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Mathematics 10 Online
OpenStudy (yeezus):

Find the parameterization for the left half of the parabola y=x^2-4x+3

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

was that helpful?

OpenStudy (yeezus):

yes thank you

OpenStudy (anonymous):

no problem (:

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