example, the function happens to be degree 10!)

well, yes, you are going to have an 11x11 matrix, and hopefully you have a good calculator or math software to solve the resulting system of linear equations (if not, wolframalpha.com makes a good substitute).