# About quartic function asap

1. May 18, 2006

### dotcom

about quartic function... asap!!!

hey everyone,

I'm stuck with a proof of quartic function.

For math, I was given a w shape 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.

The ratio was, after all, 1:1.618...:1.

Then my teacher said to prove that every w shape quartic function gives the same ratio, but I don't know how to start. He said not to deal with massive amount of calculation, so I think I should use f"(x)=(x-a)(x-b) or something.
Can anyone help me????

2. May 18, 2006

### AKG

1.618... looks like the start of the golden ratio, i.e. the positive solution to x2 - x - 1 = 0. Anyways, it doesn't look like you've started doing any work and you haven't shown us anything, so you probably won't get help. Try setting up some equations describing the distances PQ, QR, and RS.

3. Dec 1, 2006

### hamo_26

i was also given the same question for a math assignment. i as asked to find the inflection points of point q and r, and asked to determine the point p and s of a line which intersected the graph at those 4 points, hence 2 of the roots of the intersection were q and r. I was wondering if someone could provide a general proof that for every quartic function, the ratio result of the distances PQ:QR:RS will always be a golden ratio. I was wondering if this could be derived algebraicly from a general equation such as ax^4 + bx^3+cx^2+dx+e = mx + b, hence this should all boil down somehow to x^2 - x -1 =0. PLEASE SOMEONE HELP ME, I AM VERY CONFUSED!!!!!!!!!!!

4. Dec 1, 2006

### HallsofIvy

That's a good start. Any quartic can be written as y= ax4+ bx3+ cx2+ dx+ e. Now, where are the points of inflection, P and Q? You get the x coordinates, of course, by differentiating y twice and setting it equal to 0. Once you have found P and Q, in terms of a, b, c, d, e, you can write the equation of the line through them- that is you will know your m and b in terms of a, b, c, d, and e. Determine where the quartic intersects the line, the distances between the points, and the ratios.

Lots and lots of computation! As AKG pointed out, that number is the golden ratio. You don't want to find a specific number but to show that the ratio satisfies x2- x- 1= 0.

5. Feb 12, 2007

### handyman

Hello everyone, we got one more.
To quickly save everyone the pain i will summarize what I am doing, the same as everyone else with the quartif function. And yes i am stuck on the proof as well, who would have guessed that. I so that fellow intelectuals have posted some question and received some respons. So i took the advice and tried to continue. I used
ax^4+bx^3+cx'2X+dx+e
I derived the function twice and got this as a result
12ax^2 +6bx +2c
i devided this by to simplify this bit a bit
so i ended up with
6ax^2+3bx+c. Now i need to find the roots of this function. I am lost here, i am trying to figure out how to factorize this . Help anyone. Please
Greetings from here

6. Feb 12, 2007

### handyman

Hello everyone, we got one more.
To quickly save everyone the pain i will summarize what I am doing, the same as everyone else with the quartif function. And yes i am stuck on the proof as well, who would have guessed that. I so that fellow intelectuals have posted some question and received some respons. So i took the advice and tried to continue. I used
ax^4+bx^3+cx'2X+dx+e
I derived the function twice and got this as a result
12ax^2 +6bx +2c
i devided this by to simplify this bit a bit
so i ended up with
6ax^2+3bx+c. Now i need to find the roots of this function. I am lost here, i am trying to figure out how to factorize this . Help anyone. Please
Greetings from here

Last edited: Feb 12, 2007
7. Feb 12, 2007

### Doodle Bob

Looks suspiciously like a quadratic equation to me... And, if I recall correctly there is a *formula* that lets us determine the roots of *every* quadratic equation without necessarily factorizing...

8. Feb 12, 2007

### handyman

Hey there,
I finially got somewhere with my proof of quadratic function. However i am almost done, but i am stuck again. Maybe i amde a mistake or its too much and i lost track. I need Robin Hood to help the ones in need, me. This is what ive done so far:

let f(x) = ax^4 + bx³ + cx² + dx + k

f'(x) = 4ax³ + 3bx² + 2cx + d
f"(x) = 12ax² + 6bx + 2c
= 0 for points of inflection

Thus 12ax² + 6bx + 2c = 0
ie 6ax² + 3bx + c = 0
So x = (-3b ± √[9b² - 24ac])/(12a) with 9b² - 24ac > 0 for real and distinct roots ie for real and distinct points of inflexion

Let the roots be α and β with α > β
Note α + β = -3b/6a = -b/2a and αβ = c/6a

So y1 = aα^4 + bα³ + cα² + dα + k
and y2 = aβ^4 + βα³ + cβ² + dβ + k

ie W ≡(β, aβ^4 + bβ³ + cβ² + dβ + k) and X ≡(α, aα^4 + bα³ + cα² + dα + k)

Slope WX = Δy/Δx
= [(aα^4 + bα³ + cα² + dα + k) - (aβ^4 + bβ³ + cβ² + dβ + k)/(α - β)
= [a(α^4 - β^4) + b(α³ - β³) + c(α² - β²)+ d(α - β)]/(α - β)
= a(α³ + α²β + αβ² + β³] + b(α² + αβ + β²) + c(α + β) + d
= a(α³ + 3α²β + 3αβ² + β³ - 2α²β - 2αβ²) + b(α² + 2αβ + β² - αβ) + c + d
= a((α + β)³ - 2αβ(α + β)) + b((α + β)² - αβ) + c(α + β) + d
= a(b/2a)³ + 2c/6a * b/2a) + b((b/2a)² - c/6a) + c.c/6a + d
= b³/8a² + bc/6a² + b²/4a² - bc/6a + c²/6a + d
= (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)

So [y - (aα^4 + bα³ + cα² + dα + k)]/(x - α)
= (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)
so y - (aα^4 + bα³ + cα² + dα + k)
= (x - α)(3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)
= (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²) * x
- α(3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)

So y = (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²) * x
- α(3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)
+ (aα^4 + bα³ + cα² + dα + k)

9. May 8, 2007

### lunzballoons

Hey... My math current math HL portfolio is the same as the one you have solved about proving the ratio. i got stuck with the proving part two, so i was wondering if u could pleeaaaaaaase help me.

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