Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 37


The answer is $y=x^2+8 x+8$.

Work Step by Step

Since $y=x^2$ is always positive, the minimum of $y$ is zero. So, the minimum of the curve occurs at $(0,0)$. The parametric equation of $y=x^2$ is $c(t)=(t,t^2)$. To translate the minimum of $c(t)$ at $(0,0)$ to $(-4,-8)$, we replace $c(t)=(t,t^2)$ by $c(t)=(-4+t,-8+t^2)$. So, the coordinates of the translated curve are $x=t-4$ and $y=t^2-8$. Solving for $t$ from $x$ we get $t=x+4$. Substituting $t$ into $y$ gives $y=(x+4)^2-8$. So, $y=x^2+8 x+8$.
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