How do we know the degree of a root which is also a point of inflexion?

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Discussion Overview

The discussion revolves around determining the degree of a root at a point of inflexion in the context of a polynomial function. Participants explore the implications of having an odd degree at a root and how to utilize given points to find specific values in the polynomial equation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant notes that the degree of the root at the point of inflexion can be any odd number, citing their understanding of polynomial behavior.
  • Another participant agrees that the degree could be any odd number but suggests that additional information is needed to narrow down the possibilities.
  • A later reply emphasizes the importance of using all given information, specifically mentioning the point (1, -16) to solve for the coefficient 'a' in the polynomial.
  • One participant expresses frustration over the reliance on guess and check methods to determine the value of 'd', indicating a desire for a more straightforward solution.
  • Another participant introduces the idea that the point (1, -16) is not just any point but a stationary point, suggesting it may have special significance in the problem.

Areas of Agreement / Disagreement

Participants generally agree that the degree of the root can be any odd number, but there is no consensus on how to definitively determine the value of 'd' or the implications of the stationary point.

Contextual Notes

The discussion highlights limitations in the problem-solving approach, particularly regarding the use of given points and the nature of the stationary point, which may not be fully explored.

karan000
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Hey guys, I was looking at an exam I did last year and tried to solve a question, which at the time I couldn't do.

Unfortunately I'm running into the same problem I had during the exam, so hear me out on this one

Question:
The graph below has equation y =ax(x-b)(x+c)^d. Write down the values for a, b, c and d.
whatthefuuu.png
Okay, so there's an intercept at x=-1, so b=-1. There's another intercept at x=3, so c = -3

So, y = ax(x+1)(x-3)^d

Now here's the problem, the answers simply say "d = 3", and then sub in the point (1,-16) and solve for a,

But at x=3 (where it's a point on inflexion), can't the degree (d) be ANY odd number? ie. 3,5,7,9,11,...,999999?

Because that's how I learned polynomials, a point on inflexion at a root means the degree of the root is odd.

And so that's where I messed up, becausing I couldn't figure what value of 'd' I should use.

I must obviously be missing some sort of concept, so can someone please help me out?
 
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karan000 said:
But at x=3 (where it's a point on inflexion), can't the degree (d) be ANY odd number? ie. 3,5,7,9,11,...,999999?

Because that's how I learned polynomials, a point on inflexion at a root means the degree of the root is odd.

And so that's where I messed up, becausing I couldn't figure what value of 'd' I should use.

I must obviously be missing some sort of concept, so can someone please help me out?

You idea is right - it could be any odd number - so far as you have got.

You now have to ask yourself the question, one of the five or so in Polya's 'How to solve it' book

-

"Am I using all the information I'm given?"
 
epenguin said:
You idea is right - it could be any odd number - so far as you have got.

You now have to ask yourself the question, one of the five or so in Polya's 'How to solve it' book

-

"Am I using all the information I'm given?"

I'm aware of the the point (1,-16),

So as you already know that b=-1 and c=-3, sub in the point to get:

-16 = a(1)(1+1)(1-3)^d
= 2a(-2)^d

Two varibles makes this unsolvable but using guess and check you'll find 3 is the only correct solution (which just makes the whole question practically ****** and pointless).

I was expecting there to be a way to solve the question without guess and check but I guess there isn't any other way...
 
OK, you have used the fact that (1, -16) is a point on the function.

But I think you are supposed to use also the fact that it is a special kind of point, a stationary point.
 
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