A couple of general questions...

[jado818]

Thanks for taking the time to read this and help me out....

I've been trying to figure out how to graph slant assymptotes...

I have no problem doing polynomial long division or synthetic division..

ie... i can solve the slant assymptotes for

(x^3+2x^2+x+1)/(x^2-3x+2)

but i can't seem to figure out how to solve the slant assymptotes for square root functions...

ie sqrt(4x^2+9)

now i know that the slant assymptotes are y=-2x and y=2x

but i can't prove that mathematically
Any help or link to an explanation would be appreciated...

Also....

I've been looking for a general method for solving zeroes in cubics.... All I seem to be able to find on the topic are newton's method which seems like trial and error and Cardano's method which doesn't appear to work for every cubic... Is there an actual general method for solving any cubic???

Thanks again for taking the time to read this

Jack

[Jameson]

I don't have a great answer for your slant asyptote question, but I'll try to do some analytical reasoning.

A slant asymptote is a function that another function tends to at infinity or negative infinity. You'll find that all constants in this case do not matter... only the variables. So sqrt[4x^2+C] would have the same slant asymptote forr any constant C. So for simplicity let's say C=0. Then you have sqrt[4x^2]. You know that this has answers of 2x and -2x. Make sense?

Yes there is a general formula for the cubics that has a nasty derivation you might want to check out. The method you mentioned is the correct one.
http://mathworld.wolfram.com/CubicFormula.html

[HallsofIvy]

Thanks for taking the time to read this and help me out....

I've been trying to figure out how to graph slant assymptotes...

I have no problem doing polynomial long division or synthetic division..

ie... i can solve the slant assymptotes for

(x^3+2x^2+x+1)/(x^2-3x+2)

but i can't seem to figure out how to solve the slant assymptotes for square root functions...

ie sqrt(4x^2+9)

now i know that the slant assymptotes are y=-2x and y=2x
A slant asymptote is a line the function approaches for large x (positive or negative). For very, very large x, 4x will be much larger than 9 so the difference between 4x+ 9 and 4x becomes, relatively, smaller and smaller: 4x+ 9 "approaches" 4x and sqrt(4x+ 9) approaches sqrt(4x)= 2x.


but i can't prove that mathematically
Any help or link to an explanation would be appreciated...

Also....

I've been looking for a general method for solving zeroes in cubics.... All I seem to be able to find on the topic are newton's method which seems like trial and error and Cardano's method which doesn't appear to work for every cubic... Is there an actual general method for solving any cubic???

Thanks again for taking the time to read this

Jack

[jado818]

Thanks for the help with the asymptote part jameson and hallsofivy....

I understand that part better now.... but i still can't seem to make Cardano's method for solving cubics work for me....

If anybody has an example of a problem solved using Cardano's method that I can look at It would help me a lot.....

I find a lot of stuff on why the method should work but I seem to be missing something....

As always i appreciate your help and i hope to be able to take classes soon so i won't have to ask so many questions here =)

Jack

[HallsofIvy]

What in the world does Cardano's method of solving cubics have to do with slant asymptotes?

[OfficeShredder]

What in the world does Cardano's method of solving cubics have to do with slant asymptotes?

It doesn't. That's why the second half of his original post asks about it.

If you look here, it spells out how to solve cubic equations nicely

http://en.wikipedia.org/wiki/Cubic_equation

[HallsofIvy]

Thanks, Office Shredder, I need to learn to read!

Here's my take on Cardano. let a and b be any two real numbers.

(a-b)^3= a^3- 3a^2b+ 3ab^2- b^3
3ab(a-b)= 3a^2b- 3ab^2

So (a-b)^3+ 3ab(a-b)= a^3- b^3.

Let x= a-b, m= 3ab, n= a^3- b^3 so that x^3- mx= n. If we know m and n, can we find a and b (and so find x)?

Yes. From m= 3ab, b= m/3a. Putting that into n= a^3- b^3, we get
a^3- m^3/(3^3 a^3)= n. Multiply through by a^3 and we have (a^3)^2- na^3= (m/3)^3 or (a^3)^2- na^3- (m/3)^3= 0, a quadratic equation for a^3. By the quadratic formula, a^3= (-n +/- sqrt(n^2- 4(m/3)^3)/2 =
-n/2 +/- sqrt((n/2)^2- (m/3)^3).

a= cuberoot(-n/2 +/- sqrt((n/2)^2- (m/3)^3)
Since a^3- b^3= n, we have b^3= a^3- n or b^3= n/2 +/- sqrt((n/2)^2- (m/3)^3) so that

b= cuberoot(n/2 +/- sqrt((n/2)^2- (m/3)^3).

Now, x= a-b.