vertex of a quartic equation

[Acinonyx]

sorry i had to make a topic about this but i've been scouring the entire internet looking for an answer.

 

 

 

is there a formula like method for finding the vertex of a quartic equation's graph? like for a quadratic, it's (-b/2a) for the x value.

 

 

 

im doing an extra credit problem from my pre-calculus class and i need to know this. since i dont want anyone to give me the answer, i'm not going to post the problem here.

 

 

 

thanks!

 

 

 

acinonyx

[SLOWSTORM]

If I understand the question correctly, you should be able to solve it with simple substitution. Take the first derivative and plug in a zero? That's the best I can think of without knowing the equation.

[LordSoultar]

Well, this is what Wikipedia has to say about. Don't know if it will help much.

[jonstark7x8]

I'm in Pre-Calculus too...But I know you can find it by graphing it on a graphing calculator, and then going to trace --> vertex. If I remember the equation I will post here :D

[MPM]

If I remember what a quadratic equation is, then there are a couple of ways.

 

Lets start with a simple equation.

 

 

 

x^2-x-6

 

 

 

First you have to find the two equations that make this.

 

What are the factors of 6?

 

 

 

1,6

 

2,3

 

 

 

Now, you have full use of negatives, so what 2 factors of 6, adjusted however you like add up to make -1? (-x)

 

 

 

2+(-3) =-1

 

 

 

So, the two equations that make up the main equation is

 

(x-3)*(x+2)

 

 

 

You can seperate these into single equations.

 

 

 

(x-3)=0

 

(x+2)=0

 

 

 

And when you solve it out, you get the x values of 3 and -2.

 

 

 

There is another method, but this is the easiest to understand/learn/explain.

 

If you want the other, feel free to message me.

[LordSoultar]

If I remember what a quadratic equation is, then there are a couple of ways.

 

Lets start with a simple equation.

 

 

 

x^2-x-6

 

That's quadratic, though. He needs quartic. A quartic equation would look something like this:

 

 

 

y = x^4+x^3+x^2+x-3

[MPM]

If I remember what a quadratic equation is, then there are a couple of ways.

 

Lets start with a simple equation.

 

 

 

x^2-x-6

 

That's quadratic, though. He needs quartic. A quartic equation would look something like this:

 

 

 

y = x^4+x^3+x^2+x-3

 

Times like this I wish I hadn't lost my old graphing calculator. Made a program that solved those with the steps in between :(

 

Can't help you cause I can't remember exactly how it's done. I'll come back in a few hours, and mabey I'll remember.

[kido14]

Use a graphing calculator like already suggested.

 

 

 

Or, you could use the derivative graph, which is what we are learning about in AP Calculus, and I think it's there there is a zero in f'(x) there is a max or a min on the f(x) graph.

 

 

 

But yea, I'm not exactly sure what you are asking.

[Viktorkrum77]

I don't advertise sites. But I feel it is an appropiate situtation. (Hopefully :uhh:)

 

 

 

Just go to Yahoo Answers. If your question is short and sweet it's better to ask there, instead of waste a thread here.

[Quasar889]

I'm not a math genius, but here's what I'd do:

 

 

 

Do you have mathematica?

 

You can solve the quartic by expressing the roots of the function as transcendatal Riemann zeta function outputs. (Not making any of this up.)

 

 

 

http://library.wolfram.com/examples/quintic/steps.html#program

 

 

 

The complex surface generate by the quartic formula has roots where the Riemann zeta function intersects itself. This is primarily a website concerned with finding an algebraic expression of the general term quintic, but with mathematic you can solve for the quartic in the same way. This is for the roots of a quartic, you can invoke the same process (averaging the roots) to get rough approximations of the vertices, which you can then check with calculus (the dy/dx function on your calc).

 

 

 

Sorry if that made no sense, I'm still only in Algebra Two honors, but I'm a total mathematica math freak.