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Classifying Cubics into Three Basic Types

Date: 08/11/2002 at 15:42:55
From: Morris Waehlti
Subject: Classifying Cubics into three basic types


Evidently any cubic can be classified both algebraically and
graphically by applying a translation and/or a y-scaling and/or an
x-scaling to one of the following three basic types:

     CUBIC              HORIZONTAL SLOPES            FUNCTION TYPE
1) y = x^3 + x                none                     one-one
2) y = x^3                    one                      one-one
3) y = x^3 - x                two                      many-one 

For example translating y = x^3-x to the point (-1,2) yields
y-2 = (x+1)^3-(x+1) and y scaling this by 2 yields
(y-2)/2 = (x+1)^3-(x+1), which simplifies to y = 2x^3+6x^2+4x+2.

My question is: how can I classify each of y = x^3+x^2, y = x^3-2x^2,
y = x^3-2x^2+x, y = x^3-2x^2+2x, and y = -x^3+x^2-x into one of three 
basic types without actually drawing their graphs?

Thank you very much for your help. 

Best regards,

Date: 08/12/2002 at 22:46:06
From: Doctor Peterson
Subject: Re: Classifying Cubics into three basic types

Hi, Morris.

We can classify a cubic easily using the derivative. Let's take the 
general cubic

    y = ax^3 + bx^2 + cx + d

It will have a horizontal slope when

    y' = 3ax^2 + 2bx + c = 0

The number of solutions to this equation is determined by the 

    D = (2b)^2 - 4(3a)(c) = 4b^2 - 12ac = 4(b^2 - 3ac)

So when

    b^2 - 3ac < 0 there are no horizontal tangents
              = 0 there is one horizontal tangent
              > 0 there are two horizontal tangents

We can go a little farther. Taking the second derivative, we can find 
the inflection point, which is at the center of a cubic:

    y" = 6ax + 2b = 0

    x = -b/(3a)

We can translate this to the origin by expressing the equation in 
terms of a new variable u = x + b/(3a):

    y = ax^3 + bx^2 + cx + d

      = a(u - b/(3a))^3 + b(u - b/(3a))^2 + c(u - b/(3a)) + d

      = au^3 - bu^2 +  b^2/(3a) u - b^3/(27a^2)
             + bu^2 - 2b^2/(3a) u + b^3/(9a^2)
                    + cu          - bc/(3a)
                                  + d

      = au^3 + (3ac-b^2)/(3a) u + (2b^3 - 9abc)/(27a^2) + d

This clearly shows the horizontal and vertical translation you 
referred to, and the role of the discriminant (which determines the 
sign of the linear term). The role of scaling takes a little more 
work to see; I'll leave that for you to work out. In any case, I've 
shown you how to classify and translate any cubic. You will also want 
to see our FAQ on solving cubics:

   Cubic and quartic equations 

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 

Date: 08/15/2002 at 11:48:24
From: Morris Waehlti
Subject: Classifying Cubics into three basic types

Dear Doctor Peterson,

Thank you very much for your reply.

Using your method, it would seem that y=x^3+x^2 has its center at
(-1/3,2/27). Translating it back to the origin yields y = x^3-x/3, 
whose graph is not quite that of y = x^3-x. I seem to be missing a y 
and/or an x-scaling of some sort and I cannot, for the life of me, 
find what they are!

The situation is the same for the other examples I mentioned, 
namely: y = x^3-2x^2 and y = x^3-2x^2+x (which should both come from 
y = x^3-x) and also y = x^3-2x^2+2x and y = -x^3+x^2-x (which should 
both come from y = x^3+x).

Could I please ask you for a little more help?

Very best regards,

Date: 08/15/2002 at 12:21:19
From: Doctor Peterson
Subject: Re: Classifying Cubics into three basic types

Hi, Morris.

I left out the scaling part of the problem, in part because I was not 
sure whether you were using x^3+x and x^3-x merely as models of types 
of cubics and only wanted to classify all cubics, or if you wanted to 
actually do the scaling to get to those forms.

A scaling with respect to x and y takes the form of a substitution

    x = pu
    y = qv

Let's do that to y = x^3 - dx:

    qv = (pu)^3 - d(pu)

    qv = p^3 u^3 - dp u

    v = (p^3/q) u^3 - (dp/q) u

In order for this to be the same as u^3 - u, we need

    p^3/q = 1
    dp/q = 1

Solving for p and q, we get

    p^3 = q
    p = q/d

    p^3 = dp

    p^2 = d

    p = sqrt(d)
    q = d sqrt(d)

So our scaling is

    x = sqrt(d) u
    y = d sqrt(d) v

scaled equation is

    d sqrt(d) v = (sqrt(d))^3 u^3 - d sqrt(d) u

    v = u^3 - u

Try applying this to your equations.

- Doctor Peterson, The Math Forum 

Date: 08/19/2002 at 12:41:10
From: Morris Waehlti
Subject: Thank you (Classifying Cubics into three basic types)

Dear Doctor Peterson,

Your methods work! Thank you very much for your help.

Very best regards,
Associated Topics:
College Calculus
High School Calculus

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