Generalized Cubic Equation

Date: 05/22/2000 at 09:40:00
From: Samantha
Subject: Cubic graphs

Dear Dr. Math:

How do we find a generalized cubic rule for a family of cubics with 
the following characteristics: they cut through the point (1,1), have 
a gradient of 2 when x = 1, and it have 1 stationary point?

Date: 05/22/2000 at 13:25:02
From: Doctor Rob
Subject: Re: cubic graphs

Thanks for writing to Ask Dr. Math, Samantha.

Let the equation of the cubic be

     y - 1 = a*(x-1)^3 + b*(x-1)^2 + c*(x-1).

That represents all cubics passing through (1,1). Then

     dy/dx = 3*a*(x-1)^2 + 2*b*(x-1) + c.

Now consider the conditions one at a time.

1) 2 = (dy/dx)_{x=1} = c

2) The quadratic equation

     dy/dx = 3*a*(x-1)^2 + 2*b*(x-1) + c = 0

has equal roots, so its discriminant is zero,

     (2*b)^2 - 4*(3*a)*c = 0

The first equation gives us the value of c. Substitute that into the 
second equation. Now use the new second equation to solve for a in 
terms of b. Put these expressions back into the equation of the cubic, 
and you'll have a one-parameter family of cubic equations satisfying 
the conditions described. That parameter will be b. The only 
stationary point will be x = 1 - 2/b, y = -4/(3*b).

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/