Show that the center of the osculating circle for the parabola y = x2 at the point (a, a2) is located at  (-4a3, 3a2 + 1/ 2 ) .

Intermediate Algebra
19th Edition
ISBN:9780998625720
Author:Lynn Marecek
Publisher:Lynn Marecek
Chapter9: Quadratic Equations And Functions
Section9.6: Graph Quadratic Functions Using Properties
Problem 9.91TI: Find the intercepts of the parabola whose function is f(x)=3x2+4x+4.
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Show that the center of the osculating circle for the parabola y = x2 at the point (a, a2) is located at  (-4a3, 3a2 + 1/ 2 ) .

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