# Homework Help: Find Equation of a Function

1. Apr 2, 2009

### Mentallic

1. The problem statement, all variables and given/known data
Given that (visually) there is a polynomial with a root at x=5 and seems like a triple root at x=1. It cuts the y-axis at y=-10 and there is a max turning point at (3,32).

Ok so basically, find the equation of the function P(x) given:
P(0)=-10
P(5)=0
P(1)=0 (triple root, thus cubic shape)
P'(3)=0

3. The attempt at a solution
Judging by the roots of the polynomial, I'm assuming the equation will be of the form:

$$P(x)=a(x-1)^3(x-5)$$

When I substitute (0,-10) I get a=-2

Thus, $$P(x)=-2(x-1)^3(x-5)$$

I tested the point (3,32) and since it satisfied the equation when I did it in the test, I assumed it was the turning point; but when I graphed it at home, I found that the turning point wasn't at (3,32) but just seemed to pass through that point.

When I take the derivative of $$P(x)=a(x-1)^3(x-5)$$
I get $$P'(x)=4a(x-1)^2(x-4)$$ and so P'(3)=0 only when a=0, but that is obviously not the answer I want for a. Maybe for what I assumed the form of the polynomial P(x) is incorrect? Is it possible to find an equation to satisfy all these criteria?

Honestly, being that this question was only worth 2 marks out of a 75 mark exam that was 90 mins long, I believe it could've just been a parabola that was hand-drawn really badly...

Anyway, thanks for any help or advice on how to approach this.

2. Apr 2, 2009

### meiso

If the polynomial definitely has those roots and that y-intercept, then it can't have a max at (3,32). Were all of the points (roots, intercept, max) labeled on the test paper, or did you have to estimate them from grid lines??

3. Apr 2, 2009

### Mentallic

It was a hand-drawn graph that was not to scale and only had those details labelled on the graph. Thus, the max turn was specifically at (3,32) without doubt, there were roots at x=1,5 and possibly triple root at x=1. Also, without doubt there was a y-intercept at y=-10.

I also tested it for a parabola with those roots and y-intercept and again the max turn wouldn't be at that coordinate. Do you suppose they were expecting the student to state that such a polynomial is inexistent? or at least not elementary?

4. Apr 2, 2009

### meiso

I guess the answer to that question would depend on the scope of the course/ personality of the professor. Only you could answer that, but I guess it's remotely possible.

I'm betting it was a mistake. I even tested it for other odd degree root behavior at x = 1 up to the 15th power just in case they were going for that, and none of the polynomials' derivatives had zeros at 3.

5. Apr 2, 2009

### Mentallic

The max turn was at x=3 for a parabola only, but it was at (3,8) and for any positive integer n such that $$P(x)=a(x-1)^{2n+1}(x-5)$$ the turning point is 3<x<5, and as n gets large, the turn approaches 5. So searching for any polynomials passed the 3rd power is fruitless, which I just thought I'd put out there, even though you already went to the trouble

The course is last year at highschool and there is a history of teachers making mistakes in the test (and there were even two other spotted errors in the exam), so I'll wait for the results, and hopefully be able to have a plausible reason to argue if I get the answer wrong.

6. Apr 2, 2009

### meiso

Very good point! Thank you for pointing that out. Good luck with the results of your exam.

7. Apr 2, 2009

### Mentallic

Thanks

This is the only problem I had in the test, so I'm sure I went well, but hey, the marker always gets me in some screwed up way and I never quite hit that 100%... hehe I'll stop boring you with my life story now