# Determining the second order polynomial from the intersection points

1. Oct 13, 2012

### Cinitiator

1. The problem statement, all variables and given/known data
Let's say that we have a second order polynomial function, and we know all of the points where it intersects with the x and y axis. Ex: (-2; 0), (0; 2), (1; 0)

How does on determine the ax^2+bx+c polynomial form based on that?

2. Relevant equations
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3. The attempt at a solution

Tried searching for it on Google without any luck.

2. Oct 13, 2012

### sjb-2812

What is a second order polynomial? What are both coordinates of the intersections?

3. Oct 13, 2012

### HallsofIvy

I assume that you are saying that this is a general second order polynomial so you have answered sbj-2812's first question. You should also know that the point (a, b) lies on the graph of y= f(x) if and only if b= f(a). If (-2, 0), (0, 2), and (1, 0) are on the graph of $y= ax^2+ bx+ c$ then we must have $0= a(-2)^2+ b(-2)+ c$ or 4a- 2b+ c= 0, $2= a(0)^2+ b(0)+ c$ or $c= 2$, and $0= a(1)^2+ b(1)+ c$ or $a+ b+ c= 0$
Solve the equations a+ b+ c= 0, c= 0, and 4a- 2b+ c= 0 for a, b, and c.

4. Oct 13, 2012

### Ray Vickson

Do you know the relation between factors of a polynomial and the roots of the polynomial? If you do not, see http://www.sosmath.com/algebra/factor/fac02/fac02.html . Using the relationship makes your problem very easy. That is material well worth knowing.

RGV

5. Oct 13, 2012

### SammyS

Staff Emeritus
A second order (second degree) polynomial having the form ax2+bx+c and having two real root may be written as a(x-D)(x-F).

From this it should be easy to solve your problem.