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

1. Apr 17, 2012

### karan000

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.

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 learnt 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?

2. Apr 17, 2012

### epenguin

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

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

3. Apr 17, 2012

### karan000

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...

4. Apr 17, 2012

### epenguin

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: Apr 17, 2012