Classifying Cubics into Three Basic TypesDate: 08/11/2002 at 15:42:55 From: Morris Waehlti Subject: Classifying Cubics into three basic types Hi, 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, Morris 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 discriminant 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 http://mathforum.org/dr.math/faq/faq.cubic.equations.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ 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, Morris 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 http://mathforum.org/dr.math/ 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, Morris |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/