# Quartic Function Explanation

1. Jan 29, 2006

### rocketboy

Hey everyone,

For math, I was given a quartic function. I was asked to find the coordinates of the points of inflection Q and R. Then to determine the points P and S, where the line QR intersects the quartic function again, and calculate the ratio PQ:QR:RS.

So I found the points of inflection. Easy.

Then I found the equation of the line through these 2 points. Very easy.

Then I set the quartic and this line equation equal, rearanging to get another quartic. I knew that there would be 4 roots to this quartic, 2 of which I have found as the points of infleciton, and 2 of which are the intersecting points I am looking for.

So I found this website, and went through the work of finding the points Q and R. I finally did, and checked my work using a graphing program, as well as the calculator on the site, and I didn't make any mistakes.

The problem is, I don't know how to explain what I've done. Finding all those variables and those strange formula's....can somebody PLEASE help me? I can't just say I used an online calculator to find the points, I had to show my work. The problem is I don't really know what I did according to that site.

http://www.1728.com/quartic2.htm

Thank you.

2. Jan 29, 2006

### shmoe

It's not necessary to use methods for solving a general quartic as you already know two of the roots, Q and R. This means (x-Q) and (x-R) are factors of your quartic, so remove them (polynomial division say) and you're left with a quadratic which you can solve easily for P and S.

3. Jan 29, 2006

### rocketboy

Oh, that would be so much easier...and makes more sense.

So I have the following:

$$f(x) = x^4 - 8x^3 + 18x^2 - 12x + 24$$

as my quartic. The second derivative is:

$$f''(x) = 12x^2 - 48x + 36$$

which gives me (x-3) and (x-1) as factors of the quartic. So do I divide the original quartic by both of these...which is essentially dividing it by my second derivative?

4. Jan 29, 2006

### shmoe

essentially yes, since the second derivative has 3 and 1 as roots. However it's a little simpler to divide by (x-3)*(x-1) instead of your second derivative 12*(x-3)*(x-1), but either works just fine (the factor of 12 has no bearing on the zeros).

The general method for quartics (and cubics) from that link are swell to have, but it's usually preferable to avoid them if possible!

5. Jan 29, 2006

### rocketboy

Awesome, thanks for your help! I would have had a difficult time trying to figure out what they were doing from the link.

6. Jan 29, 2006

### HallsofIvy

Staff Emeritus
which quartic is this? As I understand it, you were given a quartic and asked to find P and Q, the points of inflection. You do that, of course, by setting the second derivative equal to 0 and solving that quadratic equation to find P and Q. It does NOT follow that (x- Px) and (x- Qx) are factors of the original quartic! (Strictly speaking, P and Q are points, not values of x: by Px and Qx I mean the x- coordinates of P and Q.)

Then you found the equation of the line through P and Q and Solved that simultaneously with the original quartic, resulting in another quartic for the x coordinate. Okay, now you already KNOW that Px and Qy satisfy that new quartic, not the original one! Divide by (x-Px) and (x- Py) to get another quadratic for the other x values.

Last edited: Jan 29, 2006
7. Jan 29, 2006

### shmoe

Sorry, I missed this word "original"! Halls is correct, these will be roots of your new quartic, the one you get by equating f(x) with the line you found.

I was also careless in not distinguishing between the points and their coordinates, apologies again. (I think Halls has a P_y where he should have a Q_x in the last line though)

8. Jan 29, 2006

### rocketboy

OK, yea that makes sense. Thanks guys!

9. Feb 6, 2006

### rocketboy

Oh no...not good, definately not good. I divided the quadratic formed by equating the line with the original quartic by (x-3) and then (x-1) and I got the quadratic x^2 - 4x - 7

This gives me the wrong x values for P and S.

10. Feb 6, 2006

### rocketboy

here is the information:

Original Quartic:
$$f(x) = x^4 - 8x^3 + 18x^2 - 12x + 24$$

Points Q and R (points of inflection):

Q(3, 15)
R(1, 23)

Equation of line passing through Q and R:

$$y= -4x + 27$$

New quartic (combined line with original quartic):

$$f(x) = x^4 - 8x^3 + 18x^2 - 8x - 3$$

I used the methods in the site above and got the correct answers, but this way doesn't work. I must be dividing wrong, or dividing into the wrong quartic.

11. Feb 6, 2006

### rocketboy

sorry, nevermind, i made an error doing synthetic division, had a sign wrong on the first 3 which messed up my final answer. The correct final quadratic is x^2 - 4x - 1.

12. Feb 17, 2006

### JohnNova

Hi,

I actually came across this site accidentally but am doing a piece of coursework the same as the person who has been asking about this. However, I have done everything already which has been discussed here I believe. however I now need to prove that for a general "W" shaped quartic function, or rather any function other than (X^4) the ratio of the distance between the points PQ:QR:RS is always equal. Actually I have to prove they are 1:1.62:1. Now, I have no idea how to go about doing this in an algebraic method. I have tried hours on end to prove it using algebra and finding the second derivative which is huge using [(x+a)(x+b)(x+c)(x+d)] as the form for the quartic and expanding etc etc etc. but this leads to an immense infinity of loopholes which I can't seem to overcome. Any indication of how to procede with this would be extremely appreciated. Thanks in advance people.

13. Jun 6, 2007

### forgetmenot

*sigh. seeing that the last post was a year ago, i really do hope that someone will help answer this questions as i am stuck too!

thank you all in advance =)

14. Jun 4, 2009

### thinkgreen95

ok, so I have data and I have three inflection points and I need to find a quartic with this information...how do I do that? If anyone could explain it to me or point me to a site with helpful material I would really appreciate that.