Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Mid-Chapter Check Point - Page 1000: 13

Answer

See graph and explanations.
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Work Step by Step

Step 1. Graph the equation as shown in the figure. Step 2. Rewrite the equation as $y^2-2y+1=2x+5+1$ or $(y-1)^2=2(x+3)$; we have $p=\frac{2}{4}=\frac{1}{2}$. Step 3. As the vertex is at $(-3,1)$, the focus is at $(-3+\frac{1}{2}, 1)$ or $(-\frac{5}{2},1)$ and the directrix is $x=-3-\frac{1}{2}=-\frac{7}{2}$ as shown in the figure.
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