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

  • Context: Undergrad 
  • Thread starter Thread starter karan000
  • Start date Start date
  • Tags Tags
    Degree Point Root
Click For Summary
SUMMARY

The discussion centers on determining the degree of a root at a point of inflexion for the polynomial equation y = ax(x+1)(x-3)^d. The values for the parameters are established as b = -1 and c = -3, with the point (1, -16) used to solve for 'a'. While the degree 'd' can indeed be any odd number, the specific context of the problem and the additional constraint of the stationary point at x=3 necessitate that d = 3 to satisfy the equation. This leads to the conclusion that the problem requires both the point of inflexion and the stationary point to arrive at the correct solution.

PREREQUISITES
  • Understanding of polynomial functions and their properties
  • Familiarity with points of inflexion and stationary points
  • Knowledge of substituting points into polynomial equations
  • Experience with solving equations involving multiple variables
NEXT STEPS
  • Study the characteristics of polynomial roots and their degrees
  • Learn about stationary points and their significance in calculus
  • Explore the method of substitution in polynomial equations
  • Review Polya's problem-solving strategies for mathematical problems
USEFUL FOR

Students preparing for mathematics exams, educators teaching polynomial functions, and anyone interested in advanced algebraic concepts related to roots and inflexion points.

karan000
Messages
8
Reaction score
1
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?
 
Physics news on Phys.org
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.
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K