A problem involving polynomials and degrees and confusing wording

[fresh]

Suppose you have n=40 points in the plane, each with a distinct x coordinate. What is the smallest degree d such that there is always a polynomial of degree d whose graph passes through all n points?

the options for the answer are what confuse me: d= 39, d=40, d=41 , d=42

initially, i thought that the answer would be d=0 (which would make for a horizontal line)

f(x)=cx^0 = c
so you'd have the same y coordinate but distinct x coordinates.

the wording of this question, and the selection of answers confuses me. any help would be much appreciated!

 

[mmm4444bot]

Suppose you have n=40 points in the plane, each with a distinct x coordinate. What is the smallest degree d such that there is always a polynomial of degree d whose graph passes through all n points?

… initially, i thought that the answer would be d=0 (which would make for a horizontal line)

… the wording of this question … confuses me …

The key is the word 'always'.

Yes, it's true that a zero-degree polynomial will pass through 40 points on a horizontal line, but what if one (or more) of those points is not on the line? The word 'always' infers that they're talking about any possible set of 40 points (with distinct x-coordinates).

What is the degree of the polynomial for which we're guaranteed existence, passing through any arbitrary list of 40 points (with distinct x-coordinates) that someone might give you?

 

[Deleted member 4993]

Suppose you have n=40 points in the plane, each with a distinct x coordinate. What is the smallest degree d such that there is always a polynomial of degree d whose graph passes through all n points?

the options for the answer are what confuse me: d= 39, d=40, d=41 , d=42

initially, i thought that the answer would be d=0 (which would make for a horizontal line)

f(x)=cx^0 = c
so you'd have the same y coordinate but distinct x coordinates.

the wording of this question, and the selection of answers confuses me. any help would be much appreciated!

For a start - think about the problem with only two points.

How about 3 points?

 

[fresh]

For a start - think about the problem with only two points.

How about 3 points?

can i assume this case for the problem: the first point is below the x-axis, the second is above, the third is below and so on

then 2 points would mean a degree of 1: it's just a line crossing the x-axis once
3 points would mean a degree of 2: function has to cross x-axis twice
4 points would be a degree of 3 (crossing the x-axis 3 times)

continuing with this logic, a function of 40 points, would have the smallest possible degree 39?

 

[tkhunny]

No, you're not seeing the idea at all. We need ALWAYS. If you make ANY assumption, you cannot get ALWAYS.

What does it take to hit ANY one point? x = c. Done. One point, one parameter. Degree 0
What does it take to hit ANY two points? y = mx+b Done. Two points, two Parameters. Degree 1.
What does it take to hit ANY three points? y = ax^2 + bx + c Done. Three points, three parameters. Degree 2.

Can you extrapolate the solution from this?

Keep in mind a VERY IMPORTANT piece of information. I'd be a little surprised if most beginning students noticed or understood its importance. It is this: "each with a distinct x coordinate". This is a game changer. Without this, it's a different story.

 

[mmm4444bot]

can i assume this case for the problem: the first point is below the x-axis, the second is above, the third is below and so on

then 2 points would mean a degree of 1: it's just a line crossing the x-axis once
3 points would mean a degree of 2: function has to cross x-axis twice
4 points would be a degree of 3 (crossing the x-axis 3 times)

continuing with this logic, a function of 40 points, would have the smallest possible degree 39?

Your conclusion is correct, but the logic is not.

You must not make any assumptions about the list of 40 points because that list could contain any 40 points in the xy-plane. (The only requirement is that each point on the list has a different x-coordinate).

The graph of a polynomial passing through those 40 points does not need to cross the x-axis. It's also possible that all of the points lie within a single Quadrant.

We could reword the first sentence in the exercise to read:

"Suppose you are given a list containing any possible set of 40 points in the plane, each with a distinct x coordinate."

I think the point of this exercise is for students to visualize the general pattern. If you have n points (where no two points lie on a vertical line), then you're guaranteed that a polynomial of degree n-1 exists, passing through each point.

For some of these possible lists of n points, there might exist one or more polynomials of degree less than n-1, but it's not guaranteed. This exercise requires that you can always find a polynomial, regardless of the given points.

Wikipedia states that this is proven by the Unisolvence Theorem, but I've never seen or heard of that theorem in any algebra classroom. Instead, a pattern is usually invoked, and the conclusion is simply given as a fact to memorize.

What is your class currently studying? Are you solving systems of polynomial equations? :cool:

 

[fresh]

Your conclusion is correct, but the logic is not.

You must not make any assumptions about the list of 40 points because that list could contain any 40 points in the xy-plane. (The only requirement is that each point on the list has a different x-coordinate).

The graph of a polynomial passing through those 40 points does not need to cross the x-axis. It's also possible that all of the points lie within a single Quadrant.

We could reword the first sentence in the exercise to read:

"Suppose you are given a list containing any possible set of 40 points in the plane, each with a distinct x coordinate."

I think the point of this exercise is for students to visualize the general pattern. If you have n points (where no two points lie on a vertical line), then you're guaranteed that a polynomial of degree n-1 exists, passing through each point.

For some of these possible lists of n points, there might exist one or more polynomials of degree less than n-1, but it's not guaranteed. This exercise requires that you can always find a polynomial, regardless of the given points.

Wikipedia states that this is proven by the Unisolvence Theorem, but I've never seen or heard of that theorem in any algebra classroom. Instead, a pattern is usually invoked, and the conclusion is simply given as a fact to memorize.

What is your class currently studying? Are you solving systems of polynomial equations? :cool:

That makes sense, and I remember learning about the general n-1 pattern. Thank you!

This question was on a list of review questions on a 'calculus readiness assessment' suffice to say, I don't think I'm quite ready.
Luckily forums like these are handy.