Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - Review Exercises - Page 51: 10

Answer

$(-\sqrt 2,2), (\sqrt 2,2), (-2,4), (2,4)$

Work Step by Step

We are given the equations: $\begin{cases} y=x^2\\ x^2+y^2-7y+8=0 \end{cases}$ Substitute $y$ for $x^2$ in the second equation and solve it: $y+y^2-7y+8=0$ $y^2-6y+8=0$ $y^2-2y-4y+8=0$ $y(y-2)-4(y-2)=0$ $(y-2)(y-4)=0$ $y-2=0\Rightarrow y=2$ $y-4=0\Rightarrow y=4$ Determine $x$: $y=2\Rightarrow x^2=2\Rightarrow x=\pm\sqrt 2$ $y=4\Rightarrow x^2=4\Rightarrow x=\pm\sqrt 4=\pm 2$ There are 4 intersection points: $(-\sqrt 2,2), (\sqrt 2,2), (-2,4), (2,4)$ Draw the curves:
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