University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter Appendices - Section A.3 - Lines and Circles - Exercises - Page AP-16: 35

Answer

Intersection points: $(-\displaystyle \frac{\sqrt{3}}{3},-\frac{1}{3})$ and $(\displaystyle \frac{\sqrt{3}}{3},-\frac{1}{3})$ .

Work Step by Step

See attached image (desmos.com.) To find the intersection points algebraically, substitute $y=-x^{2} $ into the second equation: $-x^{2}=2x^{2}-1$ $0=3x^{2}-1$ $3x^{2}=1$ $x^{2}=\displaystyle \frac{1}{3}$ $x=\displaystyle \pm\sqrt{\frac{1}{3}}=\pm\frac{\sqrt{3}}{3}\approx\pm 0.577$ Back-substitute: $y=-x^{2}=-\displaystyle \frac{1}{3}$ Intersection points: $(-\displaystyle \frac{\sqrt{3}}{3},-\frac{1}{3})$ and $(\displaystyle \frac{\sqrt{3}}{3},-\frac{1}{3})$
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