Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 730: 6

Answer

$y=\pm x \sqrt {x+1}$ Also, see the attached graph for $y=\pm x \sqrt {x+1}$.
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Work Step by Step

Calculate $f(t)$. This shows that it is a vertical parabola with zeroes at $t=\pm 1$ $f(t)=k(t-1)(t+1)=k(t^2-1)$ Refer graph, we get $f(0)=-1$ hence $k=1$ $$f(t)=(t^2-1)=x ... (1)$$ we can also write $t=\pm \sqrt {x+1}$ ... (2) Now we will calculate $g(t)$ which shows that it is a cubical polynomial with zeroes at $t=0,\pm 1$ $g(t)=k(t-1)(t+1)=k(t^2-1)$ $g(0.5)=-0.4$ ( from graph) Thus, $$k=1$$ $g(t)=(t^2-1)=y$ It can be written as:$y=\pm x \sqrt {x+1}$ Also, see the attached graph for $y=\pm x \sqrt {x+1}$.
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