# Polynomial testing

1. May 15, 2013

### jasonlr82794

Hey guys, I know what polynomials are but what I really don't understand is the way you are able to find the equation to a set amount of points. I don't understand why you have to have a certain amount of points to find different degrees of functions. For example, why are only three points needed to find a second degree function? And how would I find the equation to a graph if I randomly insert a random point. For example, I have 0,0 1,1 2,4 and then im suppose to have 3,9 but I changed it to 3,16? How would I find this function?

2. May 16, 2013

### Staff: Mentor

You didn't ask about linear functions, but I'll include them. Lines that aren't vertical have an equation of the form y = ax + b. To determine the constants a and b you need two points, which will give you two equations in the unknowns a and b.

A quadratic function has the form y = ax2 + bx + c. To determine a, b, and c, you need three equations, which you can get from three points.

If you had the points (0, 0), (1, 1), (2, 4), and (3, 9), you could use these points to find the equation of the third-degree polynomial (a cubic) of the form y = ax3 + bx2 + cx + d.

If you change one of the points that you're working with, you'll get different values for a, b, c, and d.

3. May 16, 2013

### jasonlr82794

Ok, that does help but why do I only need two for a linear, three for a quadratic, and so on?

4. May 16, 2013

### Staff: Mentor

I explained that. A linear equation (y = ax + b) has two unknown constants, a and b. You need two points on the line so that you can generate two equations in the two unknowns. Similarly for a quadratic function, there are three unknown constants, so you need three points, so you can get three equations that involve a, b, and c.

In general, for an nth degree polynomial y = a0xn + a1xn-1 + ... + an-1x + an, there are n + 1 constants, so you need n + 1 points.

5. May 17, 2013

### jasonlr82794

Ok, this to an extent helps me but one my friends was talking to me and he said something to the effect that you need certain points that are the characteristic on that function. Do these constants have anything to do with these characteristics of these functions?

6. May 17, 2013

### Staff: Mentor

I don't know what that means. You'll need to ask your friend to be more specific.

7. May 17, 2013

### jasonlr82794

And also, I don't think you understand where im coming from. If you are given any amount of points, it doesn't matter how many and you don't have a graph of the function, just the points, how would you be able to determine the function and along with that the degree(im aware once you know the function you will know the degree). How would you be able to determine which polynomial equation(y=ax+b, y=ax^2+bx+c, and so on) to use to find this function? If you don't get what I mean I will try to restate it. I know on tests they give you the set amount of points that you need then you must figure out what equation to use but that's just not even intuitive especially if a teacher tells you that you only need three points to find a quadratic and you know that you need that equation. I guess im kind of thinking outside of the box.

8. May 17, 2013

### Staff: Mentor

If you have n + 1 points, you can always fit them to an n-th degree polynomial. This is sort of the converse of what I said at the end of post #4.

So, if you have two points, you can find a first degree (linear) polynomial.
And so on.

9. May 17, 2013

### jasonlr82794

The rule is that through any two points only one line can exist. With the quadratics the three points give you the vertex and the zeros. This is exactly what he said so is this true and if so must these points always be the vertex and the zeros?

10. May 17, 2013

### Staff: Mentor

With a quadratic, if you have any three points, you can determine the equation. If you are given the vertex point, all you need is one other point and you can find the equation. By symmetry, you can figure out where the third point is.

For example, if the vertex is at, say, (2, 1), and you are given the point (3, 3), the third point would have to be (1, 3). So now you have three points and you're ready to find the equation of the quadratic.
No. For an equation of degree n, any n+1 points will do.