Date: Fri, 16 Oct 1998 11:54:26 -0600 (CST)
Sender: AGRrules@juno.com (Shay)
Subject: Parametrizing the left half of a parabola
Who is asking: Student
Find the parametrized equation for the left half of the parabola with the equation: Y=x^2-4x+3
y=x^2-4x+3 can be factored as y=(x-1)(x-3) so the vertex of the parabola is at x=2, midway between 1 and 3. To get the left side of the parabola you need to force the x-coordinate to be 2 or less. That is you want x to be 2 minus something that is positive, or at least 2 minus something that is never negative. Thus let x=2-t^2 and whatever value t has, x will be less than or equal to 2. Now substitute x=2-t^2 into y=x^2-4x+3 to find y in terms of t.
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