Generalized Cubic EquationDate: 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/ |
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